title

Integration and the fundamental theorem of calculus | Chapter 8, Essence of calculus

description

Intuition for integrals, and why they are inverses of derivatives.
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Timestamps:
0:00 - Car example
8:20 - Areas under graphs
11:18 - Fundamental theorem of calculus
16:20 - Recap
17:45 - Negative area
18:55 - Outro
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{'title': 'Integration and the fundamental theorem of calculus | Chapter 8, Essence of calculus', 'heatmap': [{'end': 562.803, 'start': 538.393, 'weight': 1}, {'end': 688.446, 'start': 658.881, 'weight': 0.987}, {'end': 1049.419, 'start': 1034.054, 'weight': 0.774}], 'summary': 'Covers integrals and velocity graphs, finding areas and integrals, distance and velocity functions, and the fundamental theorem of calculus. it discusses practical applications, precise area calculation, and the relationship between velocity and distance functions, providing insights into math and science applications.', 'chapters': [{'end': 381.067, 'segs': [{'end': 53.98, 'src': 'embed', 'start': 31.368, 'weight': 0, 'content': [{'end': 38.812, 'text': 'In this video, I want to talk about integrals, and the thing that I want to become almost obvious is that they are an inverse of derivatives.', 'start': 31.368, 'duration': 7.444}, {'end': 42.194, 'text': "Here we're just going to focus on one example,", 'start': 39.892, 'duration': 2.302}, {'end': 48.117, 'text': 'which is a kind of dual to the example of a moving car that I talked about in Chapter 2 of the series Introducing Derivatives.', 'start': 42.194, 'duration': 5.923}, {'end': 53.98, 'text': "Then in the next video, we're going to see how this same idea generalizes, but to a couple other contexts.", 'start': 49.317, 'duration': 4.663}], 'summary': 'Integrals are the inverse of derivatives, illustrated with a specific example.', 'duration': 22.612, 'max_score': 31.368, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/rfG8ce4nNh0/pics/rfG8ce4nNh031368.jpg'}, {'end': 188.73, 'src': 'embed', 'start': 147.914, 'weight': 1, 'content': [{'end': 154.979, 'text': 'But first, I want to spend the bulk of this video showing how this question is related to finding the area bounded by the velocity graph.', 'start': 147.914, 'duration': 7.065}, {'end': 162.065, 'text': 'Because that helps to build an intuition for a whole class of problems, things called integral problems in math and science.', 'start': 155.58, 'duration': 6.485}, {'end': 168.708, 'text': 'To start off, notice that this question would be a lot easier if the car was just moving at a constant velocity right?', 'start': 162.864, 'duration': 5.844}, {'end': 176.633, 'text': 'In that case, you could just multiply the velocity in meters per second times the amount of time that has passed in seconds,', 'start': 169.408, 'duration': 7.225}, {'end': 178.875, 'text': 'and that would give you the number of meters traveled.', 'start': 176.633, 'duration': 2.242}, {'end': 184.187, 'text': 'And notice, you can visualize that product, that distance, as an area.', 'start': 179.963, 'duration': 4.224}, {'end': 188.73, 'text': "And if visualizing distance as area seems kinda weird, I'm right there with you.", 'start': 184.967, 'duration': 3.763}], 'summary': 'Relating question to finding area of velocity graph, integral problems in math and science, visualization of distance as area.', 'duration': 40.816, 'max_score': 147.914, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/rfG8ce4nNh0/pics/rfG8ce4nNh0147914.jpg'}, {'end': 261.416, 'src': 'embed', 'start': 231.803, 'weight': 3, 'content': [{'end': 236.829, 'text': "In fact, it's actually physically impossible, but it would make your calculations a lot more straightforward.", 'start': 231.803, 'duration': 5.026}, {'end': 245.98, 'text': 'You could just compute the distance traveled on each interval by multiplying the constant velocity on that interval by the change in time,', 'start': 237.59, 'duration': 8.39}, {'end': 247.562, 'text': 'and then just add all of those up.', 'start': 245.98, 'duration': 1.582}, {'end': 254.365, 'text': "So what we're going to do is approximate the velocity function as if it was constant on a bunch of intervals.", 'start': 248.956, 'duration': 5.409}, {'end': 261.416, 'text': "And then, as is common in calculus, we'll see how refining that approximation leads us to something more precise.", 'start': 255.086, 'duration': 6.33}], 'summary': 'Approximate velocity as constant on intervals, refine for precision.', 'duration': 29.613, 'max_score': 231.803, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/rfG8ce4nNh0/pics/rfG8ce4nNh0231803.jpg'}, {'end': 353.252, 'src': 'embed', 'start': 328.674, 'weight': 4, 'content': [{'end': 339.877, 'text': "That treating this car's journey as a bunch of discontinuous jumps in speed between portions of constant velocity becomes a less wrong reflection of reality as we decrease the time between those jumps.", 'start': 328.674, 'duration': 11.203}, {'end': 345.468, 'text': 'So, for convenience, on an interval like this,', 'start': 342.887, 'duration': 2.581}, {'end': 353.252, 'text': "let's just approximate the speed with whatever the true car's velocity is at the start of that interval, the height of the graph above the left side,", 'start': 345.468, 'duration': 7.784}], 