title

Calculus 2 - Full College Course

description

Learn Calculus 2 in this full college course.
This course was created by Dr. Linda Green, a lecturer at the University of North Carolina at Chapel Hill. Check out her YouTube channel: https://www.youtube.com/channel/UCkyLJh6hQS1TlhUZxOMjTFw
⭐️ Prerequisites ⭐️
🎥 Algebra: https://www.youtube.com/watch?v=LwCRRUa8yTU
🎥 Precalculus: https://www.youtube.com/watch?v=eI4an8aSsgw
🎥 Calculus 1: https://www.youtube.com/watch?v=HfACrKJ_Y2w
⭐️ Lecture Notes ⭐️
🔗 Calculus 2 Notes: http://lindagreen.web.unc.edu/files/2020/08/courseNotes_m232_2018F.pdf
⭐️ Course Contents ⭐️
⌨️ (0:00:00) Area Between Curves
⌨️ (0:11:11) Volumes of Solids of Revolution
⌨️ (0:21:33) Volumes Using Cross-Sections
⌨️ (0:31:34) Arclength
⌨️ (0:39:50) Work as an Integral
⌨️ (0:51:30) Average Value of a Function
⌨️ (0:59:05) Proof of the Mean Value Theorem for Integrals
⌨️ (1:04:55) Integration by Parts
⌨️ (1:13:03) Trig Identities
⌨️ (1:25:01) Proof of the Angle Sum Formulas
⌨️ (1:29:37) Integrals Involving Odd Powers of Sine and Cosine
⌨️ (1:36:21) Integrals Involving Even Powers of Sine and Cosine
⌨️ (1:44:59) Special Trig Integrals
⌨️ (1:47:36) Integration Using Trig Substitution
⌨️ (1:56:21) Integrals of Rational Functions
⌨️ (2:03:59) Improper Integrals - Type 1
⌨️ (2:10:13) Improper Integrals - Type 2
⌨️ (2:14:20) The Comparison Theorem for Integrals
⌨️ (2:19:43) Sequences - Definitions and Notation
⌨️ (2:31:05) Series Definitions
⌨️ (2:39:41) Sequences - More Definitions
⌨️ (2:46:06) Monotonic and Bounded Sequences Extra
⌨️ (2:49:46) L'Hospital's Rule
⌨️ (2:57:02) L'Hospital's Rule on Other Indeterminate Forms
⌨️ (3:06:48) Convergence of Sequences
⌨️ (3:30:38) Geometric Series
⌨️ (3:43:32) The Integral Test
⌨️ (3:56:48) Comparison Test for Series
⌨️ (4:03:29) The Limit Comparison Test
⌨️ (4:09:24) Proof of the Limit Comparison Test
⌨️ (4:16:00) Absolute Convergence
⌨️ (4:25:03) The Ratio Test
⌨️ (4:30:22) Proof of the Ratio Test
⌨️ (4:39:34) Series Convergence Test Strategy
⌨️ (4:50:26) Taylor Series Introduction
⌨️ (5:02:28) Power Series
⌨️ (5:05:42) Convergence of Power Series
⌨️ (5:21:07) Power Series Interval of Convergence Example
⌨️ (5:30:46) Proofs of Facts about Convergence of Power Series
⌨️ (5:36:19) Power Series as Functions
⌨️ (5:43:36) Representing Functions with Power Series
⌨️ (5:52:07) Using Taylor Series to find Sums of Series
⌨️ (6:02:18) Taylor Series Theory and Remainder
⌨️ (6:15:38) Parametric Equations
⌨️ (6:27:06) Slopes of Parametric Curves
⌨️ (6:32:47) Area under a Parametric Curve
⌨️ (6:38:07) Arclength of Parametric Curves
⌨️ (6:45:20) Polar Coordinates

detail

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explanations.", 'chapters': [{'end': 570.855, 'segs': [{'end': 177.406, 'src': 'embed', 'start': 118.052, 'weight': 0, 'content': [{'end': 123.396, 'text': 'delta x times its height, f of x i star, and so on.', 'start': 118.052, 'duration': 5.344}, {'end': 133.344, 'text': 'If there are n rectangles, then the last rectangle will have base delta x and height f of x n star.', 'start': 125.017, 'duration': 8.327}, {'end': 139.829, 'text': 'So the approximate area under the curve is given by adding up all these areas of all these rectangles.', 'start': 135.105, 'duration': 4.724}, {'end': 143.288, 'text': 'In sigma notation.', 'start': 141.747, 'duration': 1.541}, {'end': 159.56, 'text': 'this can be written as sigma the sum from i equals 1 to n, the number of rectangles of the area of the ith rectangle delta x times f of x i star.', 'start': 143.288, 'duration': 16.272}, {'end': 169.188, 'text': 'The exact area is then given by the limit of these approximating areas as the number of rectangles goes to infinity.', 'start': 161.562, 'duration': 7.626}, {'end': 171.029, 'text': "That's the limit.", 'start': 169.208, 'duration': 1.821}, {'end': 177.406, 'text': 'as n goes to infinity of this Riemann sum of areas.', 'start': 171.522, 'duration': 5.884}], 'summary': 'The area under the curve is approximated by summing the areas of n rectangles, with the exact area given by the limit as n goes to infinity.', 'duration': 59.354, 'max_score': 118.052, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe8118052.jpg'}, {'end': 411.315, 'src': 'embed', 'start': 380.779, 'weight': 3, 'content': [{'end': 389.683, 'text': 'Actually, this formula only works if f of x is greater than or equal to g of x on the interval from a to b.', 'start': 380.779, 'duration': 8.904}, {'end': 400.331, 'text': 'That inequality guarantees that this expression f of x i star minus g of x i star will be a positive number.', 'start': 391.988, 'duration': 8.343}, {'end': 404.533, 'text': "We want a positive height for our rectangle so that we'll get a positive area.", 'start': 400.691, 'duration': 3.842}, {'end': 411.315, 'text': 'If, instead, f of x is less than or equal to g of x,', 'start': 406.794, 'duration': 4.521}], 'summary': 'Formula works if f(x) >= g(x) on interval [a, b] for positive area.', 'duration': 30.536, 'max_score': 380.779, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe8380779.jpg'}, {'end': 470.46, 'src': 'embed', 'start': 439.408, 'weight': 1, 'content': [{'end': 445.632, 'text': "Remembering that you'll need to replace the top y value and the bottom y value with functions of x before you can integrate.", 'start': 439.408, 'duration': 6.224}, {'end': 452.195, 'text': "Let's look at an example.", 'start': 451.054, 'duration': 1.141}, {'end': 459.788, 'text': 'We want to find the area between the curve y equals x squared plus x and y equals 3 minus x squared.', 'start': 453.541, 'duration': 6.247}, {'end': 470.46, 'text': 'x squared plus x is a parabola pointing upwards, so that must be the red curve here, while 3 minus x squared is a parabola pointing downwards,', 'start': 461.29, 'duration': 9.17}], 'summary': 'Integration example to find area between parabolas.', 'duration': 31.052, 'max_score': 439.408, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe8439408.jpg'}], 'start': 0.209, 'title': 'Finding area between curves', 'summary': 'Introduces the concept of finding the area between two curves using riemann sum and integral notation, emphasizing the necessity of ensuring the top curve is greater than the bottom curve to obtain a positive area. an example with specific equations and bounds is illustrated.', 'chapters': [{'end': 208.711, 'start': 0.209, 'title': 'Area between two curves', 'summary': 'Introduces the concept of finding the area between two curves by using riemann sum and converting it to integral notation, with the example of calculating the area under a curve using sigma notation and limits.', 'duration': 208.502, 'highlights': ['The exact area is given by the limit of these approximating areas as the number of rectangles goes to infinity, represented by the integral of f of x dx from x equals a to x equals b.', 'The area of a rectangle is base times height, where the approximate area under the curve is given by adding up all these areas of all these rectangles in sigma notation.', 'In Calculus 1, the area between a curve and the x-axis was approximated by dividing it up into tall, skinny rectangles, represented by delta x, a small distance along the x-axis or a small change in x values.']}, {'end': 570.855, 'start': 208.711, 'title': 'Area between two curves', 'summary': 'Discusses the method of calculating the area between two curves using the riemann sum and the integral, emphasizing the necessity of ensuring the top curve is greater than the bottom curve to obtain a positive area, illustrated through an example with specific equations and bounds.', 'duration': 362.144, 'highlights': ['The integral for calculating the area between two curves is obtained by finding the difference between the top y value and the bottom y value integrated with respect to x, with the necessity of ensuring the top curve is greater than the bottom curve to obtain a positive area, demonstrated through the example with equations y equals x squared plus x and y equals 3 minus x squared.', 'The necessity of ensuring that f of x is greater than or equal to g of x on the interval from a to b is emphasized to guarantee a positive height for the rectangle and, consequently, a positive area, particularly highlighted by the requirement to switch around the subtraction and take the integral of g of x minus f of x dx if f of x is less than or equal to g of x.', 'The concept of setting the top curve equal to the bottom curve to find the bounds of integration, as illustrated through the example equations 3 minus x squared and x squared plus x, is demonstrated to determine the values of a and b, thereby facilitating the setup for calculating the area between the curves.']}], 'duration': 570.646, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe8209.jpg', 'highlights': ['The exact area is given by the limit of these approximating areas as the number of rectangles goes to infinity, represented by the integral of f of x dx from x equals a to x equals b.', 'The integral for calculating the area between two curves is obtained by finding the difference between the top y value and the bottom y value integrated with respect to x, with the necessity of ensuring the top curve is greater than the bottom curve to obtain a positive area, demonstrated through the example with equations y equals x squared plus x and y equals 3 minus x squared.', 'The area of a rectangle is base times height, where the approximate area under the curve is given by adding up all these areas of all these rectangles in sigma notation.', 'The necessity of ensuring that f of x is greater than or equal to g of x on the interval from a to b is emphasized to guarantee a positive height for the rectangle and, consequently, a positive area, particularly highlighted by the requirement to switch around the subtraction and take the integral of g of x minus f of x dx if f of x is less than or equal to g of x.']}, {'end': 3047.605, 'segs': [{'end': 676.175, 'src': 'embed', 'start': 641.233, 'weight': 2, 'content': [{'end': 651.455, 'text': 'minus the bottom y values dx, where the top y values and bottom y values are functions of x.', 'start': 641.233, 'duration': 10.222}, {'end': 660.636, 'text': 'More specifically, if the top curve is given by y equals f of x and the bottom curve is y equals g of x,', 'start': 651.455, 'duration': 9.181}, {'end': 668.913, 'text': 'then this integral is the integral from a to b of f of x, minus g of x dx.', 'start': 660.636, 'duration': 8.277}, {'end': 676.175, 'text': "In this video, we'll calculate the volumes of solids of revolution.", 'start': 672.314, 'duration': 3.861}], 'summary': 'Calculate volumes of solids of revolution using integral from a to b of f(x) - g(x) dx', 'duration': 34.942, 'max_score': 641.233, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe8641233.jpg'}, {'end': 780.063, 'src': 'embed', 'start': 756.326, 'weight': 4, 'content': [{'end': 762.631, 'text': 'The area of a disk is given by the familiar formula pi r squared, where r is the radius.', 'start': 756.326, 'duration': 6.305}, {'end': 777.062, 'text': 'And the area of a washer can be written as pi times r outer squared minus pi times r inner squared, where r outer is the radius of the big circle.', 'start': 763.191, 'duration': 13.871}, {'end': 780.063, 'text': 'and our inner is the radius of the little circle.', 'start': 777.682, 'duration': 2.381}], 'summary': 'Area of a disk is pi r squared, and area of a washer is pi r outer squared minus pi r inner squared.', 'duration': 23.737, 'max_score': 756.326, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe8756326.jpg'}, {'end': 1241.142, 'src': 'heatmap', 'start': 986.254, 'weight': 1, 'content': [{'end': 997.903, 'text': 'I can pull out the pi and rewrite this integral using exponents and integrate and then evaluate using bounds of integration.', 'start': 986.254, 'duration': 11.649}, {'end': 1006.393, 'text': 'to get three fifths pi times eight to the five thirds minus zero.', 'start': 999.727, 'duration': 6.666}, {'end': 1019.384, 'text': 'Now eight to the five thirds means eight to the one third raised to the fifth power h, the one third is two and two to the fifth is 32.', 'start': 1007.934, 'duration': 11.45}, {'end': 1024.808, 'text': 'So this expression simplifies to three fifths pi times 32, or 96 fifths pi.', 'start': 1019.384, 'duration': 5.424}, {'end': 1037.034, 'text': "As our next example, let's consider the region in the first quadrant bounded by two curves.", 'start': 1031.589, 'duration': 5.445}, {'end': 1043.819, 'text': 'the curve y equals the cube root of x and y equals 1 fourth times x.', 'start': 1037.034, 'duration': 6.785}, {'end': 1050.185, 'text': "We'll start by rotating this region around the x-axis to get this sort of hollow vase shape.", 'start': 1043.819, 'duration': 6.366}, {'end': 1056.73, 'text': 'Notice that our cross sections this time are shaped like washers,', 'start': 1051.786, 'duration': 4.944}, {'end': 1065.131, 'text': 'where the outer circle of the washer is swept out by the curve y equals cube root of x and the inner circle is swept out by the curve y equals 1,', 'start': 1056.73, 'duration': 8.401}, {'end': 1067.573, 'text': 'fourth x.', 'start': 1065.131, 'duration': 2.442}, {'end': 1076.418, 'text': 'We know our volume is given by the integral of pi times our outer squared minus pi times our inner squared.', 'start': 1067.573, 'duration': 8.845}, {'end': 1082.642, 'text': "And since our washers are thin in the x direction, we know we'll need to integrate dx.", 'start': 1077.679, 'duration': 4.963}, {'end': 1092.738, 'text': 'Our bounds of integration are just our lowest x value of 0 and our largest x value, which is where these two curves intersect,', 'start': 1083.742, 'duration': 8.996}, {'end': 1096.621, 'text': 'which is an x value of 8..', 'start': 1092.738, 'duration': 3.883}, {'end': 1106.329, 'text': 'We can confirm that the two curves intersect when x equals 8 by setting them equal to each other and solving for x.', 'start': 1096.621, 'duration': 9.708}, {'end': 1115.036, 'text': 'Dividing both sides by x to the 1 third and multiplying both sides by 4 gives us 4 equals x to the 2 thirds.', 'start': 1106.329, 'duration': 8.707}, {'end': 1125.412, 'text': 'And raising both sides to the 3 halves power gives us x equals 4 to the 3 halves or x equals 8.', 'start': 1116.457, 'duration': 8.955}, {'end': 1128.054, 'text': 'This confirms we have the correct bound of integration here.', 'start': 1125.412, 'duration': 2.642}, {'end': 1137.223, 'text': 'Now we need to figure out a formula for the outer radius as a function of x and a formula for the inner radius also.', 'start': 1130.176, 'duration': 7.047}, {'end': 1143.409, 'text': 'Since the outer circle is swept out by the curve y equals cube root of x.', 'start': 1138.965, 'duration': 4.444}, {'end': 1149.668, 'text': 'The outer radius is just given by the y coordinate of this curve as a function of x.', 'start': 1143.885, 'duration': 5.783}, {'end': 1152.71, 'text': "That's the cube root of x.", 'start': 1149.668, 'duration': 3.042}, {'end': 1157.013, 'text': 'Now the inner radius is given by the y coordinate of this line.', 'start': 1152.71, 'duration': 4.303}, {'end': 1162.936, 'text': 'And the y coordinate of this line as a function of x is 1 fourth x.', 'start': 1157.693, 'duration': 5.243}, {'end': 1165.998, 'text': "So we've set up an equation for the volume of our solid.", 'start': 1162.936, 'duration': 3.062}, {'end': 1174.55, 'text': 'And a routine computation gives us a volume of 128 15th times pi.', 'start': 1167.419, 'duration': 7.131}, {'end': 1181.335, 'text': "Now let's switch gears and rotate this region around the y-axis instead.", 'start': 1177.372, 'duration': 3.963}, {'end': 1188.58, 'text': 'Our cross sections are still washers, but this time the washers are thin in the y direction.', 'start': 1183.216, 'duration': 5.364}, {'end': 1193.523, 'text': "So we're going to be integrating with respect to y.", 'start': 1188.94, 'duration': 4.583}, {'end': 1204.394, 'text': 'Our bounds of integration are also y values and start at the minimum y value of 0 and the maximum y value of 2 that corresponds to this intersection,', 'start': 1193.523, 'duration': 10.871}, {'end': 1214.897, 'text': 'where x equals 8 and y, which is the cube root of x or 1, fourth times, x is then equal to 2..', 'start': 1204.394, 'duration': 10.503}, {'end': 1220.499, 'text': 'For this problem, we need our outer and our inner to be functions of y.', 'start': 1214.897, 'duration': 5.602}, {'end': 1225.901, 'text': 'From the picture, we see that our outer is actually the x-coordinate on this line.', 'start': 1220.499, 'duration': 5.402}, {'end': 1231.123, 'text': 'The line has the equation y equals 1 fourth x.', 'start': 1228.182, 'duration': 2.941}, {'end': 1234.86, 'text': 'And so x is equal to 4y.', 'start': 1232.219, 'duration': 2.641}, {'end': 1241.142, 'text': 'So that gives us our outer as a function of y.', 'start': 1235.4, 'duration': 5.742}], 'summary': 'Integration results in 96/5 pi and volume of 128/15 pi.', 'duration': 254.888, 'max_score': 986.254, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe8986254.jpg'}, {'end': 1741.503, 'src': 'heatmap', 'start': 1479.145, 'weight': 0.885, 'content': [{'end': 1490.093, 'text': 'And therefore, we can rewrite it as an integral, where the xi star becomes our variable x and the delta x becomes our dx.', 'start': 1479.145, 'duration': 10.948}, {'end': 1497.19, 'text': "For the bounds of integration, we'll just use the abstract x values of a and b.", 'start': 1491.988, 'duration': 5.202}, {'end': 1501.192, 'text': 'For a real problem, we would fill these in based on the context of the problem.', 'start': 1497.19, 'duration': 4.002}, {'end': 1508.116, 'text': 'This gives us an abstract expression for the volume of a three-dimensional object.', 'start': 1502.893, 'duration': 5.223}, {'end': 1519.421, 'text': "But in practice, in order to compute volumes like this, we'll first need a formula for a of x, the cross-sectional area, as a function of x.", 'start': 1509.696, 'duration': 9.725}, {'end': 1522.171, 'text': 'As an example,', 'start': 1521.01, 'duration': 1.161}, {'end': 1532.379, 'text': "let's try to find the volume of a solid whose base is an ellipse given by this equation and whose cross-sections perpendicular to the x-axis are squares.", 'start': 1522.171, 'duration': 10.208}, {'end': 1535.221, 'text': 'First, let me graph the base.', 'start': 1533.56, 'duration': 1.661}, {'end': 1541.486, 'text': "It looks like an ellipse that's thinner in the x direction than in the y direction, so something like this.", 'start': 1535.602, 'duration': 5.884}, {'end': 1552.152, 'text': "Now, sitting above this base are a bunch of squares, and the squares are oriented in such a way that they're perpendicular to the x-axis.", 'start': 1542.767, 'duration': 9.385}, {'end': 1554.793, 'text': "so they're oriented sort of like this.", 'start': 1552.152, 'duration': 2.641}, {'end': 1556.274, 'text': "I'll try to draw a square here.", 'start': 1554.793, 'duration': 1.481}, {'end': 1563.736, 'text': "that's supposed to be coming out of the picture here and here's another square again coming out of the picture.", 'start': 1556.274, 'duration': 7.462}, {'end': 1567.938, 'text': "here's a slightly better picture that I drew using Mathematica.", 'start': 1563.736, 'duration': 4.202}, {'end': 1568.618, 'text': "it's tilted.", 'start': 1567.938, 'duration': 0.68}, {'end': 1576.301, 'text': "so we're looking at it from below, where you can see the ellipse here and you can see the square cross sections.", 'start': 1568.618, 'duration': 7.683}, {'end': 1578.992, 'text': 'the x-axis is going in this direction.', 'start': 1576.301, 'duration': 2.691}, {'end': 1583.795, 'text': "That's x and the y-axis is in that direction.", 'start': 1580.693, 'duration': 3.102}, {'end': 1593.963, 'text': "This picture is actually an approximation of the solid where they're only about eight or ten slices.", 'start': 1587.018, 'duration': 6.945}, {'end': 1597.625, 'text': "Each one's kind of thick and has the same area on the front and the back.", 'start': 1594.363, 'duration': 3.262}, {'end': 1601.828, 'text': 'A better picture of the solid is this one.', 'start': 1598.646, 'duration': 3.182}, {'end': 1607.172, 'text': "Here the slices are infinitely thin but they're still square shaped.", 'start': 1603.85, 'duration': 3.322}, {'end': 1611.911, 'text': "And they're still oriented in such a way that they're perpendicular to the x-axis.", 'start': 1607.588, 'duration': 4.323}, {'end': 1623.097, 'text': 'Now we know that volume is given by the integral from a to b of area as a function of x dx.', 'start': 1614.312, 'duration': 8.785}, {'end': 1633.183, 'text': 'On our ellipse, the minimum x value is negative 2 and the maximum x value is 2.', 'start': 1625.159, 'duration': 8.024}, {'end': 1635.585, 'text': 'So I can write those in for my bounds of integration.', 'start': 1633.183, 'duration': 2.402}, {'end': 1643.065, 'text': 'Also, I know that the area of a square is just the side length squared.', 'start': 1638.402, 'duration': 4.663}, {'end': 1657.254, 'text': 'So I can write my cross-sectional area as s of x squared, where s of x is the side length of the square as a function of x.', 'start': 1644.906, 'duration': 12.348}, {'end': 1666.24, 'text': 'Notice that for different x values, my side lengths are different, but my side length is always twice the y value.', 'start': 1657.254, 'duration': 8.986}, {'end': 1670.305, 'text': 'twice the distance from the x-axis to the y value on the ellipse.', 'start': 1666.744, 'duration': 3.561}, {'end': 1680.37, 'text': "So I'll rewrite my formula as 2 times y as a function of x squared.", 'start': 1672.426, 'duration': 7.944}, {'end': 1690.954, 'text': 'And I can simplify this a little bit as the integral of 4 times y of x squared dx.', 'start': 1680.39, 'duration': 10.564}, {'end': 1697.16, 'text': 'Now all I need to do is find a formula for y as a function of x.', 'start': 1692.918, 'duration': 4.242}, {'end': 1703.824, 'text': "And since I've got this equation up here relating y and x, all I have to do is solve for y in terms of x.", 'start': 1697.16, 'duration': 6.664}, {'end': 1709.847, 'text': "In fact, I can get by solving for y squared since I've really got y squared in my formula.", 'start': 1703.824, 'duration': 6.023}, {'end': 1713.909, 'text': 'Solving for y squared I have.', 'start': 1711.568, 'duration': 2.341}, {'end': 1728.071, 'text': 'y squared over 9 is equal to 1 minus x squared over 4, which means that y squared is equal to 9 times 1 minus x squared over 4..', 'start': 1713.909, 'duration': 14.162}, {'end': 1741.503, 'text': "Now I'll plug this into my volume equation, and I get the integral from negative 2 to 2 of 4 times 9, 1 minus x squared over 4 dx.", 'start': 1728.071, 'duration': 13.432}], 'summary': 'Using integrals to find volume of a solid with ellipse base and square cross-sections perpendicular to x-axis.', 'duration': 262.358, 'max_score': 1479.145, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe81479145.jpg'}, {'end': 1633.183, 'src': 'embed', 'start': 1598.646, 'weight': 3, 'content': [{'end': 1601.828, 'text': 'A better picture of the solid is this one.', 'start': 1598.646, 'duration': 3.182}, {'end': 1607.172, 'text': "Here the slices are infinitely thin but they're still square shaped.", 'start': 1603.85, 'duration': 3.322}, {'end': 1611.911, 'text': "And they're still oriented in such a way that they're perpendicular to the x-axis.", 'start': 1607.588, 'duration': 4.323}, {'end': 1623.097, 'text': 'Now we know that volume is given by the integral from a to b of area as a function of x dx.', 'start': 1614.312, 'duration': 8.785}, {'end': 1633.183, 'text': 'On our ellipse, the minimum x value is negative 2 and the maximum x value is 2.', 'start': 1625.159, 'duration': 8.024}], 'summary': 'Volume of solid: integral from -2 to 2 of area as a function of x', 'duration': 34.537, 'max_score': 1598.646, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe81598646.jpg'}, {'end': 1909.404, 'src': 'embed', 'start': 1878.536, 'weight': 0, 'content': [{'end': 1883.458, 'text': 'In this video we saw that if we divide a three-dimensional object into slices,', 'start': 1878.536, 'duration': 4.922}, {'end': 1892.301, 'text': 'then the volume of the three-dimensional object is the integral of the cross-sectional area dx.', 'start': 1883.458, 'duration': 8.843}, {'end': 1900.405, 'text': "In this video I'll derive a formula to calculate the length of a curve.", 'start': 1895.643, 'duration': 4.762}, {'end': 1905.5, 'text': 'given as a function, y equals f.', 'start': 1900.405, 'duration': 5.095}, {'end': 1909.404, 'text': "For a curve like this one that's made up of a bunch of straight line segments,", 'start': 1905.5, 'duration': 3.904}], 'summary': 'Derivation of formula to calculate length of curve given as function y=f', 'duration': 30.868, 'max_score': 1878.536, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe81878536.jpg'}, {'end': 2407.113, 'src': 'embed', 'start': 2373.996, 'weight': 1, 'content': [{'end': 2385.563, 'text': 'If the curve stretches from x equals a to x equals b, then the arc length is given by the integral from a to b of the square root of one plus f,', 'start': 2373.996, 'duration': 11.567}, {'end': 2388.985, 'text': 'prime of x squared dx.', 'start': 2385.563, 'duration': 3.422}, {'end': 2398.771, 'text': 'This video introduces the idea of work from physics and the key role of integration in doing work calculations.', 'start': 2391.146, 'duration': 7.625}, {'end': 2407.113, 'text': 'If a constant force F is applied to move an object a distance d,', 'start': 2400.911, 'duration': 6.202}], 'summary': 'Introduction to using integration for work calculations in physics.', 'duration': 33.117, 'max_score': 2373.996, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe82373996.jpg'}], 'start': 570.855, 'title': 'Calculating volumes and areas', 'summary': 'Covers the calculation of areas between two curves resulting in 125/24, introduces finding volumes of solids using integrals, explains solids of revolution, formulas for cross-sectional areas, volume calculation by integrating cross-sectional areas, application in rotation around x-axis and y-axis, and demonstrates volume calculation for solids of revolution. it also includes calculating volumes of solids using disk and washer methods resulting in volumes of 128/15π and 512/21π, and introduces volume calculation using cross sections. additionally, it explains volume calculation of three-dimensional objects using cross-sectional areas and integrals, and covers arc length and work calculations, deriving the formula for arc length, illustrating its application, demonstrating the key role of integration in work calculations, providing examples of work calculation, and demonstrating the application of integration in determining the work required to lift a satellite.', 'chapters': [{'end': 676.175, 'start': 570.855, 'title': 'Area between two curves', 'summary': 'Demonstrates the calculation of the area between two curves using definite integrals, resulting in an area of 125/24, and introduces the concept of finding volumes of solids of revolution using integrals.', 'duration': 105.32, 'highlights': ['The area between two curves in between the x values of A and B can be given by the integral from a to b of the top y values, minus the bottom y values dx, where the top y values and bottom y values are functions of x.', 'The integral from negative three halves to one of minus two x squared minus x plus three dx integrates to 125 20 fourths as the area.', 'The volumes of solids of revolution are introduced for calculation using integrals.']}, {'end': 1043.819, 'start': 676.955, 'title': 'Solids of revolution and volumes', 'summary': 'Explains the concept of solids of revolution, the formulas for the area of cross sections, and the volume calculation by integrating the area of cross sections. it also discusses the application of these concepts in rotating around the x-axis and y-axis, with examples demonstrating the volume calculation for solids of revolution.', 'duration': 366.864, 'highlights': ['The volume of any solid that can be sliced into cross sections using planes perpendicular to the x-axis is given by v = ∫[a to b] A(x) dx, where A(x) represents the area of the cross section at point x integrated with respect to x.', 'The formula for the area of a disk is A = πr^2, and for a washer, it is A = π(r_outer^2 - r_inner^2), where r_outer and r_inner represent the radii of the outer and inner circles respectively.', 'Example calculation: Finding the volume of the solid of revolution formed by rotating the region bounded by the curve y = ∛x, the x-axis, and the line x = 8 around the x-axis results in a volume of 96/5π units^3.']}, {'end': 1403.688, 'start': 1043.819, 'title': 'Volumes of solids of revolution', 'summary': 'Covers calculating volumes of solids using disk and washer methods, resulting in a volume of 128/15π and 512/21π, and introduces the concept of calculating volume using cross sections.', 'duration': 359.869, 'highlights': ['The chapter covers calculating volumes of solids using disk and washer methods, resulting in a volume of 128/15π and 512/21π.', 'Introduces the concept of calculating volume using cross sections.']}, {'end': 2168.927, 'start': 1404.562, 'title': 'Volume of three-dimensional objects', 'summary': 'Explains how to calculate the volume of a three-dimensional object by using cross-sectional areas and integrals, and demonstrates the process with examples of finding the volume of a solid with different cross sections, yielding different volumes.', 'duration': 764.365, 'highlights': ["The volume of a three-dimensional object is calculated using the integral of the cross-sectional area with respect to 'x' or 'y', as demonstrated by finding the volume of a solid with different cross sections perpendicular to the x-axis and y-axis.", 'The process of approximating the length of a curve by dividing it into small pieces and using the distance formula to calculate the length of line segments, which can be generalized into a Riemann sum.']}, {'end': 3047.605, 'start': 2169.848, 'title': 'Arc length and work calculations', 'summary': "Introduces the concept of arc length and work calculations, deriving the formula for arc length and illustrating its application to find the length of a specific curve. it also explains the key role of integration in work calculations and provides examples of calculating work in both metric and english units, as well as deriving the formula for work done in moving a particle with varying force. furthermore, it demonstrates the application of integration in determining the work required to lift a satellite from the earth's surface to a specified altitude.", 'duration': 877.757, 'highlights': ['The chapter introduces the idea of work from physics and the key role of integration in doing work calculations, providing insight into the concept of arc length and its formula derivation (arc length is given by the integral of the square root of 1 plus f, prime of x squared dx, starting from the first x value of a to the last x value of b).', "The chapter demonstrates the application of integration in determining the work required to lift a satellite from the Earth's surface to a specified altitude, explaining the varying force of gravity with distance and the need to use integration to calculate work in this scenario.", "The chapter illustrates the application of the arc length formula to find the length of the curve y equals x to the three halves between x equals one and x equals four, providing a detailed step-by-step calculation and arriving at an approximate length of 7.6 units based on the graph's scale.", 'The chapter explains the formula for work (w equals force times distance) and provides examples of calculating work in both metric and English units, as well as deriving the formula for work done in moving a particle with varying force, emphasizing the need for integration to determine the total work done.']}], 'duration': 2476.75, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe8570855.jpg', 'highlights': ["The volume of a three-dimensional object is calculated using the integral of the cross-sectional area with respect to 'x' or 'y', as demonstrated by finding the volume of a solid with different cross sections perpendicular to the x-axis and y-axis.", 'The chapter introduces the idea of work from physics and the key role of integration in doing work calculations, providing insight into the concept of arc length and its formula derivation (arc length is given by the integral of the square root of 1 plus f, prime of x squared dx, starting from the first x value of a to the last x value of b).', 'The area between two curves in between the x values of A and B can be given by the integral from a to b of the top y values, minus the bottom y values dx, where the top y values and bottom y values are functions of x.', 'The volume of any solid that can be sliced into cross sections using planes perpendicular to the x-axis is given by v = ∫[a to b] A(x) dx, where A(x) represents the area of the cross section at point x integrated with respect to x.', 'The formula for the area of a disk is A = πr^2, and for a washer, it is A = π(r_outer^2 - r_inner^2), where r_outer and r_inner represent the radii of the outer and inner circles respectively.']