'summary': 'Modeling car journey as discontinuous jumps in speed reflects reality as time between jumps decreases.', 'duration': 24.578, 'max_score': 328.674, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/rfG8ce4nNh0/pics/rfG8ce4nNh0328674.jpg'}], 'start': 1.031, 'title': 'Integrals and velocity graphs', 'summary': 'Explains integrals as inverses of derivatives using the example of v=t(8-t), and discusses how the area bounded by the velocity graph relates to integral problems, providing insights into math and science applications.', 'chapters': [{'end': 147.174, 'start': 1.031, 'title': 'Understanding integrals as inverses of derivatives', 'summary': 'Explains the concept of integrals as inverses of derivatives, using the example of finding a distance function from velocity, with a specific function v=t(8-t), and highlights the process of finding the antiderivative of the function.', 'duration': 146.143, 'highlights': ['The chapter explains the concept of integrals as inverses of derivatives, using the example of finding a distance function from velocity, with a specific function v=t(8-t).', 'It discusses the process of finding the antiderivative of the function, which is often described as finding the antiderivative of a function.']}, {'end': 381.067, 'start': 147.914, 'title': 'Finding area bounded by velocity graph', 'summary': "Explains how the area bounded by the velocity graph relates to integral problems in math and science, using the example of a car's changing velocity to illustrate the concept, and how approximating the velocity as constant on small intervals leads to a more precise understanding.", 'duration': 233.153, 'highlights': ["The concept of finding the area bounded by the velocity graph is related to integral problems in math and science, with the example of a car's changing velocity illustrating the concept (relevance: 5)", 'Illustrates how, in the case of constant velocity, multiplying velocity by time gives the distance traveled, visualized as an area, and explains how this visualization corresponds to meters (relevance: 4)', 'Explains the challenge of dealing with changing velocity, and how approximating the velocity as constant on intervals can simplify calculations, providing a conceptual understanding of the process (relevance: 3)', "Describes the process of approximating the car's motion by treating its velocity as constant on intervals, emphasizing the importance of refining the approximation for more precision (relevance: 2)", "Provides a concrete example by using numbers to illustrate how the car's velocity changes over intervals and how approximating it as constant on each interval can help calculate the distance traveled, emphasizing the importance of decreasing the time between jumps for a more accurate reflection of reality (relevance: 1)"]}], 'duration': 380.036, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/rfG8ce4nNh0/pics/rfG8ce4nNh01031.jpg', 'highlights': ['The chapter explains the concept of integrals as inverses of derivatives, using the example of finding a distance function from velocity, with a specific function v=t(8-t).', "The concept of finding the area bounded by the velocity graph is related to integral problems in math and science, with the example of a car's changing velocity illustrating the concept (relevance: 5)", 'Illustrates how, in the case of constant velocity, multiplying velocity by time gives the distance traveled, visualized as an area, and explains how this visualization corresponds to meters (relevance: 4)', 'Explains the challenge of dealing with changing velocity, and how approximating the velocity as constant on intervals can simplify calculations, providing a conceptual understanding of the process (relevance: 3)', "Describes the process of approximating the car's motion by treating its velocity as constant on intervals, emphasizing the importance of refining the approximation for more precision (relevance: 2)", "Provides a concrete example by using numbers to illustrate how the car's velocity changes over intervals and how approximating it as constant on each interval can help calculate the distance traveled, emphasizing the importance of decreasing the time between jumps for a more accurate reflection of reality (relevance: 1)"]}, {'end': 573.352, 'segs': [{'end': 407.161, 'src': 'embed', 'start': 381.647, 'weight': 3, 'content': [{'end': 387.792, 'text': "It's just that you'd be plugging in a different value for t at each one of these, giving a different height for each rectangle.", 'start': 381.647, 'duration': 6.145}, {'end': 395.193, 'text': "I'm going to write out an expression for the sum of the areas of all those rectangles in kind of a funny way.", 'start': 389.949, 'duration': 5.244}, {'end': 400.276, 'text': 'Take this symbol here, which looks like a stretched S for sum,', 'start': 395.973, 'duration': 4.303}, {'end': 407.161, 'text': "and then put a 0 at its bottom and an 8 at its top to indicate that we'll be ranging over time steps between 0 and 8 seconds.", 'start': 400.276, 'duration': 6.885}], 'summary': 'Expression for sum of rectangle areas from 0 to 8 seconds.', 'duration': 25.514, 'max_score': 381.647, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/rfG8ce4nNh0/pics/rfG8ce4nNh0381647.jpg'}, {'end': 453.225, 'src': 'embed', 'start': 421.935, 'weight': 1, 'content': [{'end': 428.66, 