}, {'end': 4382.771, 'segs': [{'end': 3080.731, 'src': 'embed', 'start': 3050.386, 'weight': 3, 'content': [{'end': 3055.848, 'text': 'Multiplying all these numbers together gives us a final answer of approximately 1.5 times 10 to the 10th joules.', 'start': 3050.386, 'duration': 5.462}, {'end': 3070.263, 'text': 'To put this number in perspective, this is about the same amount of work done by a car in a year or by the human heart beating for about 400 years.', 'start': 3060.936, 'duration': 9.327}, {'end': 3080.731, 'text': 'In this video, we saw that for a constant force, work is just equal to the force times distance.', 'start': 3074.206, 'duration': 6.525}], 'summary': 'Final answer: 1.5 x 10^10 joules, equivalent to car work in a year or human heart beating for 400 years.', 'duration': 30.345, 'max_score': 3050.386, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe83050386.jpg'}, {'end': 3262.286, 'src': 'embed', 'start': 3231.997, 'weight': 0, 'content': [{'end': 3241.462, 'text': 'And so the average value of the function is given by the integral on the interval from A to B divided by the length of the interval.', 'start': 3231.997, 'duration': 9.465}, {'end': 3250.567, 'text': 'Notice the similarity between the formula for the average value of a function and the formula for the average value of a list of numbers.', 'start': 3243.484, 'duration': 7.083}, {'end': 3262.286, 'text': 'The integral for the function corresponds to the summation sign for the list of numbers and the length of the interval B minus A for the function corresponds to n,', 'start': 3251.328, 'duration': 10.958}], 'summary': 'Average function value is integral from a to b divided by interval length', 'duration': 30.289, 'max_score': 3231.997, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe83231997.jpg'}, {'end': 3525.397, 'src': 'embed', 'start': 3491.741, 'weight': 1, 'content': [{'end': 3501.384, 'text': 'Namely, for any continuous function, f of x, on an interval from a to b, there has to be at least one number c between a and b,', 'start': 3491.741, 'duration': 9.643}, {'end': 3514.09, 'text': 'such that f of c equals its average value, or, in symbols, f of c equals the integral from a to b of f of x.', 'start': 3501.384, 'duration': 12.706}, {'end': 3518.513, 'text': 'dx divided by b, minus a.', 'start': 3514.09, 'duration': 4.423}, {'end': 3525.397, 'text': 'This video gave the definition of an average value of a function and stated the mean value theorem for integrals.', 'start': 3518.513, 'duration': 6.884}], 'summary': 'Mean value theorem for integrals states f(c) = average value of f(x) on [a, b]', 'duration': 33.656, 'max_score': 3491.741, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe83491741.jpg'}, {'end': 4072.282, 'src': 'embed', 'start': 4033.283, 'weight': 2, 'content': [{'end': 4045.347, 'text': 'In this notation we can rewrite the formula as the integral of u times dv is equal to u times.', 'start': 4033.283, 'duration': 12.064}, {'end': 4056.731, 'text': 'v minus the integral of v, du, since v is our g of x and du is our f, prime of x, dx.', 'start': 4045.347, 'duration': 11.384}, {'end': 4062.654, 'text': "Again, We can include bounds of integration if we're working with definite integrals.", 'start': 4057.811, 'duration': 4.843}, {'end': 4067.678, 'text': 'This will be our key formula for this section.', 'start': 4065.476, 'duration': 2.202}, {'end': 4072.282, 'text': 'And integrating using this formula is called integration by parts.', 'start': 4068.238, 'duration': 4.044}], 'summary': 'Integration by parts: u times v minus the integral of v times du.', 'duration': 38.999, 'max_score': 4033.283, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe84033283.jpg'}], 'start': 3050.386, 'title': 'Calculus concepts and techniques', 'summary': 'Covers topics such as average value and riemann sum, mean value theorem for integrals, and integration by parts technique, providing insights and examples to understand these concepts thoroughly.', 'chapters': [{'end': 3229.485, 'start': 3050.386, 'title': 'Average value and riemann sum', 'summary': 'Discusses the concept of average value of a function, which is approximated using finite sample points and eventually defined as the limit of a riemann sum, leading to the calculation of the integral from a to b of f dx.', 'duration': 179.099, 'highlights': ['The final answer of the multiplication of numbers is approximately 1.5 times 10 to the 10th joules, equivalent to the work done by a car in a year or by the human heart beating for about 400 years.', 'For a variable force, work is equal to the integral of force with respect to distance, introducing the idea of an average value of a function.', 'Defining the average value of a continuous function involves estimating it by sampling it at finitely many evenly spaced x values, with the approximation improving as the number of sample points increases.', 'The average value of f at sample points is the sum of the values of f divided by n, which can be defined as the limit as n goes to infinity of the sample average, resembling a Riemann sum.', 'The limit of the Riemann sum in the numerator is the integral from a to b of f dx, indicating the calculation of the integral using the Riemann sum approach.']}, {'end': 3436.713, 'start': 3231.997, 'title': 'Average value of a function', 'summary': 'Explains the concept of average value of a function using an example, illustrating the use of u substitution and determining if the function achieves its average value within the given interval.', 'duration': 204.716, 'highlights': ['The average value of a function is given by the integral on the interval from A to B divided by the length of the interval, highlighting the similarity between the formula for the average value of a function and the average value of a list of numbers.', 'Illustration of an example for the function g of x equals 1 over 1 minus 5x on the interval from 2 to 5, demonstrating the process of finding the average value using u substitution and evaluating the integral.', 'Determination of whether the function g ever achieves its average value within the interval from 2 to 5 by setting g of c equal to its average value and solving for c using mathematical manipulations.']}, {'end': 3709.223, 'start': 3436.713, 'title': 'Mean value theorem for integrals', 'summary': 'Explains the mean value theorem for integrals, stating that for any continuous function f of x on an interval from a to b, there has to be at least one number c between a and b such that f of c equals the average value, or, in symbols, f of c equals the integral from a to b of f of x. dx divided by b, minus a. it also provides a geometric interpretation for average value and gives two proofs of the mean value theorem for integrals.', 'duration': 272.51, 'highlights': ['The chapter explains the Mean Value Theorem for Integrals, stating that for any continuous function f of x on an interval from a to b, there has to be at least one number c between a and b such that f of c equals the average value, or, in symbols, f of c equals the integral from a to b of f of x. dx divided by b, minus a.', 'The chapter provides a geometric interpretation for average value, stating that the area of the box with height the average value is the same as the area under the curve.', 'The chapter gives two proofs of the Mean Value Theorem for Integrals, one using the intermediate value theorem and the other addressing the scenario of a non-constant continuous function on a closed interval.']}, {'end': 3894.852, 'start': 3710.899, 'title': 'Mean value theorem for integrals', 'summary': 'Explains the mean value theorem for integrals, showing that the intermediate value theorem is achieved for some c in the interval a b, and provides a second proof as a corollary to the regular mean value theorem for functions.', 'duration': 183.953, 'highlights': ['The intermediate value theorem says that f average is achieved by f of c for some c in between my x1 and x2, and therefore for some c in my interval a b. And that proves the mean value theorem for integrals. (Relevance Score: 5)', "I'm going to define a function g of x, to be the integral, from a to x, of f of t dt, where f is the function given to us in the statement of the mean value theorem for integrals. (Relevance Score: 4)", 'we know that g prime of c has to equal g of b minus g of a over b, minus a for some number c in the interval a b. If we substitute in the three facts above into our equation below, we get f of c is equal to the integral from a to b of f of t dt minus 0 over b minus a, which is exactly the conclusion that we wanted to reach. (Relevance Score: 3)']}, {'end': 4382.771, 'start': 3896.813, 'title': 'Integration by parts technique', 'summary': 'Explores the technique of integration by parts, explaining the formula and its application, and using examples to illustrate its effectiveness in solving integrals, with emphasis on choosing the appropriate u and dv. it also highlights the connection between integration by parts and the product rule for derivatives.', 'duration': 485.958, 'highlights': ['The chapter explores the technique of integration by parts, explaining the formula and its application, and using examples to illustrate its effectiveness in solving integrals, with emphasis on choosing the appropriate u and dv.', 'It also highlights the connection between integration by parts and the product rule for derivatives.', 'The integral of x e to the x dx is computed using integration by parts, showcasing the process and its effectiveness.']}], 'duration': 1332.385, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe83050386.jpg', 'highlights': ['The average value of a function is given by the integral on the interval from A to B divided by the length of the interval, highlighting the similarity between the formula for the average value of a function and the average value of a list of numbers.', 'The chapter explains the Mean Value Theorem for Integrals, stating that for any continuous function f of x on an interval from a to b, there has to be at least one number c between a and b such that f of c equals the average value, or, in symbols, f of c equals the integral from a to b of f of x. dx divided by b, minus a.', 'The chapter explores the technique of integration by parts, explaining the formula and its application, and using examples to illustrate its effectiveness in solving integrals, with emphasis on choosing the appropriate u and dv.', 'The final answer of the multiplication of numbers is approximately 1.5 times 10 to the 10th joules, equivalent to the work done by a car in a year or by the human heart beating for about 400 years.']}, {'end': 7429.638, 'segs': [{'end': 4432.815, 'src': 'embed', 'start': 4405.07, 'weight': 0, 'content': [{'end': 4408.512, 'text': "So I'm going to tell you the three trig identities that I think everybody should know.", 'start': 4405.07, 'duration': 3.442}, {'end': 4411.633, 'text': 'Fortunately, all three of them are very easy to remember.', 'start': 4409.192, 'duration': 2.441}, {'end': 4415.455, 'text': 'The first one is the Pythagorean identity.', 'start': 4413.074, 'duration': 2.381}, {'end': 4426.23, 'text': "That's the identity that says sine squared of theta plus cosine squared of theta is equal to 1.", 'start': 4417.284, 'duration': 8.946}, {'end': 4432.815, 'text': "That one's easy to remember because it's really just the Pythagorean theorem for triangles in the context of the unit circle.", 'start': 4426.23, 'duration': 6.585}], 'summary': 'Three essential trig identities: pythagorean, sine, and cosine squared equals 1.', 'duration': 27.745, 'max_score': 4405.07, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe84405070.jpg'}, {'end': 4511.847, 'src': 'embed', 'start': 4482.774, 'weight': 2, 'content': [{'end': 4486.877, 'text': 'The second identities that everybody should know are the even and odd identities.', 'start': 4482.774, 'duration': 4.103}, {'end': 4494.262, 'text': "I guess technically these are two different identities, but they go together, so I'm counting them as 1.", 'start': 4488.378, 'duration': 5.884}, {'end': 4501.145, 'text': 'The even identity says that cosine of negative theta equal to cosine of theta.', 'start': 4494.262, 'duration': 6.883}, {'end': 4504.305, 'text': 'in other words, cosine is an even function.', 'start': 4501.145, 'duration': 3.16}, {'end': 4511.847, 'text': 'and the odd identity says that sine of negative theta is equal to negative sine of theta.', 'start': 4504.305, 'duration': 7.542}], 'summary': 'Even and odd identities: cos(-θ) = cos(θ), sin(-θ) = -sin(θ)', 'duration': 29.073, 'max_score': 4482.774, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe84482774.jpg'}, {'end': 5154.006, 'src': 'embed', 'start': 5061.361, 'weight': 1, 'content': [{'end': 5069.043, 'text': 'While all these formulas are very useful, the ones that are of particular importance for techniques of integration are the first one,', 'start': 5061.361, 'duration': 7.682}, {'end': 5074.365, 'text': 'the Pythagorean identity and its various forms, and the last two.', 'start': 5069.043, 'duration': 5.322}, {'end': 5090.563, 'text': 'In this video I told you my three favorite trig identities the Pythagorean identity, the even and odd identities and the angle sum formulas.', 'start': 5079.335, 'duration': 11.228}, {'end': 5100.47, 'text': 'From these, I derived a bunch of other identities, including a handful that will be particularly useful as we do techniques of integration.', 'start': 5092.445, 'duration': 8.025}, {'end': 5111.3, 'text': 'Remember the angle sum formulas? Those are the formulas for computing sine of a plus b and cosine of A plus B.', 'start': 5102.192, 'duration': 9.108}, {'end': 5126.764, 'text': 'I like to sing them sine cosine cosine sine cosine cosine minus sine sine.', 'start': 5111.3, 'duration': 15.464}, {'end': 5131.546, 'text': 'This video gives a geometric proof of those formulas.', 'start': 5128.285, 'duration': 3.261}, {'end': 5138.654, 'text': "There are many great proofs of the angle sum formulas, but I'd like to share with you one of my favorites for those who are interested.", 'start': 5132.91, 'duration': 5.744}, {'end': 5143.658, 'text': "I'll write the angle sum formulas up here, so we'll know what we're trying to prove.", 'start': 5140.135, 'duration': 3.523}, {'end': 5154.006, 'text': 'To prove these formulas, let me start by drawing an angle A and an angle B on top of that.', 'start': 5145.019, 'duration': 8.987}], 'summary': 'The video discusses important trigonometric identities and their applications in integration techniques.', 'duration': 92.645, 'max_score': 5061.361, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe85061361.jpg'}, {'end': 5776.042, 'src': 'embed', 'start': 5743.689, 'weight': 5, 'content': [{'end': 5755.773, 'text': 'In this video, we use u substitution and the Pythagorean identity to evaluate integrals with at least one odd power of sine or cosine.', 'start': 5743.689, 'duration': 12.084}, {'end': 5760.554, 'text': 'The idea is to separate off one copy from the odd power.', 'start': 5756.673, 'duration': 3.881}, {'end': 5770.998, 'text': 'That one copy becomes part of our du and the remaining even power gets converted using the Pythagorean identity.', 'start': 5761.915, 'duration': 9.083}, {'end': 5776.042, 'text': 'In this case, we convert sine squared into one minus cosine squared.', 'start': 5771.538, 'duration': 4.504}], 'summary': 'Using u substitution and pythagorean identity to evaluate integrals with odd power of sine or cosine.', 'duration': 32.353, 'max_score': 5743.689, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe85743689.jpg'}, {'end': 6442.3, 'src': 'heatmap', 'start': 6188.977, 'weight': 0.743, 'content': [{'end': 6193.699, 'text': 'And so du is 2 cosine of 2x.', 'start': 6188.977, 'duration': 4.722}, {'end': 6205.924, 'text': 'So we can replace this integral here with minus the integral of 1 minus u squared times 1 half du.', 'start': 6195.259, 'duration': 10.665}, {'end': 6209.345, 'text': "I'll copy the first part down again.", 'start': 6207.764, 'duration': 1.581}, {'end': 6222.3, 'text': "and integrate that last part, I'll pull out the negative a half, and I get u minus one third u cubed, I need a plus C now.", 'start': 6211.03, 'duration': 11.27}, {'end': 6227.284, 'text': "I'll plug back in for you and simplify a little bit.", 'start': 6224.462, 'duration': 2.822}, {'end': 6252.921, 'text': 'And I get a final answer of five sixteenths x minus one fourth sine of two x plus three 64th sine of four x plus one 48th sine cubed of two x.', 'start': 6228.445, 'duration': 24.476}, {'end': 6254.842, 'text': 'Oh, and we need a plus C here.', 'start': 6252.921, 'duration': 1.921}, {'end': 6257.624, 'text': "That's a complicated answer.", 'start': 6256.504, 'duration': 1.12}, {'end': 6259.986, 'text': 'And it was a complicated computation.', 'start': 6258.245, 'duration': 1.741}, {'end': 6263.709, 'text': 'But it was all based on the same sort of tricks as before.', 'start': 6261.047, 'duration': 2.662}, {'end': 6274.692, 'text': 'Using identities like this one, to handle even powers of sine and cosine, and using the Pythagorean identity to handle odd powers.', 'start': 6265.25, 'duration': 9.442}, {'end': 6287.237, 'text': 'In this video, we use these last two identities to rewrite even powers of cosine and sine in terms of lower powers of cosine.', 'start': 6276.953, 'duration': 10.284}, {'end': 6294.559, 'text': 'This trick can be used to compute the integrals of complicated even powers of sine and cosine.', 'start': 6288.677, 'duration': 5.882}, {'end': 6298.31, 'text': 'provided that you have a lot of time and patience on hand.', 'start': 6295.188, 'duration': 3.122}, {'end': 6311.339, 'text': 'This video is about some special trig integrals, namely the integral of tangent squared of x and the integral of secant of x.', 'start': 6300.492, 'duration': 10.847}, {'end': 6315.823, 'text': "These integrals are special only in the sense that there's some special tricks required to integrate them.", 'start': 6311.339, 'duration': 4.484}, {'end': 6324.729, 'text': 'When I come across the integral of tangent squared x, I find myself wishing that I could integrate secant squared x instead.', 'start': 6318.064, 'duration': 6.665}, {'end': 6327.523, 'text': 'because integrating secant squared x is easy.', 'start': 6325.401, 'duration': 2.122}, {'end': 6332.106, 'text': "It's just tangent x plus c, since the derivative of tangent is secant squared.", 'start': 6327.543, 'duration': 4.563}, {'end': 6344.216, 'text': 'But happily, I know how to rewrite tangent squared in terms of secant, because tangent squared is secant squared minus 1.', 'start': 6333.528, 'duration': 10.688}, {'end': 6357.499, 'text': "So I'll just rewrite this integral as the integral of secant squared minus 1, and that integrates easily to tangent of x minus x plus c.", 'start': 6344.216, 'duration': 13.283}, {'end': 6365.585, 'text': "The same sort of trick works to integrate cotangent squared of x, since there's a similar identity relating cotangent and cosecant.", 'start': 6357.499, 'duration': 8.086}, {'end': 6372.009, 'text': 'Every now and then, you have to integrate secant of x.', 'start': 6367.907, 'duration': 4.102}, {'end': 6374.411, 'text': "There's several possible tricks that can be used to do this.", 'start': 6372.009, 'duration': 2.402}, {'end': 6384.111, 'text': 'One of them is to multiply secant x by secant x plus tangent x in the numerator and denominator.', 'start': 6375.112, 'duration': 8.999}, {'end': 6398.354, 'text': 'Now distributing, we get secant squared x plus secant x tangent x in the numerator and secant x plus tangent x in the denominator.', 'start': 6386.552, 'duration': 11.802}, {'end': 6408.876, 'text': 'Since secant squared is the derivative of tangent and secant tangent is the derivative of secant,', 'start': 6400.474, 'duration': 8.402}, {'end': 6419.315, 'text': 'we can set u equal to secant x plus tangent x and have du sitting right where we want it in the integrand.', 'start': 6408.876, 'duration': 10.439}, {'end': 6429.699, 'text': 'So now we just have to integrate 1 over u du, which is the ln of u plus c.', 'start': 6420.836, 'duration': 8.863}, {'end': 6442.3, 'text': 'And now plugging back in for u, we get that our original integral of secant x evaluates to the natural log of secant x plus tangent x plus a constant.', 'start': 6429.699, 'duration': 12.601}], 'summary': 'Techniques for integrating special trigonometric functions demonstrated, including results for specific functions and identities used.', 'duration': 253.323, 'max_score': 6188.977, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe86188977.jpg'}, {'end': 6469.558, 'src': 'embed', 'start': 6443.621, 'weight': 6, 'content': [{'end': 6450.447, 'text': 'A similar trick can be used to evaluate the integral of cosecant x.', 'start': 6443.621, 'duration': 6.826}, {'end': 6455.011, 'text': "And that's all for the integral of tangent squared and the integral of secant.", 'start': 6450.447, 'duration': 4.564}, {'end': 6464.84, 'text': 'This video introduces the technique of trig substitution to evaluate integrals involving square root signs, like this one.', 'start': 6457.574, 'duration': 7.266}, {'end': 6469.558, 'text': 'There are a few trig identities that are especially useful for this technique.', 'start': 6466.496, 'duration': 3.062}], 'summary': 'Introduces trig substitution technique for evaluating integrals with square roots.', 'duration': 25.937, 'max_score': 6443.621, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe86443621.jpg'}, {'end': 6519.265, 'src': 'embed', 'start': 6491.027, 'weight': 9, 'content': [{'end': 6497.671, 'text': "As our first example, let's look at the integral of x squared over the square root of 49 minus x squared.", 'start': 6491.027, 'duration': 6.644}, {'end': 6511.78, 'text': 'According to Wolfram alpha, this integral evaluates to this expression involving a square root expression, kind of as expected, and a sine inverse,', 'start': 6497.691, 'duration': 14.089}, {'end': 6514.342, 'text': 'which, just sort of come, seems to come out of the blue here.', 'start': 6511.78, 'duration': 2.562}, {'end': 6519.265, 'text': "Let's see where this answer comes from using a trig substitution.", 'start': 6516.143, 'duration': 3.122}], 'summary': 'Integral of x squared over square root of 49 minus x squared yields a result involving a square root expression and a sine inverse, which can be derived using a trig substitution.', 'duration': 28.238, 'max_score': 6491.027, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe86491027.jpg'}, {'end': 7006.394, 'src': 'embed', 'start': 6975.506, 'weight': 10, 'content': [{'end': 6981.377, 'text': 'In this video, we used trig substitution to evaluate an integral with a square root in it.', 'start': 6975.506, 'duration': 5.871}, {'end': 6988.18, 'text': 'This video introduces the method of partial fractions as a way to integrate many rational functions.', 'start': 6982.717, 'duration': 5.463}, {'end': 6996.963, 'text': 'Recall that a rational function is a function of the form of polynomial divided by another polynomial.', 'start': 6989.28, 'duration': 7.683}, {'end': 7006.394, 'text': "Let's work out the integral of 3x plus 2 divided by x squared plus 2x minus 3.", 'start': 6999.945, 'duration': 6.449}], 'summary': 'Introducing trig substitution and partial fractions for integrating rational functions.', 'duration': 30.888, 'max_score': 6975.506, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe86975506.jpg'}], 'start': 4384.712, 'title': 'Trigonometric identities and integrals', 'summary': 'Covers essential trig identities, their derivation, geometric proofs, and techniques for trigonometric integrals, including trig substitution and partial fractions. it emphasizes the significance of these identities for integration techniques and includes specific examples of trigonometric integrals and their solutions.', 'chapters': [{'end': 4683.352, 'start': 4384.712, 'title': 'Trig identities and derivation', 'summary': 'Discusses three essential trig identities: the pythagorean identity, the even and odd identities, and the angle sum formula, emphasizing their significance and ease of memorization, and highlights the process of deriving additional trig identities from these foundational ones.', 'duration': 298.64, 'highlights': ['The Pythagorean identity, sine squared of theta plus cosine squared of theta equals 1, is easy to remember as it relates to the Pythagorean theorem and the unit circle, providing a foundational understanding of trigonometric relationships.', 'The even and odd identities, including cosine of negative theta equals cosine of theta and sine of negative theta equals negative sine of theta, are crucial for understanding the symmetry and function properties of sine and cosine.', 'The angle sum formula, containing separate identities for sine and cosine, is emphasized through a mnemonic song, providing an easy and memorable way to recall these trigonometric relationships.']}, {'end': 5154.006, 'start': 4684.656, 'title': 'Favorite trig identities and their derivations', 'summary': 'Discusses the derivation of several trigonometric identities, including pythagorean, angle sum, and double angle formulas, with a particular focus on the pythagorean identity and its various forms, as well as the significance of these identities for techniques of integration.', 'duration': 469.35, 'highlights': ['The Pythagorean identity and its variations involving tangent, secant, cotangent, and cosecant are derived from dividing both sides of the Pythagorean identity by cosine squared theta and sine squared theta, respectively.', 'The angle sum formulas are used to derive the angle difference formula for sine and cosine, as well as the double angle formulas for sine and cosine.', 'The chapter emphasizes the significance of the derived trig identities, particularly the Pythagorean identity and its variations, as well as the angle sum formulas and their geometric proof, for techniques of integration.']}, {'end': 5741.496, 'start': 5156.986, 'title': 'Geometric proof & trigonometric integrals', 'summary': 'Discusses a geometric proof for angle sum formulas, highlighting a rectangle divided into right triangles, followed by the integration of trig functions with odd powers of sine or cosine using u substitution.', 'duration': 584.51, 'highlights': ['The chapter discusses a geometric proof for angle sum formulas, highlighting a rectangle divided into right triangles.', 'The integration of trig functions with odd powers of sine or cosine using u substitution is demonstrated.']}, {'end': 6488.572, 'start': 5743.689, 'title': 'Special trig integrals and trig substitution', 'summary': 'Covers using u substitution and the pythagorean identity to evaluate integrals with odd powers of sine or cosine, and introduces the technique of trig substitution to evaluate integrals involving square root signs, along with special tricks for integrating tangent squared x and secant x.', 'duration': 744.883, 'highlights': ['The chapter covers using u substitution and the Pythagorean identity to evaluate integrals with odd powers of sine or cosine.', 'Introduces the technique of trig substitution to evaluate integrals involving square root signs.', 'Special tricks for integrating tangent squared x and secant x are explained.']}, {'end': 7429.638, 'start': 6491.027, 'title': 'Trig substitution and partial fractions', 'summary': 'Discusses evaluating the integral of x squared over the square root of 49 minus x squared using trig substitution, leading to the expression 49/4 * sin^(-1)(x/7) - x/2 * sqrt(49 - x^2) + c. it then explores the method of partial fractions to integrate 3x + 2 divided by x^2 + 2x - 3, resulting in the expression ln| x - 1 | + ln| x + 3 |. the key points include the use of trig substitution, solving for partial fractions, and the conditions for the method to work.', 'duration': 938.611, 'highlights': ['The chapter discusses evaluating the integral of x squared over the square root of 49 minus x squared using trig substitution, leading to the expression 49/4 * sin^(-1)(x/7) - x/2 * sqrt(49 - x^2) + C, providing insights into the process of trig substitution and demonstrating the resulting expression from the integral (49/4 * sin^(-1)(x/7) - x/2 * sqrt(49 - x^2) + C).', 'It then explores the method of partial fractions to integrate 3x + 2 divided by x^2 + 2x - 3, resulting in the expression ln| x - 1 | + ln| x + 3 |, showcasing the application of partial fractions and demonstrating the resulting expression from the integral (ln| x - 1 | + ln| x + 3 |).', 'The key points include the use of trig substitution, solving for partial fractions, and the conditions for the method to work, emphasizing the importance of these techniques and outlining the necessary conditions for successful application.']}], 'duration': 3044.926, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe84384712.jpg', 'highlights': ['The Pythagorean identity, sine squared of theta plus cosine squared of theta equals 1, is foundational for trigonometric relationships.', 'The angle sum formula is emphasized through a mnemonic song, providing an easy way to recall trigonometric relationships.', 'The even and odd identities are crucial for understanding the symmetry and function properties of sine and cosine.', 'The derived trig identities, particularly the Pythagorean identity and its variations, are significant for integration techniques.', 'The angle sum formulas and their geometric proof are emphasized for techniques of integration.', 'The chapter covers using u substitution and the Pythagorean identity to evaluate integrals with odd powers of sine or cosine.', 'Introduces the technique of trig substitution to evaluate integrals involving square root signs.', 'The chapter discusses a geometric proof for angle sum formulas, highlighting a rectangle divided into right triangles.', 'The integration of trig functions with odd powers of sine or cosine using u substitution is demonstrated.', 'The chapter discusses evaluating the integral of x squared over the square root of 49 minus x squared using trig substitution.', 'It explores the method of partial fractions to integrate 3x + 2 divided by x^2 + 2x - 3, showcasing the application of partial fractions.']}, {'end': 8382.855, 'segs': [{'end': 7543.035, 'src': 'embed', 'start': 7490.93, 'weight': 1, 'content': [{'end': 7496.833, 'text': "It's called a type 1 improper integral if we're integrating over an infinite interval.", 'start': 7490.93, 'duration': 5.903}, {'end': 7504.157, 'text': "In other words, there's an infinity or a negative infinity somewhere in the bounds of integration.", 'start': 7498.154, 'duration': 6.003}, {'end': 7509.466, 'text': 'So the first example is a type one improper integral.', 'start': 7505.978, 'duration': 3.488}, {'end': 7520.511, 'text': "A type two improper integral occurs when the function that we're integrating itself has an infinite discontinuity on the interval.", 'start': 7510.806, 'duration': 9.705}, {'end': 7526.934, 'text': 'By an infinite discontinuity, I mean the function is going to infinity or negative infinity.', 'start': 7522.352, 'duration': 4.582}, {'end': 7529.555, 'text': 'This is also called a vertical asymptote.', 'start': 7527.354, 'duration': 2.201}, {'end': 7538.951, 'text': "This vertical asymptote could occur in the interior of the interval we're integrating over or, as in this example,", 'start': 7531.104, 'duration': 7.847}, {'end': 7543.035, 'text': 'it could occur on the endpoint of that interval of integration.', 'start': 7538.951, 'duration': 4.084}], 'summary': 'Improper integrals can be type 1 or type 2, with infinite intervals or discontinuities.', 'duration': 52.105, 'max_score': 7490.93, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe87490930.jpg'}, {'end': 7707.147, 'src': 'embed', 'start': 7680.448, 'weight': 3, 'content': [{'end': 7685.77, 'text': 'We say that the integral converges if this limit exists as a finite number.', 'start': 7680.448, 'duration': 5.322}, {'end': 7695.441, 'text': "and we say that the integral diverges if the limit is infinity or negative infinity or if it doesn't exist.", 'start': 7687.136, 'duration': 8.305}, {'end': 7707.147, 'text': 'Similarly, we evaluate the integral from negative infinity to some finite number by taking bigger and bigger intervals that extend off to negative infinity.', 'start': 7697.141, 'duration': 10.006}], 'summary': 'Integral converges if limit is finite, diverges if infinite or non-existent.', 'duration': 26.699, 'max_score': 7680.448, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe87680448.jpg'}, {'end': 7944.374, 'src': 'heatmap', 'start': 7680.448, 'weight': 4, 'content': [{'end': 7685.77, 'text': 'We say that the integral converges if this limit exists as a finite number.', 'start': 7680.448, 'duration': 5.322}, {'end': 7695.441, 'text': "and we say that the integral diverges if the limit is infinity or negative infinity or if it doesn't exist.", 'start': 7687.136, 'duration': 8.305}, {'end': 7707.147, 'text': 'Similarly, we evaluate the integral from negative infinity to some finite number by taking bigger and bigger intervals that extend off to negative infinity.', 'start': 7697.141, 'duration': 10.006}, {'end': 7720.841, 'text': 'That is, this integral is defined as the limit as the left endpoint t goes to negative infinity of the integral from t to b of f of x dx.', 'start': 7708.107, 'duration': 12.734}, {'end': 7729.467, 'text': 'We say that this integral converges if the limit exists as a finite number and diverges otherwise.', 'start': 7722.622, 'duration': 6.845}, {'end': 7742.735, 'text': 'So to evaluate the integral from negative infinity to negative one of one over x dx, we take the limit as t goes to negative infinity.', 'start': 7732.389, 'duration': 10.346}, {'end': 7749.055, 'text': 'of the integral from t to negative one of one over x dx.', 'start': 7744.454, 'duration': 4.601}, {'end': 7760.279, 'text': 'Now the integral of one over x is ln of the absolute value of x, which will need to integrate between t and negative one and take a limit.', 'start': 7750.476, 'duration': 9.803}, {'end': 7769.322, 'text': "If we evaluate here, we know that ln of the absolute value of negative one, that's ln of one, which is zero.", 'start': 7761.559, 'duration': 7.763}, {'end': 7775.097, 'text': 'a graph of ln is helpful for evaluating the rest of this expression.', 'start': 7771.235, 'duration': 3.862}, {'end': 7784.343, 'text': 'As t goes to negative infinity the absolute value of t is going to infinity and so ln is also going to infinity.', 'start': 7776.778, 'duration': 7.565}, {'end': 7792.568, 'text': 'Therefore our limit is actually negative infinity and so the integral diverges.', 'start': 7786.044, 'duration': 6.524}, {'end': 7805.67, 'text': "In this video, we evaluated improper integrals, in which the interval that we're integrating over is infinite,", 'start': 7795.89, 'duration': 9.78}, {'end': 7812.033, 'text': 'by looking at the integrals over larger and larger finite intervals and taking a limit.', 'start': 7805.67, 'duration': 6.363}, {'end': 7816.714, 'text': 'This video is about type II improper integrals.', 'start': 7814.093, 'duration': 2.621}, {'end': 7822.637, 'text': "These are integrals, for which the interval that we're integrating over is finite,", 'start': 7817.355, 'duration': 5.282}, {'end': 7826.458, 'text': "but the function that we're integrating goes to infinity on that interval.", 'start': 7822.637, 'duration': 3.821}, {'end': 7834.413, 'text': 'To integrate a type II improper integral like this one or this one.', 'start': 7828.851, 'duration': 5.562}, {'end': 7840.476, 'text': 'we integrate our function over larger and larger subintervals on which the function is finite.', 'start': 7834.413, 'duration': 6.063}, {'end': 7845.058, 'text': 'And then we take a limit.', 'start': 7843.997, 'duration': 1.061}, {'end': 7856.963, 'text': 'So in this first picture, where the function approaches infinity as x goes to b from the left, we can call that moving endpoint t.', 'start': 7846.498, 'duration': 10.465}, {'end': 7870.794, 'text': 'and evaluate the integral as the limit as that right endpoint t approaches b from the left of the integral from a to t of f dx.', 'start': 7858.427, 'duration': 12.367}, {'end': 7878.139, 'text': 'The same definition works if f is going to negative infinity instead of infinity.', 'start': 7873.136, 'duration': 5.003}, {'end': 7889.424, 'text': 'In the second picture here, we again want to take a limit of integrals over subintervals on which the function is finite.', 'start': 7880.14, 'duration': 9.284}, {'end': 7901.692, 'text': 'So in symbols that says that if f goes to infinity or negative infinity, as x goes to a from the positive side,', 'start': 7890.405, 'duration': 11.287}, {'end': 7917.132, 'text': 'then the integral from a to b of f of x, dx is going to be the limit, as t goes to a from the right of the integral from t to b of f, dx.', 'start': 7901.692, 'duration': 15.44}, {'end': 7926.222, 'text': "As an example, let's find the area under the curve y equals x over the square root of x squared minus 1,", 'start': 7919.675, 'duration': 6.547}, {'end': 7931.868, 'text': 'between the lines of x equals 1 and x equals 2..', 'start': 7926.222, 'duration': 5.646}, {'end': 7936.552, 'text': 'That area can be described as the integral from 1 to 2 of our function.', 'start': 7931.868, 'duration': 4.684}, {'end': 7944.374, 'text': 'But this is an improper integral because the function is going to infinity as x goes to 1 from the right.', 'start': 7937.691, 'duration': 6.683}], 'summary': 'The video discusses improper integrals, evaluating convergence and divergence as limits, and type ii improper integrals.', 'duration': 24.699, 'max_score': 7680.448, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe87680448.jpg'}, {'end': 8090.067, 'src': 'embed', 'start': 8061.471, 'weight': 0, 'content': [{'end': 8064.694, 'text': "Sometimes we don't really care what the value of an improper integral is.", 'start': 8061.471, 'duration': 3.223}, {'end': 8069.939, 'text': "We just want to know whether it's finite or infinite, whether it converges or diverges.", 'start': 8065.275, 'duration': 4.664}, {'end': 8074.863, 'text': 'In this situation, the comparison theorem can be very handy.', 'start': 8071.82, 'duration': 3.043}, {'end': 8083.583, 'text': 'the comparison theorem can allow us to determine if an integral converges or diverges without actually having to evaluate the integral.', 'start': 8076.738, 'duration': 6.845}, {'end': 8090.067, 'text': 'Instead, by comparing it to the integral of a function that we know converges or diverges.', 'start': 8084.303, 'duration': 5.764}], 'summary': 'Comparison theorem helps determine if an integral converges or diverges without evaluating it.', 'duration': 28.596, 'max_score': 8061.471, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe88061471.jpg'}], 'start': 7431.761, 'title': 'Improper integrals and convergence', 'summary': 'Discusses type i and type ii improper integrals, demonstrating the convergence and divergence of integrals over infinite intervals, and applying the comparison theorem to determine convergence or divergence.', 'chapters': [{'end': 7543.035, 'start': 7431.761, 'title': 'Improper integrals and partial fractions', 'summary': 'Discusses improper integrals, particularly type 1 and type 2, with examples involving infinity in the bounds and functions approaching infinity, emphasizing the concept of vertical asymptotes.', 'duration': 111.274, 'highlights': ['Improper integrals are defined by either having infinity in the bound of integration or integrating a function with an infinite discontinuity, known as a vertical asymptote.', 'Type 1 improper integrals involve integrating over an infinite interval, while type 2 improper integrals occur when the function being integrated has an infinite discontinuity on the interval.', 'Examples of improper integrals include integrating 1 over x squared from 1 to infinity and integrating tan x from 0 to pi over 2, both cases demonstrating the concept of improper integrals.']}, {'end': 8060.01, 'start': 7544.697, 'title': 'Improper integrals and convergence', 'summary': 'Discusses type i and type ii improper integrals, demonstrating the convergence and divergence of integrals over infinite intervals, and computing the area under a curve as the limit of integrals over larger subintervals where the function is finite.', 'duration': 515.313, 'highlights': ['The video focuses on type I improper integrals, where the integral asks to integrate over an infinite interval, demonstrating convergence by taking the limit as t goes to infinity of the integral from a to t of f(x) dx.', 'Explains the definition of type I improper integrals, stating that the integral converges if the limit exists as a finite number, and diverges if the limit is infinity, negative infinity, or does not exist.', 'Discusses the evaluation of the integral from negative infinity to some finite number by taking bigger and bigger intervals that extend off to negative infinity, and provides an example where the integral diverges.', 'Introduces type II improper integrals, explaining that these integrals involve finite intervals where the function goes to infinity, and demonstrates the computation of these integrals as the limit of integrals over larger subintervals where the function is finite.', 'Provides an example of computing the area under a curve as the limit of integrals over larger subintervals on which the function is finite, using u substitution and demonstrating the convergence of the integral.']}, {'end': 8382.855, 'start': 8061.471, 'title': 'Comparison theorem for integrals', 'summary': 'Explains the comparison theorem for integrals, demonstrating how to determine if an integral converges or diverges by comparing it to the integral of a known function, and provides an example of applying the theorem to determine the convergence or divergence of a specific integral.', 'duration': 321.384, 'highlights': ['The comparison theorem allows us to determine if an integral converges or diverges without actually having to evaluate the integral, by comparing it to the integral of a function that we know converges or diverges.', 'If the integral of the bigger function converges, then the integral of the smaller function also has to converge, and if the integral of the smaller function diverges, the integral of the bigger function also has to diverge.', 'When comparing functions for convergence or divergence, if the integral of the smaller function diverges, the integral of the bigger function also has to diverge, and if the integral of the bigger function converges, the integral of the smaller function also has to converge.']}], 'duration': 951.094, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe87431761.jpg', 'highlights': ['The comparison theorem allows us to determine if an integral converges or diverges without actually having to evaluate the integral, by comparing it to the integral of a function that we know converges or diverges.', 'Type 1 improper integrals involve integrating over an infinite interval, while type 2 improper integrals occur when the function being integrated has an infinite discontinuity on the interval.', 'Improper integrals are defined by either having infinity in the bound of integration or integrating a function with an infinite discontinuity, known as a vertical asymptote.', 'Explains the definition of type I improper integrals, stating that the integral converges if the limit exists as a finite number, and diverges if the limit is infinity, negative infinity, or does not exist.', 'Provides an example of computing the area under a curve as the limit of integrals over larger subintervals on which the function is finite, using u substitution and demonstrating the convergence of the integral.']}, {'end': 9531.727, 'segs': [{'end': 8584.846, 'src': 'embed', 'start': 8554.372, 'weight': 1, 'content': [{'end': 8560.075, 'text': 'Sometimes the nth term of a sequence is defined indirectly in terms of previous terms.', 'start': 8554.372, 'duration': 5.703}, {'end': 8562.816, 'text': 'This is called a recursive formula.', 'start': 8561.155, 'duration': 1.661}, {'end': 8572.06, 'text': "To write out the first few terms of this recursive sequence, we're told that a sub 1 equals 2.", 'start': 8565.377, 'duration': 6.683}, {'end': 8580.123, 'text': 'To find a sub 2, we just use the recursive formula 4 minus 1 over a sub 1.', 'start': 8572.06, 'duration': 8.063}, {'end': 8584.846, 'text': "Since a sub 1 is 2, That's 4 minus 1 half or 7 halves.", 'start': 8580.123, 'duration': 4.723}], 'summary': 'Recursive formula generates sequence with a sub 1 = 2. a sub 2 = 7/2.', 'duration': 30.474, 'max_score': 8554.372, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe88554372.jpg'}, {'end': 8748.428, 'src': 'embed', 'start': 8711.363, 'weight': 4, 'content': [{'end': 8717.931, 'text': 'This is an example of an arithmetic sequence, a sequence for which consecutive terms have the same common difference.', 'start': 8711.363, 'duration': 6.568}, {'end': 8730.926, 'text': 'And in general, if A is the first term and D is the common difference, then an arithmetic sequence has the form D times K plus A.', 'start': 8718.451, 'duration': 12.475}, {'end': 8748.428, 'text': "if our index is k and starts at 0, like it did over here, or if we'd rather start with an index of 1, we can rewrite that as d times n minus 1 plus a.", 'start': 8731.724, 'duration': 16.704}], 'summary': 'Arithmetic sequence: consecutive terms with common difference d.', 'duration': 37.065, 'max_score': 8711.363, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe88711363.jpg'}, {'end': 8973.118, 'src': 'embed', 'start': 8943.015, 'weight': 3, 'content': [{'end': 8951.361, 'text': 'These three sequences are all examples of geometric sequences, which are sequences where consecutive terms have the same common ratio.', 'start': 8943.015, 'duration': 8.346}, {'end': 8963.896, 'text': 'And in general, if a is the first term and r is the common ratio, then a geometric sequence can be written in the form of a times r to the n,', 'start': 8952.022, 'duration': 11.874}, {'end': 8973.118, 'text': 'where n starts at 0, or as a times r to the n minus 1, where n starts at 1..', 'start': 8963.896, 'duration': 9.222}], 'summary': 'Geometric sequences have a common ratio, and can be expressed as a times r to the power of n.', 'duration': 30.103, 'max_score': 8943.015, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe88943015.jpg'}, {'end': 9416.197, 'src': 'heatmap', 'start': 9157.697, 'weight': 0.897, 'content': [{'end': 9167.954, 'text': "If I add 1⁸ to that, My sum goes up to 7 eighths, and then I'll add the next one.", 'start': 9157.697, 'duration': 10.257}, {'end': 9174.96, 'text': 'I get 15 sixteenths, and so on, just adding one more term each time.', 'start': 9167.974, 'duration': 6.986}, {'end': 9183.746, 'text': 'This process of repeated addition gives me a new sequence down at the bottom.', 'start': 9178.963, 'duration': 4.783}, {'end': 9188.77, 'text': "that's called the sequence of partial sums, and it's usually denoted by s sub n.", 'start': 9183.746, 'duration': 5.024}, {'end': 9196.884, 'text': 'So the first term in the sequence of partial sums is S sub 1, just adding together the first term.', 'start': 9190.72, 'duration': 6.164}, {'end': 9201.928, 'text': "Here's a second partial sum, S sub 2, adding together the first two terms.", 'start': 9197.465, 'duration': 4.463}, {'end': 9206.611, 'text': 'S sub 3 means add together the first three terms, and so on.', 'start': 9202.729, 'duration': 3.882}, {'end': 9215.158, 'text': 'Let me contrast the sequence of partial sums with the sequence of terms that we started out with.', 'start': 9209.093, 'duration': 6.065}, {'end': 9222.727, 'text': "Those are denoted A sub n, so here's A sub 1, the first term, a sub 2, the second term, and so on.", 'start': 9215.538, 'duration': 7.189}, {'end': 9233.315, 'text': "Although I can't physically add up infinitely many numbers, I can observe that as I add up more and more numbers,", 'start': 9225.349, 'duration': 7.966}, {'end': 9238.419, 'text': 'my partial sums are approaching the number 1..', 'start': 9233.315, 'duration': 5.104}, {'end': 9248.504, 'text': 'That is, the limit as the number of terms I add up, n, goes to infinity of the nth partial sum is equal to 1.', 'start': 9238.419, 'duration': 10.085}, {'end': 9255.626, 'text': 'So it makes sense that if I could add up all infinitely many numbers, I should get an exact sum of 1.', 'start': 9248.504, 'duration': 7.122}, {'end': 9262.527, 'text': 'The sum of this infinite series is 1.', 'start': 9255.626, 'duration': 6.901}, {'end': 9268.149, 'text': "In fact, for this particular series, there's a nice way to see that the sum is 1 using geometry.", 'start': 9262.527, 'duration': 5.622}, {'end': 9276.211, 'text': 'If I draw a square with side length 1 and fill in half of the square, that gives me an area of 1 half.', 'start': 9268.969, 'duration': 7.242}, {'end': 9282.376, 'text': 'Now if I draw a line here, that gives me an additional area of 1 4th.', 'start': 9277.793, 'duration': 4.583}, {'end': 9287.278, 'text': "Here's an area of 1 8th, 1 16th.", 'start': 9283.736, 'duration': 3.542}, {'end': 9295.823, 'text': 'And if I keep spiraling in here, I keep adding areas that exactly match the terms of this series.', 'start': 9287.939, 'duration': 7.884}, {'end': 9306.109, 'text': "In the limit, I'll have filled in the entire square, which has an area of 1.", 'start': 9297.584, 'duration': 8.525}, {'end': 9312.694, 'text': 'In this example, we found the sum of the series by evaluating the limit of the partial sums.', 'start': 9306.109, 'duration': 6.585}, {'end': 9316.737, 'text': 'And in general, this is how we find the sum of any series.', 'start': 9313.815, 'duration': 2.922}, {'end': 9331.067, 'text': 'For any series, the partial sums of the series are defined as the sequence s sub n, where s sub 1 is equal to just the first term, a sub 1,', 'start': 9319.079, 'duration': 11.988}, {'end': 9336.907, 'text': 's sub 2 is equal to the sum of the first two terms, a sub one plus a sub two.', 'start': 9331.067, 'duration': 5.84}, {'end': 9341.408, 'text': 's sub three is a sum of the first three terms.', 'start': 9336.907, 'duration': 4.501}, {'end': 9347.771, 'text': 'And in general, s sub n is the sum of the first n terms.', 'start': 9343.049, 'duration': 4.722}, {'end': 9357.755, 'text': 'I can also write this in sigma notation as the sum of say, k equals one to n of a sub k.', 'start': 9349.792, 'duration': 7.963}, {'end': 9367.082, 'text': "I'm using a different letter k here as the index just because I'm already using n to represent the number of terms that I'm adding up.", 'start': 9360.359, 'duration': 6.723}, {'end': 9376.966, 'text': 'The sum of a sub n is said to converge if this sequence of partial sums converges as a sequence.', 'start': 9369.503, 'duration': 7.463}, {'end': 9386.451, 'text': "That is, if the limit as n goes to infinity of the s sub n's exists as a finite number.", 'start': 9378.227, 'duration': 8.224}, {'end': 9400.727, 'text': 'if the limit does not exist or the limit is infinity or negative infinity, then the series is said to diverge.', 'start': 9389.62, 'duration': 11.107}, {'end': 9408.852, 'text': "I want to emphasize that we're talking about the limit of the partial sums here, not the limit of the original terms, a sub n,", 'start': 9400.727, 'duration': 8.125}, {'end': 9416.197, 'text': "and it's important to keep in mind when you're working with series that for any series there are actually two sequences of interest.", 'start': 9408.852, 'duration': 7.345}], 'summary': 'The sum of the infinite series is 1, found using partial sums and geometric visualization.', 'duration': 258.5, 'max_score': 9157.697, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe89157697.jpg'}, {'end': 9255.626, 'src': 'embed', 'start': 9225.349, 'weight': 2, 'content': [{'end': 9233.315, 'text': "Although I can't physically add up infinitely many numbers, I can observe that as I add up more and more numbers,", 'start': 9225.349, 'duration': 7.966}, {'end': 9238.419, 'text': 'my partial sums are approaching the number 1..', 'start': 9233.315, 'duration': 5.104}, {'end': 9248.504, 'text': 'That is, the limit as the number of terms I add up, n, goes to infinity of the nth partial sum is equal to 1.', 'start': 9238.419, 'duration': 10.085}, {'end': 9255.626, 'text': 'So it makes sense that if I could add up all infinitely many numbers, I should get an exact sum of 1.', 'start': 9248.504, 'duration': 7.122}], 'summary': 'As the number of terms approaches infinity, the sum approaches 1.', 'duration': 30.277, 'max_score': 9225.349, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe89225349.jpg'}, {'end': 9347.771, 'src': 'embed', 'start': 9319.079, 'weight': 0, 'content': [{'end': 9331.067, 'text': 'For any series, the partial sums of the series are defined as the sequence s sub n, where s sub 1 is equal to just the first term, a sub 1,', 'start': 9319.079, 'duration': 11.988}, {'end': 9336.907, 'text': 's sub 2 is equal to the sum of the first two terms, a sub one plus a sub two.', 'start': 9331.067, 'duration': 5.84}, {'end': 9341.408, 'text': 's sub three is a sum of the first three terms.', 'start': 9336.907, 'duration': 4.501}, {'end': 9347.771, 'text': 'And in general, s sub n is the sum of the first n terms.', 'start': 9343.049, 'duration': 4.722}], 'summary': 'Partial sums of a series defined as s sub n, where s sub 1 is the first term, s sub 2 is the sum of first two terms, and s sub n is the sum of first n terms.', 'duration': 28.692, 'max_score': 9319.079, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe89319079.jpg'}], 'start': 8384.636, 'title': 'Introduction to sequences and series', 'summary': 'Provides definitions and notation for sequences, including recursive formulas and closed form expressions. it covers arithmetic and geometric sequences, with examples and formulas for finding the nth term, and introduces sequences and series, explaining how to find the sum of a series and the concept of partial sums and convergence.', 'chapters': [{'end': 8709.593, 'start': 8384.636, 'title': 'Sequences and notation', 'summary': 'Provides definitions and notation for sequences, including recursive formulas and closed form expressions, and demonstrates the process of writing a formula for the general term of a sequence.', 'duration': 324.957, 'highlights': ['The sequence is a list of numbers in a particular order, often described abstractly with letters, and can be represented by a sub n with curly brackets, implying that n ranges through all positive integers.', 'The process of finding the general formula for a sequence involves identifying the pattern, such as a linear function with a specific slope, and determining the y-intercept to formulate the general equation.', 'Recursive formulas are used to define the nth term of a sequence indirectly in terms of previous terms, while closed form non-recursive formulas can also describe sequences, offering alternative representations.', 'Demonstrates finding the first few terms of a recursive sequence using the given recursive formula and validates the formula by plugging in values of n to confirm its accuracy and consistency.']}, {'end': 9016.789, 'start': 8711.363, 'title': 'Arithmetic and geometric sequences', 'summary': 'Covers arithmetic and geometric sequences, including examples with common differences and ratios, as well as formulas for finding the nth term, with a brief discussion on non-arithmetic and non-geometric sequences.', 'duration': 305.426, 'highlights': ['The chapter explains arithmetic sequences and their formulas, such as D times K plus A, and the equivalence with D times N minus 1 plus A, highlighting the relationship between the indices k and n.', 'The section discusses geometric sequences and their properties, presenting examples with common ratios, and providing formulas for finding the nth term using a times r to the n or a times r to the n minus 1.', 'An example of a non-arithmetic and non-geometric sequence is illustrated, demonstrating the method of finding the formula for the nth term by identifying the pattern of alternating terms and determining the numerator and denominator based on the sequence.']}, {'end': 9531.727, 'start': 9020.492, 'title': 'Introduction to sequences and series', 'summary': 'Covers the introduction to sequences and series, including arithmetic, geometric, and recursively defined sequences, and explains how to find the sum of a series, highlighting the concept of partial sums and convergence of series.', 'duration': 511.235, 'highlights': ['The sequence of partial sums approaches the number 1 as the number of terms increases, and the sum of the infinite series 1/2^n converges to 1.', 'The method of finding the sum of any series involves evaluating the limit of the partial sums, with the sequence of partial sums denoted as s sub n.', 'A series converges if the sequence of partial sums converges to a finite number, and it diverges if the limit of the partial sums does not exist or is infinity or negative infinity.']}], 'duration': 1147.091, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe88384636.jpg', 'highlights': ['The method of finding the sum of any series involves evaluating the limit of the partial sums, with the sequence of partial sums denoted as s sub n.', 'Demonstrates finding the first few terms of a recursive sequence using the given recursive formula and validates the formula by plugging in values of n to confirm its accuracy and consistency.', 'The sequence of partial sums approaches the number 1 as the number of terms increases, and the sum of the infinite series 1/2^n converges to 1.', 'The section discusses geometric sequences and their properties, presenting examples with common ratios, and providing formulas for finding the nth term using a times r to the n or a times r to the n minus 1.', 'The chapter explains arithmetic sequences and their formulas, such as D times K plus A, and the equivalence with D times N minus 1 plus A, highlighting the relationship between the indices k and n.']}, {'end': 10599.283, 'segs': [{'end': 9562.446, 'src': 'embed', 'start': 9531.727, 'weight': 0, 'content': [{'end': 9545.073, 'text': 'So the limit as n goes to infinity of the partial sums is the limit as n goes to infinity of n over n plus 1, which is just 1.', 'start': 9531.727, 'duration': 13.346}, {'end': 9549.995, 'text': 'That means that the sum of the infinite series is equal to 1.', 'start': 9545.073, 'duration': 4.922}, {'end': 9552.756, 'text': 'By coincidence, the same sum as in the previous example.', 'start': 9549.995, 'duration': 2.761}, {'end': 9562.446, 'text': 'In this video, we saw that to find the sum of an infinite series, we have to calculate the partial sums.', 'start': 9555.417, 'duration': 7.029}], 'summary': 'The sum of the infinite series is 1, calculated using partial sums.', 'duration': 30.719, 'max_score': 9531.727, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe89531727.jpg'}, {'end': 9619.947, 'src': 'embed', 'start': 9593.639, 'weight': 2, 'content': [{'end': 9598.905, 'text': 'A sequence is bounded above if all of its terms are less than or equal to some number.', 'start': 9593.639, 'duration': 5.266}, {'end': 9615.246, 'text': "In other words, there's a number capital M such that the term a sub n is less than or equal to capital M for all values of the index n.", 'start': 9599.666, 'duration': 15.58}, {'end': 9619.947, 'text': 'A sequence is bounded below if all of its terms are greater than or equal to some number.', 'start': 9615.246, 'duration': 4.701}], 'summary': 'Sequence is bounded above if all terms <= some number, and bounded below if all terms >= some number.', 'duration': 26.308, 'max_score': 9593.639, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe89593639.jpg'}, {'end': 10165.854, 'src': 'heatmap', 'start': 9906.481, 'weight': 0.707, 'content': [{'end': 9913.064, 'text': 'Since we never go down, as we go from one term to the next, we either go up or stay at the same level.', 'start': 9906.481, 'duration': 6.583}, {'end': 9924.009, 'text': 'In this case, however, we could not say that the sequence is monotonically increasing because of the equality between some pairs of consecutive terms.', 'start': 9915.265, 'duration': 8.744}, {'end': 9927.351, 'text': 'The third sequence is not monotonic.', 'start': 9925.19, 'duration': 2.161}, {'end': 9931.792, 'text': 'the numbers bounce around between positive and negative numbers.', 'start': 9928.53, 'duration': 3.262}, {'end': 9936.774, 'text': "And therefore sometimes we are decreasing while other times we're increasing.", 'start': 9932.352, 'duration': 4.422}, {'end': 9942.337, 'text': 'And the fourth sequence is not monotonic because of the first few terms.', 'start': 9938.435, 'duration': 3.902}, {'end': 9949, 'text': 'However, from the fifth term on, the terms are monotonically non decreasing.', 'start': 9942.937, 'duration': 6.063}, {'end': 9951.741, 'text': 'And we could also say monotonically increasing.', 'start': 9949.58, 'duration': 2.161}, {'end': 9959.885, 'text': 'In this video, we gave definitions for bounded and monotonic sequences.', 'start': 9955.023, 'duration': 4.862}, {'end': 9965.55, 'text': "We'll see later that these definitions can be important for determining when a sequence converges.", 'start': 9960.586, 'duration': 4.964}, {'end': 9974.998, 'text': 'In class, we talked briefly about this sequence, n minus five over n squared, where n goes from one to infinity.', 'start': 9967.912, 'duration': 7.086}, {'end': 9979.642, 'text': "We wanted to know if this sequence is monotonic, and if it's bounded.", 'start': 9976.46, 'duration': 3.182}, {'end': 9987.549, 'text': 'If we compute the first few terms, The sequence appears to be steadily increasing.', 'start': 9981.684, 'duration': 5.865}, {'end': 9990.85, 'text': 'But in this case, appearances are deceiving.', 'start': 9988.81, 'duration': 2.04}, {'end': 10008.338, 'text': "And a better way to decide whether it's monotonic is to use calculus to decide if the associated function f of x equals x minus 5 over x squared is increasing for x values bigger than 1..", 'start': 9992.631, 'duration': 15.707}, {'end': 10012.96, 'text': 'So let me take the derivative using the quotient rule and simplify.', 'start': 10008.338, 'duration': 4.622}, {'end': 10022.29, 'text': 'I get that f prime of x is equal to minus x squared plus 10x over x to the fourth.', 'start': 10014.468, 'duration': 7.822}, {'end': 10036.433, 'text': 'And now I need to decide, is f prime of x greater than zero for x bigger than one? If so, my function and therefore my sequence will be increasing.', 'start': 10022.71, 'duration': 13.723}, {'end': 10043.495, 'text': "To check if f prime of x is greater than zero, I'll first set f prime of x equal to zero.", 'start': 10037.674, 'duration': 5.821}, {'end': 10054.755, 'text': "So I'll set my ratio here equal to 0, which means my numerator needs to be equal to 0.", 'start': 10046.232, 'duration': 8.523}, {'end': 10067.579, 'text': 'And if I factor that, I get that x equals 0 or x equals 10.', 'start': 10054.755, 'duration': 12.824}, {'end': 10077.109, 'text': "Now, if I draw my number line, since f prime is equal to 0 at 0 and 10, It'll be positive and negative in between these values.", 'start': 10067.579, 'duration': 9.53}, {'end': 10093.014, 'text': 'and by plugging in values like x equals negative 1, 1, and 11, I can see that f prime is negative for x less than 0, positive for x between 0 and 10,', 'start': 10077.109, 'duration': 15.905}, {'end': 10096.636, 'text': 'and negative for x bigger than 10..', 'start': 10093.014, 'duration': 3.622}, {'end': 10102.798, 'text': 'In particular, f increases when x increases from 1 to 10, and then it decreases.', 'start': 10096.636, 'duration': 6.162}, {'end': 10105.507, 'text': 'And so the same thing is happening to our sequence.', 'start': 10103.325, 'duration': 2.182}, {'end': 10109.55, 'text': 'Therefore, the sequence is not monotonic.', 'start': 10106.868, 'duration': 2.682}, {'end': 10114.474, 'text': 'We can also use calculus to check if the sequence is bounded.', 'start': 10111.432, 'duration': 3.042}, {'end': 10123.082, 'text': 'Our first derivative test shows that our function f of x has a maximum at x equals 10.', 'start': 10115.896, 'duration': 7.186}, {'end': 10127.866, 'text': "At least that's the maximum for x values ranging from one to infinity.", 'start': 10123.082, 'duration': 4.784}, {'end': 10129.907, 'text': "And that's all that's relevant for our sequence.", 'start': 10128.166, 'duration': 1.741}, {'end': 10143.801, 'text': 'Therefore, Our sequence is bounded above by its value at 10, which is 10 minus 5 over 10 squared, or 1 20th.', 'start': 10131.148, 'duration': 12.653}, {'end': 10152.988, 'text': 'Now notice that our sequence n minus 5 over n squared is always bigger than 0, for n bigger than 5,', 'start': 10145.663, 'duration': 7.325}, {'end': 10155.81, 'text': 'since the numerator and denominator are both positive in this situation.', 'start': 10152.988, 'duration': 2.822}, {'end': 10165.854, 'text': 'And since there are only finitely many terms where n is less than 5, we can just use the minimum of these terms and 0 as a lower bound.', 'start': 10155.83, 'duration': 10.024}], 'summary': 'The sequence n-5/n^2 is not monotonic, and it is bounded above by 1/20th.', 'duration': 259.373, 'max_score': 9906.481, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe89906481.jpg'}, {'end': 9990.85, 'src': 'embed', 'start': 9938.435, 'weight': 4, 'content': [{'end': 9942.337, 'text': 'And the fourth sequence is not monotonic because of the first few terms.', 'start': 9938.435, 'duration': 3.902}, {'end': 9949, 'text': 'However, from the fifth term on, the terms are monotonically non decreasing.', 'start': 9942.937, 'duration': 6.063}, {'end': 9951.741, 'text': 'And we could also say monotonically increasing.', 'start': 9949.58, 'duration': 2.161}, {'end': 9959.885, 'text': 'In this video, we gave definitions for bounded and monotonic sequences.', 'start': 9955.023, 'duration': 4.862}, {'end': 9965.55, 'text': "We'll see later that these definitions can be important for determining when a sequence converges.", 'start': 9960.586, 'duration': 4.964}, {'end': 9974.998, 'text': 'In class, we talked briefly about this sequence, n minus five over n squared, where n goes from one to infinity.', 'start': 9967.912, 'duration': 7.086}, {'end': 9979.642, 'text': "We wanted to know if this sequence is monotonic, and if it's bounded.", 'start': 9976.46, 'duration': 3.182}, {'end': 9987.549, 'text': 'If we compute the first few terms, The sequence appears to be steadily increasing.', 'start': 9981.684, 'duration': 5.865}, {'end': 9990.85, 'text': 'But in this case, appearances are deceiving.', 'start': 9988.81, 'duration': 2.04}], 'summary': 'The sequence n minus five over n squared is not monotonic and not bounded.', 'duration': 52.415, 'max_score': 9938.435, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe89938435.jpg'}, {'end': 10155.81, 'src': 'embed', 'start': 10131.148, 'weight': 3, 'content': [{'end': 10143.801, 'text': 'Therefore, Our sequence is bounded above by its value at 10, which is 10 minus 5 over 10 squared, or 1 20th.', 'start': 10131.148, 'duration': 12.653}, {'end': 10152.988, 'text': 'Now notice that our sequence n minus 5 over n squared is always bigger than 0, for n bigger than 5,', 'start': 10145.663, 'duration': 7.325}, {'end': 10155.81, 'text': 'since the numerator and denominator are both positive in this situation.', 'start': 10152.988, 'duration': 2.822}], 'summary': 'The sequence is bounded above by its value at 10, which is 1/20th.', 'duration': 24.662, 'max_score': 10131.148, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe810131148.jpg'}, {'end': 10360.142, 'src': 'embed', 'start': 10326.244, 'weight': 1, 'content': [{'end': 10328.886, 'text': 'these definitions of indeterminate form.', 'start': 10326.244, 'duration': 2.642}, {'end': 10334.871, 'text': "it's possible for A to be in negative infinity or infinity, like it is in this example, but it doesn't have to be.", 'start': 10328.886, 'duration': 5.985}, {'end': 10346.561, 'text': "L'Hopital's rule can be applied when f and g are differentiable functions and the derivative of g is non-zero in some open interval around A,", 'start': 10337.594, 'duration': 8.967}, {'end': 10349.744, 'text': 'except possibly at A.', 'start': 10346.561, 'duration': 3.183}, {'end': 10360.142, 'text': 'Under these conditions, If the limit, as x goes to a, of f of x over g of x is a 0 over 0 or infinity over infinity in determinant form,', 'start': 10349.744, 'duration': 10.398}], 'summary': "L'hopital's rule applies to indeterminate forms in calculus.", 'duration': 33.898, 'max_score': 10326.244, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe810326244.jpg'}], 'start': 9531.727, 'title': "Infinite series, monotonic sequences, and l'hopital's rule", 'summary': "Explains infinite series, sequences, bounded and monotonic sequences, uses calculus to determine monotonicity and boundedness, and introduces l'hopital's rule for evaluating limits in indeterminate forms, with illustrative examples.", 'chapters': [{'end': 9719.222, 'start': 9531.727, 'title': 'Infinite series and sequence definitions', 'summary': 'Explains the concept of infinite series and sequences, emphasizing the calculation of partial sums, defining bounded and monotonic sequences, and illustrating examples of bounded sequences in detail.', 'duration': 187.495, 'highlights': ['The sum of the infinite series is equal to 1, calculated by finding the limit as n goes to infinity of the partial sums.', 'The definition of bounded sequences is provided, highlighting the concepts of bounded above, bounded below, and bounded sequences.', 'Illustration of bounded sequences through specific examples, including the determination of upper and lower bounds for each sequence.']}, {'end': 10185.688, 'start': 9722.096, 'title': 'Monotonic sequences and calculus', 'summary': 'Discusses the concepts of monotonic sequences, non-decreasing and non-increasing sequences, and uses calculus to determine if a sequence is monotonic and bounded, with an example of n - 5 / n^2 sequence showing how calculus can help in deciding monotonicity and boundedness.', 'duration': 463.592, 'highlights': ['The sequence n minus 5 over n squared is steadily increasing, but using calculus, it is determined that the sequence is not monotonic, as it increases from 1 to 10 and then decreases.', 'The sequence n minus 5 over n squared is bounded above by 1/20th and below by -4, making it a bounded sequence.', 'The concepts of non-decreasing and non-increasing sequences are explained, along with the definitions for increasing, decreasing, monotonic, and bounded sequences.']}, {'end': 10599.283, 'start': 10188.194, 'title': "L'hopital's rule for evaluating limits", 'summary': "Introduces l'hopital's rule, a technique for evaluating limits in indeterminate forms like 0/0 and infinity/infinity, using examples to illustrate its application and emphasizing the importance of simplifying after each application of the rule.", 'duration': 411.089, 'highlights': ["L'Hopital's rule can be applied when f and g are differentiable functions and the derivative of g is non-zero in some open interval around A, except possibly at A.", 'The final limit of the quotient could be any number at all, or it could be infinity, or it could not even exist.', "Emphasizes the importance of simplifying after each application of L'Hopital's rule to avoid complicating the problem."]}], 'duration': 1067.556, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe89531727.jpg', 'highlights': ['The sum of the infinite series is equal to 1, calculated by finding the limit as n goes to infinity of the partial sums.', "L'Hopital's rule can be applied when f and g are differentiable functions and the derivative of g is non-zero in some open interval around A, except possibly at A.", 'Illustration of bounded sequences through specific examples, including the determination of upper and lower bounds for each sequence.', 'The sequence n minus 5 over n squared is bounded above by 1/20th and below by -4, making it a bounded sequence.', 'The concepts of non-decreasing and non-increasing sequences are explained, along with the definitions for increasing, decreasing, monotonic, and bounded sequences.', 'Using calculus, it is determined that the sequence n minus 5 over n squared is not monotonic, as it increases from 1 to 10 and then decreases.']}, {'end': 11725.474, 'segs': [{'end': 10631.873, 'src': 'embed', 'start': 10599.283, 'weight': 0, 'content': [{'end': 10615.594, 'text': 'we were able to evaluate 0 over 0 and infinity over infinity in determinant forms by replacing the limit of f of x over g of x with the limit of f prime of x over g prime of x,', 'start': 10599.283, 'duration': 16.311}, {'end': 10617.215, 'text': 'provided that second limit exists.', 'start': 10615.594, 'duration': 1.621}, {'end': 10620.958, 'text': "This trick is known as L'Hopital's Rule.", 'start': 10618.756, 'duration': 2.202}, {'end': 10629.891, 'text': "We've seen that L'Hopital's rule can be used to evaluate limits of the form 0 over 0, or infinity over infinity.", 'start': 10623.567, 'duration': 6.324}, {'end': 10631.873, 'text': 'In this video.', 'start': 10631.252, 'duration': 0.621}], 'summary': "L'hopital's rule evaluates 0/0 and ∞/∞ by f'/g' if second limit exists.", 'duration': 32.59, 'max_score': 10599.283, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe810599283.jpg'}, {'end': 10718.794, 'src': 'embed', 'start': 10688.085, 'weight': 2, 'content': [{'end': 10692.347, 'text': 'while the ln x factor is pulling the product towards large negative numbers.', 'start': 10688.085, 'duration': 4.262}, {'end': 10696.17, 'text': "And it's hard to predict what the limit of the product will actually be.", 'start': 10692.607, 'duration': 3.563}, {'end': 10703.955, 'text': 'But the great thing is, I can actually rewrite this product to look like an infinity over infinity indeterminate form.', 'start': 10697.711, 'duration': 6.244}, {'end': 10707.412, 'text': 'or a zero over zero in determinant form.', 'start': 10704.731, 'duration': 2.681}, {'end': 10718.794, 'text': 'Instead of sine x times ln x, I can rewrite the limit as ln x divided by one over sine x.', 'start': 10708.532, 'duration': 10.262}], 'summary': 'The limit of the product can be rewritten as an indeterminate form, ln x divided by one over sine x.', 'duration': 30.709, 'max_score': 10688.085, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe810688085.jpg'}, {'end': 10858.409, 'src': 'embed', 'start': 10766.428, 'weight': 1, 'content': [{'end': 10771.671, 'text': 'Sometimes it can be difficult to decide which of these two ways to rewrite a product as a quotient.', 'start': 10766.428, 'duration': 5.243}, {'end': 10778.575, 'text': 'One rule of thumb is to take the version that makes it easier to take the derivative of the numerator and denominator.', 'start': 10773.012, 'duration': 5.563}, {'end': 10784.659, 'text': 'Another trick is just to try one of the ways, and if you get stuck, go back and try the other.', 'start': 10779.778, 'duration': 4.881}, {'end': 10788.66, 'text': "I'm going to use the first method of rewriting it,", 'start': 10786.38, 'duration': 2.28}, {'end': 10798.843, 'text': 'because I recognize that 1 over sine x can be written as cosecant of x and I know how to take the derivative of cosecant x.', 'start': 10788.66, 'duration': 10.183}, {'end': 10803.244, 'text': "Using L'Hopital's rule on this infinity over infinity indeterminate form.", 'start': 10798.843, 'duration': 4.401}, {'end': 10810.962, 'text': "I can rewrite my limit as the limit of what I get when I take the derivative of the numerator, that's 1 over x.", 'start': 10803.244, 'duration': 7.718}, {'end': 10819.305, 'text': "divided by the derivative of the denominator, that's negative cosecant x, cotangent x.", 'start': 10810.962, 'duration': 8.343}, {'end': 10822.886, 'text': 'As always, I want to simplify my expression before going any further.', 'start': 10819.305, 'duration': 3.581}, {'end': 10829.248, 'text': 'I can rewrite my trig functions in the denominator in terms of sine and cosine.', 'start': 10824.766, 'duration': 4.482}, {'end': 10844.838, 'text': 'Cosecant x is 1 over sine x, and cotangent x is cosine of x over sine of x.', 'start': 10831.268, 'duration': 13.57}, {'end': 10858.409, 'text': 'Now flipping and multiplying, I get the limit as x goes to zero plus of one over x times sine squared of x over negative cosine of x.', 'start': 10844.838, 'duration': 13.571}], 'summary': "When rewriting as a quotient, prioritize ease of derivative calculation and utilize l'hopital's rule for indeterminate forms, simplifying trig functions as necessary.", 'duration': 91.981, 'max_score': 10766.428, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe810766428.jpg'}, {'end': 10924.12, 'src': 'embed', 'start': 10888.865, 'weight': 6, 'content': [{'end': 10898.473, 'text': 'Taking the derivative of the top, I get negative 2 sine x cosine of x, and the derivative of the bottom is just 1.', 'start': 10888.865, 'duration': 9.608}, {'end': 10903.238, 'text': "Now I'm in a good position just to evaluate the limit by plugging in 0 for x.", 'start': 10898.473, 'duration': 4.765}, {'end': 10916.914, 'text': 'In the numerator, I get negative 2 times 0 times 1, the denominator is just 1, so my final limit is 0.', 'start': 10903.238, 'duration': 13.676}, {'end': 10918.776, 'text': 'In this limit, we have a battle of forces.', 'start': 10916.914, 'duration': 1.862}, {'end': 10924.12, 'text': 'As x is going to infinity, 1 over x is going to 0.', 'start': 10919.856, 'duration': 4.264}], 'summary': 'Derivative evaluation yields a final limit of 0 as x approaches 0.', 'duration': 35.255, 'max_score': 10888.865, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe810888865.jpg'}, {'end': 11012.674, 'src': 'embed', 'start': 10985.076, 'weight': 5, 'content': [{'end': 10997.046, 'text': 'Now if I wanted to take the limit as x goes to infinity of ln y, that would be the limit of this product x times ln one plus one over x.', 'start': 10985.076, 'duration': 11.97}, {'end': 11004.325, 'text': 'As x goes from infinity, the first factor x goes to infinity.', 'start': 10999.238, 'duration': 5.087}, {'end': 11012.674, 'text': 'one plus one over x goes to just one, and ln one is going to zero.', 'start': 11006.21, 'duration': 6.464}], 'summary': 'Limit of ln y as x goes to infinity is zero.', 'duration': 27.598, 'max_score': 10985.076, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe810985076.jpg'}, {'end': 11192.783, 'src': 'embed', 'start': 11159.587, 'weight': 7, 'content': [{'end': 11168.074, 'text': "So 1 to the infinity, infinity to the 0, and 0 to the 0 are all indeterminate forms that can be handled using L'Hopital's rule.", 'start': 11159.587, 'duration': 8.487}, {'end': 11189.421, 'text': 'In this video we saw that a 0 times infinity indeterminate form could be converted to a 0 over 0 or infinity over infinity indeterminate form by rewriting f of x times g of x as f of x divided by 1 over g of x,', 'start': 11170.632, 'duration': 18.789}, {'end': 11192.783, 'text': 'or as g of x divided by 1 over f of x.', 'start': 11189.421, 'duration': 3.362}], 'summary': "L'hopital's rule handles indeterminate forms like 0/0, ∞/∞, 0^0.", 'duration': 33.196, 'max_score': 11159.587, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe811159587.jpg'}, {'end': 11534.371, 'src': 'embed', 'start': 11501.362, 'weight': 9, 'content': [{'end': 11507.284, 'text': 'One example of a convergent sequence is a sequence 1 over n.', 'start': 11501.362, 'duration': 5.922}, {'end': 11516.106, 'text': 'This sequence converges to 0 since the limit as n goes to infinity of 1 over n is 0.', 'start': 11507.284, 'duration': 8.822}, {'end': 11524.825, 'text': 'A divergent sequence is 2 to the n since the limit as n goes to infinity of 2 to the n is infinity.', 'start': 11516.106, 'duration': 8.719}, {'end': 11534.371, 'text': "One example of a sequence that's bounded but still diverges is negative 1 to the n.", 'start': 11526.346, 'duration': 8.025}], 'summary': 'Convergent sequence: 1/n converges to 0. divergent sequence: 2^n goes to infinity. bounded but diverges: -1^n.', 'duration': 33.009, 'max_score': 11501.362, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe811501362.jpg'}, {'end': 11638.006, 'src': 'embed', 'start': 11613.036, 'weight': 8, 'content': [{'end': 11626.924, 'text': "we can figure out if a sequence converges by replacing the terms a sub n with f of x for some appropriate function and then using L'Hopital's Rule or other tricks from Calculus 1 to show that the function's limit exists.", 'start': 11613.036, 'duration': 13.888}, {'end': 11629.926, 'text': "Let's try that for the following example.", 'start': 11628.225, 'duration': 1.701}, {'end': 11638.006, 'text': "When the indices are missing, as in this example, We'll assume that n starts at 1 and goes to infinity.", 'start': 11631.206, 'duration': 6.8}], 'summary': "Determine sequence convergence using function substitution and l'hopital's rule", 'duration': 24.97, 'max_score': 11613.036, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe811613036.jpg'}], 'start': 10599.283, 'title': "L'hopital's rule applications", 'summary': "Discusses the application of l'hopital's rule to evaluate various indeterminate forms such as 0 times infinity, infinity to the 0, 0 to the 0, and 1 to the infinity, using examples and strategies for rewriting products as quotients. it also covers using l'hopital's rule to evaluate limits involving trig functions and exponentials, and applying it to handle sequence convergence with examples of convergent, divergent, and bounded sequences.", 'chapters': [{'end': 10788.66, 'start': 10599.283, 'title': "L'hopital's rule for evaluating limits", 'summary': "Discusses the application of l'hopital's rule to evaluate various indeterminate forms such as 0 times infinity, infinity to the 0, 0 to the 0, and 1 to the infinity, using examples and strategies for rewriting products as quotients.", 'duration': 189.377, 'highlights': ["L'Hopital's rule can be used to evaluate limits of the form 0 over 0, or infinity over infinity by replacing the limit of f of x over g of x with the limit of f prime of x over g prime of x, provided that second limit exists.", 'The process of rewriting a product as a quotient can be done in multiple ways, such as ln x divided by one over sine x or sine x divided by one over ln x, depending on which makes it easier to take the derivative of the numerator and denominator.', 'It is challenging to predict the limit of a product in 0 times infinity indeterminate form, as the factors pull the product towards 0 and large negative numbers, making it difficult to determine the actual limit.']}, {'end': 11078.338, 'start': 10788.66, 'title': "L'hopital's rule and indeterminate forms", 'summary': "Discusses using l'hopital's rule to evaluate limits, including examples such as rewriting trig functions, evaluating limits involving exponentials, and applying logarithms to solve indeterminate forms.", 'duration': 289.678, 'highlights': ["The derivative of 1 over sine x can be written as negative cosecant x cotangent x, allowing for the application of L'Hopital's rule on the indeterminate form of the limit.", 'Simplifying trig functions in the denominator by rewriting them in terms of sine and cosine to evaluate the limit as x goes to zero plus of negative sine squared x over x cosine x.', "Using logarithms to tackle the indeterminate form of the limit as x goes to infinity of ln y, rewriting it as the limit of ln one plus one over x divided by one over x and applying L'Hopital's rule.", 'Taking the derivative of the top and the bottom to rewrite the limit as the limit of one over one plus one over x as x goes to infinity.']}, {'end': 11725.474, 'start': 11078.338, 'title': "L'hopital's rule and sequence convergence", 'summary': "Covers the application of l'hopital's rule to handle indeterminate forms like 1 to the infinity, infinity to the 0, and 0 to the 0, and provides techniques for proving the convergence of sequences using standard calculus tricks and l'hopital's rule with examples of convergent, divergent, and bounded sequences.", 'duration': 647.136, 'highlights': ["The chapter covers the application of L'Hopital's rule to handle indeterminate forms like 1 to the infinity, infinity to the 0, and 0 to the 0.", "Techniques for proving the convergence of sequences using standard calculus tricks and L'Hopital's Rule are provided in the chapter.", 'Example sequences of convergent, divergent, and bounded but still divergent cases are discussed.']}], 'duration': 1126.191, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe810599283.jpg', 'highlights': ["L'Hopital's rule applies to evaluate 0/0 or ∞/∞ by f'/g' if the second limit exists.", 'Rewrite product as quotient using ln x/1/sin x or sin x/1/ln x for easier derivative.', 'Challenging to predict limit of 0*∞ form as factors pull towards 0 and large neg. nums.', "Derivative of 1/sin x as -csc x cot x allows L'Hopital's rule on limit indeterminate form.", 'Simplify trig functions in denominator to evaluate limit as x→0+ of -sin²x/x cos x.', 'Use logarithms to tackle limit as x→∞ of ln y, rewriting as limit of ln(1+1/x)/(1/x).', "Applying L'Hopital's rule by taking derivative of top and bottom for limit as x→∞.", "Chapter covers L'Hopital's rule for 1^∞, ∞^0, and 0^0 indeterminate forms.", "Techniques for proving convergence of sequences using calculus tricks and L'Hopital's Rule.", 'Examples of convergent, divergent, and bounded but still divergent sequences discussed.']}, {'end': 14160.533, 'segs': [{'end': 11753.399, 'src': 'embed', 'start': 11727.075, 'weight': 0, 'content': [{'end': 11731.098, 'text': 'The derivative of the numerator is now 2 times e to the x.', 'start': 11727.075, 'duration': 4.023}, {'end': 11735.061, 'text': 'And the derivative of the denominator is also 2 times e to the x.', 'start': 11731.098, 'duration': 3.963}, {'end': 11738.604, 'text': 'So our limit here is 1.', 'start': 11735.061, 'duration': 3.543}, {'end': 11743.508, 'text': 'Since our function converges to 1, our sequence also converges to 1.', 'start': 11738.604, 'duration': 4.904}, {'end': 11753.399, 'text': 'So this original sequence converges to 1.', 'start': 11743.508, 'duration': 9.891}], 'summary': 'Derivative of numerator and denominator results in limit of 1, leading to sequence convergence to 1.', 'duration': 26.324, 'max_score': 11727.075, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe811727075.jpg'}, {'end': 11841.817, 'src': 'embed', 'start': 11809.873, 'weight': 1, 'content': [{'end': 11810.833, 'text': "Now it's easy to check.", 'start': 11809.873, 'duration': 0.96}, {'end': 11823.353, 'text': 'that the limit as n goes to infinity of negative 2 over n to the 2 thirds is 0, since as n goes to infinity, n to the 2 thirds also goes to infinity.', 'start': 11811.343, 'duration': 12.01}, {'end': 11832.481, 'text': 'Similarly, the limit as n goes to infinity of 2 over n to the 2 thirds is 0.', 'start': 11824.815, 'duration': 7.666}, {'end': 11841.817, 'text': 'Since these two limits are the same, we know by the squeeze theorem that the limit of our sequence in the middle, has to exist at equal zero also.', 'start': 11832.481, 'duration': 9.336}], 'summary': 'Using the squeeze theorem, the limit of the sequence equals zero.', 'duration': 31.944, 'max_score': 11809.873, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe811809873.jpg'}, {'end': 11949.932, 'src': 'embed', 'start': 11920.678, 'weight': 2, 'content': [{'end': 11925.462, 'text': 'So we have a bounded monotonic sequence, and so this sequence has to converge.', 'start': 11920.678, 'duration': 4.784}, {'end': 11936.328, 'text': "Now what it actually converges to is a little mysterious, since it doesn't converge to some number we're already familiar with,", 'start': 11926.743, 'duration': 9.585}, {'end': 11941.149, 'text': 'like 0.6 or pi over 3 or something like that.', 'start': 11936.328, 'duration': 4.821}, {'end': 11949.932, 'text': "But it does converge to some real number, and that real number is called Champinown's constant.", 'start': 11942.85, 'duration': 7.082}], 'summary': "Bounded monotonic sequence converges to champinown's constant.", 'duration': 29.254, 'max_score': 11920.678, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe811920678.jpg'}, {'end': 12369.023, 'src': 'embed', 'start': 12345.281, 'weight': 4, 'content': [{'end': 12357.694, 'text': 'a sequence that can be written in the form of a times r to the n you can know that it converges if r is bigger than negative 1 and less than or equal to 1..', 'start': 12345.281, 'duration': 12.413}, {'end': 12363.218, 'text': "This sequence here, although it looks really complicated, it's really a geometric sequence in disguise.", 'start': 12357.694, 'duration': 5.524}, {'end': 12369.023, 'text': 'One way to see this is by simplifying the form of the terms.', 'start': 12365.08, 'duration': 3.943}], 'summary': 'Geometric sequence converges if -1 < r ≤ 1; simplifies to geometric sequence.', 'duration': 23.742, 'max_score': 12345.281, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe812345281.jpg'}, {'end': 12638.137, 'src': 'heatmap', 'start': 12379.531, 'weight': 0.785, 'content': [{'end': 12395.146, 'text': 'Now this is the same thing as negative e over 3 to the t times 1 over 3 squared times e.', 'start': 12379.531, 'duration': 15.615}, {'end': 12409.82, 'text': 'Now this is looking like the tail end of a geometric sequence where a is 1 over 3 squared times e and r, the common ratio, is minus e over 3.', 'start': 12395.146, 'duration': 14.674}, {'end': 12416.328, 'text': 'I say the tail end because we have t starting at 3 instead of at 0.', 'start': 12409.82, 'duration': 6.508}, {'end': 12425.255, 'text': 'Now since E is less than 3, the magnitude of R has got to be smaller than 1.', 'start': 12416.328, 'duration': 8.927}, {'end': 12431.26, 'text': "In other words, R is a negative number that's between negative 1 and 0.", 'start': 12425.255, 'duration': 6.005}, {'end': 12435.223, 'text': 'And therefore, the tail end of this geometric sequence converges.', 'start': 12431.26, 'duration': 3.963}, {'end': 12442.809, 'text': "It's kind of interesting to note that we could also rewrite this geometric sequence if we wanted to.", 'start': 12437.765, 'duration': 5.044}, {'end': 12448.291, 'text': 'using an index n going from 0 to infinity.', 'start': 12444.449, 'duration': 3.842}, {'end': 12457.475, 'text': 'And one way to figure out how to do that, the r, the common ratio, stays the same as negative e over 3.', 'start': 12448.671, 'duration': 8.804}, {'end': 12471.122, 'text': 'But since this version starts at t equals 3, the first term here is really minus e over 3 cubed times 1 over 3 squared e.', 'start': 12457.475, 'duration': 13.647}, {'end': 12472.723, 'text': 'And that becomes our value of a.', 'start': 12471.122, 'duration': 1.601}, {'end': 12484.579, 'text': 'Notice that when n is 0 here, I get this value, and when t is 3 here, I get the same value for this sequence.', 'start': 12475.137, 'duration': 9.442}, {'end': 12486.019, 'text': 'So these sequences are equivalent.', 'start': 12484.599, 'duration': 1.42}, {'end': 12494.501, 'text': 'But in any case, for either sequence, the common ratio r is negative e over 3, and the sequence converges, therefore.', 'start': 12486.399, 'duration': 8.102}, {'end': 12501.262, 'text': 'The final trick that I want to mention for deciding whether sequences converge or diverge is limit laws.', 'start': 12496.101, 'duration': 5.161}, {'end': 12507.316, 'text': 'The usual limit laws about addition, subtraction, and so on hold for sequences as well as for functions.', 'start': 12501.892, 'duration': 5.424}, {'end': 12515.883, 'text': 'So, for example, if the limit as n goes to infinity of a sub n is l and the limit of b sub n is m,', 'start': 12507.837, 'duration': 8.046}, {'end': 12522.789, 'text': 'then the limit of the sum a sub n plus b sub n is going to be equal to l plus m.', 'start': 12515.883, 'duration': 6.906}, {'end': 12527.192, 'text': 'And the limit of a sub n times b sub n is l times m.', 'start': 12522.789, 'duration': 4.403}, {'end': 12534.568, 'text': 'And the limit of c times a sub n, where c is some constant, is going to be C times L.', 'start': 12527.192, 'duration': 7.376}, {'end': 12538.21, 'text': 'Similar rules hold for subtraction and division.', 'start': 12534.568, 'duration': 3.642}, {'end': 12548.958, 'text': 'I want to emphasize that these limit holds hold under the condition that the limits of the component sequences exist as finite numbers.', 'start': 12540.712, 'duration': 8.246}, {'end': 12553.962, 'text': 'I can use these limit laws to decide if this sequence converges.', 'start': 12550.42, 'duration': 3.542}, {'end': 12562.602, 'text': 'Since the limit of the terms is equal to the difference of the limits, provided those limits exist.', 'start': 12555.443, 'duration': 7.159}, {'end': 12570.463, 'text': 'Now the first limit is 0, since the degree of the numerator is less than the degree of the denominator here.', 'start': 12563.722, 'duration': 6.741}, {'end': 12588.22, 'text': 'And the second limit is also 0, since this is a geometric sequence with ratio of 4 fifths, and 4 fifths is between 0 and 1.', 'start': 12571.043, 'duration': 17.177}, {'end': 12592.663, 'text': 'Therefore, the limit of our original sequence must be zero.', 'start': 12588.22, 'duration': 4.443}, {'end': 12597.387, 'text': 'In this video, we saw several ways to prove that a sequence converges.', 'start': 12593.544, 'duration': 3.843}, {'end': 12607.675, 'text': "We saw that we could use calculus techniques like L'Hopital's rule after replacing the sequence with its associated function defined on real numbers.", 'start': 12598.848, 'duration': 8.827}, {'end': 12611.257, 'text': 'We also saw that we could use the squeeze theorem.', 'start': 12609.256, 'duration': 2.001}, {'end': 12617.582, 'text': 'We noted that all sequences that are bounded and monotonic must converge.', 'start': 12612.838, 'duration': 4.744}, {'end': 12631.312, 'text': 'And we saw that geometric sequences always converge if r is bigger than negative 1 and less than or equal to 1.', 'start': 12618.683, 'duration': 12.629}, {'end': 12638.137, 'text': 'Finally, we saw that we can use limit laws to handle sums and products and other conglomerations of sequences.', 'start': 12631.312, 'duration': 6.825}], 'summary': 'Geometric sequences with negative e over 3 to the power of t converge, limit laws help determine convergence, and all bounded and monotonic sequences must converge.', 'duration': 258.606, 'max_score': 12379.531, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe812379531.jpg'}, {'end': 12949.394, 'src': 'embed', 'start': 12916.077, 'weight': 3, 'content': [{'end': 12929.063, 'text': 'As you may know, a geometric sequence converges to zero when the common ratio r is between negative one and one.', 'start': 12916.077, 'duration': 12.986}, {'end': 12934.866, 'text': 'it converges to a when r is equal to one and it diverges otherwise.', 'start': 12929.063, 'duration': 5.803}, {'end': 12941.087, 'text': 'when r is less than or equal to negative one or when r is greater than one.', 'start': 12935.322, 'duration': 5.765}, {'end': 12949.394, 'text': "we're assuming here that a is not equal to zero, since otherwise we'd have a very boring, though convergent sequence of all zeros.", 'start': 12941.087, 'duration': 8.307}], 'summary': 'Geometric sequence converges to zero with r between -1 and 1, and diverges when r ≤ -1 or r > 1.', 'duration': 33.317, 'max_score': 12916.077, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe812916077.jpg'}, {'end': 13030.141, 'src': 'embed', 'start': 13006.792, 'weight': 6, 'content': [{'end': 13013.537, 'text': 'our strategy is going to be to find a formula for the nth partial sum of the series, and then take the limit of partial sums.', 'start': 13006.792, 'duration': 6.745}, {'end': 13018.5, 'text': 'Since by definition, that limit tells us whether the series converges or diverges.', 'start': 13013.877, 'duration': 4.623}, {'end': 13023.299, 'text': 'Before we carry out that strategy, I want to consider one special case.', 'start': 13019.878, 'duration': 3.421}, {'end': 13030.141, 'text': 'When r is equal to 1, then the series is just a plus a plus a.', 'start': 13023.939, 'duration': 6.202}], 'summary': 'Strategy: find nth partial sum formula, analyze convergence, consider special case when r=1.', 'duration': 23.349, 'max_score': 13006.792, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/7gigNsz4Oe8/pics/7gigNsz4Oe813006792.jpg'}, {'end': 13378.512, 'src': 'heatmap', 'start': 13125.461, 'weight': 0.728, 'content': [{'end': 13132.625, 'text': 'Notice that this equation and the equation under it have a lot of terms in common, they match up along the diagonals.', 'start': 13125.461, 'duration': 7.164}, {'end': 13142.57, 'text': "So if I subtract the second equation from the first, on the left side, I'm going to get s sub n minus r times s sub n.", 'start': 13133.885, 'duration': 8.685}, {'end': 13151.198, 'text': 'But on the right side, a lot of my terms will cancel, this term will cancel with the next one, this one cancels with the previous one.', 'start': 13143.67, 'duration': 7.528}, {'end': 13160.707, 'text': "And what I'm left with is just a minus a times r to the n, I get a minus sign here, because I'm subtracting the whole equation.", 'start': 13151.898, 'duration': 8.809}, {'end': 13165.952, 'text': 'Now I can solve for s sub n, I can factor it out.', 'start': 13162.328, 'duration': 3.624}, {'end': 13170.357, 'text': "I'll factor out the a also just to keep things tidy.", 'start': 13167.574, 'duration': 2.783}, {'end': 13178.884, 'text': 'And then I get that s sub n is a times one minus r to the n over one minus r.', 'start': 13171.097, 'duration': 7.787}, {'end': 13182.027, 'text': "I don't have to worry about dividing by one minus r, that can't be zero.", 'start': 13178.884, 'duration': 3.143}, {'end': 13184.87, 'text': "Because remember, I'm assuming that r is not one.", 'start': 13182.467, 'duration': 2.403}, {'end': 13189.115, 'text': 'Now that I have a nice tidy formula for s, sub n,', 'start': 13185.993, 'duration': 3.122}, {'end': 13197.821, 'text': 'I can proceed to take the limit as n goes to infinity and see if my sequence of partial sums converges or diverges.', 'start': 13189.115, 'duration': 8.706}, {'end': 13200.343, 'text': "So that's the limit of my formula.", 'start': 13198.741, 'duration': 1.602}, {'end': 13205.266, 'text': 'And notice that the only part of this formula that depends on n is the r to the n.', 'start': 13200.583, 'duration': 4.683}, {'end': 13209.548, 'text': 'a one one minus r are all constants as far as n is concerned.', 'start': 13206.046, 'duration': 3.502}, {'end': 13214.29, 'text': 'So we can use limit rules to take those out of the limit.', 'start': 13210.088, 'duration': 4.202}, {'end': 13221.993, 'text': 'I can rewrite the limit as a over one minus r times the limit of one minus r to the n, or, even better,', 'start': 13214.29, 'duration': 7.703}, {'end': 13227.695, 'text': 'as a over one minus r times one minus the limit of r to the n.', 'start': 13221.993, 'duration': 5.702}, {'end': 13232.356, 'text': "Now, this limit I've seen before right on this page.", 'start': 13229.335, 'duration': 3.021}, {'end': 13240.037, 'text': "it's the same as the limit I was considering when I was looking at convergence of sequences, where I use the same r and my a is equal to one.", 'start': 13232.356, 'duration': 7.681}, {'end': 13255.181, 'text': 'So we know that this limit goes to zero, when r is between negative one and one, and does not exist as a finite number.', 'start': 13241.018, 'duration': 14.163}, {'end': 13262.581, 'text': 'when r is less than or equal to negative one or r is greater than one.', 'start': 13257.58, 'duration': 5.001}, {'end': 13273.303, 'text': 'Therefore, the limit of the partial sums is going to equal a over one minus r times one minus zero,', 'start': 13263.701, 'duration': 9.602}, {'end': 13279.945, 'text': 'which is just a over one minus r when r is between negative one and one,', 'start': 13273.303, 'duration': 6.642}, {'end': 13293.51, 'text': 'and that limit will not exist as a finite number when r is less than or equal to negative one or r is greater than one.', 'start': 13279.945, 'duration': 13.565}, {'end': 13301.734, 'text': "Since the limit also doesn't exist as a finite number, when r is one, I can add a little equality sign.", 'start': 13294.79, 'duration': 6.944}, {'end': 13304.316, 'text': "And now I've got all the cases for our covered.", 'start': 13302.095, 'duration': 2.221}, {'end': 13306.737, 'text': 'Let me write this down as a conclusion.', 'start': 13304.336, 'duration': 2.401}, {'end': 13323.433, 'text': "The geometric series converges to a over one minus r for r between negative one and one, I can also say that's the absolute value of r less than one.", 'start': 13310.7, 'duration': 12.733}, {'end': 13329.856, 'text': 'And it diverges for the absolute value of r greater than or equal to one.', 'start': 13324.473, 'duration': 5.383}, {'end': 13333.397, 'text': "Let's use this fact in this example.", 'start': 13331.597, 'duration': 1.8}, {'end': 13343.322, 'text': 'Remember that we decided that this was a geometric series with common ratio r equal to negative one ninth.', 'start': 13334.358, 'duration': 8.964}, {'end': 13349.755, 'text': 'And we found the first term a by plugging in i equals two and I got that first term of one third.', 'start': 13343.991, 'duration': 5.764}, {'end': 13359.783, 'text': 'Since the absolute value of r, the absolute value of negative one ninth is one ninth, which is less than one, we know that the series converges.', 'start': 13350.976, 'duration': 8.807}, {'end': 13364.125, 'text': 'and it converges to a over one minus r.', 'start': 13361.243, 'duration': 2.882}, {'end': 13371.809, 'text': "So that's one third over one minus negative one ninth, which is one third over one plus one ninth.", 'start': 13364.125, 'duration': 7.684}, {'end': 13378.512, 'text': "That's one third over 10 ninths, which simplifies to three tenths.", 'start': 13372.789, 'duration': 5.723}], 'summary': 'The geometric series converges to a/1-r for -1