'text': "Not only is it a factor in each quantity that we're adding up, it also indicates the spacing between each sampled time step.", 'start': 421.935, 'duration': 6.725}, {'end': 435.041, 'text': 'So, when you make dt smaller and smaller, even though it decreases the area of each rectangle,', 'start': 429.52, 'duration': 5.521}, {'end': 441.643, 'text': "it increases the total number of rectangles whose areas we're adding up, because if they're thinner, it takes more of them to fill that space.", 'start': 435.041, 'duration': 6.602}, {'end': 443.563, 'text': 'And second,', 'start': 442.823, 'duration': 0.74}, {'end': 453.225, 'text': "the reason we don't use the usual sigma notation to indicate a sum is that this expression is technically not any particular sum for any particular choice of dt.", 'start': 443.563, 'duration': 9.662}], 'summary': 'Decreasing dt increases total number of rectangles for summing up areas.', 'duration': 31.29, 'max_score': 421.935, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/rfG8ce4nNh0/pics/rfG8ce4nNh0421935.jpg'}, {'end': 492.652, 'src': 'embed', 'start': 466.274, 'weight': 2, 'content': [{'end': 473.638, 'text': 'Remember, smaller choices of dt indicate closer approximations for the original question how far does the car actually go?', 'start': 466.274, 'duration': 7.364}, {'end': 483.243, 'text': 'So this limiting value for the sum, the area under this curve, gives us the precise answer to the question in full, unapproximated precision.', 'start': 474.558, 'duration': 8.685}, {'end': 485.565, 'text': "Now, tell me that's not surprising.", 'start': 484.144, 'duration': 1.421}, {'end': 492.652, 'text': 'We had this pretty complicated idea of approximations that can involve adding up a huge number of very tiny things.', 'start': 486.086, 'duration': 6.566}], 'summary': 'Limiting value of sum gives precise answer for approximations.', 'duration': 26.378, 'max_score': 466.274, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/rfG8ce4nNh0/pics/rfG8ce4nNh0466274.jpg'}, {'end': 573.352, 'src': 'heatmap', 'start': 538.393, 'weight': 0, 'content': [{'end': 551.118, 'text': "But finding the area between a function's graph and the horizontal axis is somewhat of a common language for many disparate problems that can be broken down and approximated as the sum of a large number of small things.", 'start': 538.393, 'duration': 12.725}, {'end': 553.879, 'text': "You'll see more in the next video,", 'start': 552.259, 'duration': 1.62}, {'end': 562.803, 'text': "but for now I'll just say in the abstract that understanding how to interpret and how to compute the area under a graph is a very general problem-solving tool.", 'start': 553.879, 'duration': 8.924}, {'end': 568.028, 'text': 'In fact, the first video of this series already covered the basics of how this works.', 'start': 564.025, 'duration': 4.003}, {'end': 573.352, 'text': 'But now that we have more of a background with derivatives, we can actually take this idea to its completion.', 'start': 568.669, 'duration': 4.683}], 'summary': 'Understanding area under a graph is a general problem-solving tool, covered in previous videos.', 'duration': 19.473, 'max_score': 538.393, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/rfG8ce4nNh0/pics/rfG8ce4nNh0538393.jpg'}], 'start': 381.647, 'title': 'Finding areas and integrals', 'summary': 'Covers finding the sum of rectangle areas over time steps, understanding integrals for precise area calculation, and the significance of smaller choices of dt in providing closer approximations. it also explores the significance of finding the area under a graph as a general problem-solving tool and a common language for various problems.', 'chapters': [{'end': 443.563, 'start': 381.647, 'title': 'Sum of rectangle areas and time steps', 'summary': "Discusses finding the sum of areas of rectangles over time steps between 0 and 8 seconds, with the value 'dt' playing a dual role and impacting the area and spacing between time steps.", 'duration': 61.916, 'highlights': ['The chapter discusses finding the sum of areas of rectangles over time steps between 0 and 8 seconds', "the value 'dt' playing a dual role and impacting the area and spacing between time steps"]}, {'end': 508.788, 'start': 443.563, 'title': 'Understanding integrals and area calculation', 'summary': 'Introduces the concept of integrals to find the precise area under a curve as the limit of a sum, highlighting the surprising simplicity in describing complex approximations and the significance of smaller choices of dt in providing closer approximations.', 'duration': 65.225, 'highlights': ["The value that those approximations approach can be described so simply. It's just the area underneath this curve.", 'Smaller choices of dt indicate closer approximations for the original question how far does the car actually go.', 'This limiting value for the sum, the area under this curve, gives us the precise answer to the question in full, unapproximated precision.', 'This expression is called an integral of v since it brings all of its values together. It integrates them.']}, {'end': 573.352, 'start': 510.461, 'title': 'Area under a graph', 'summary': 'Discusses how finding the area under a graph, although initially challenging, provides a general problem-solving tool and a common language for various problems, which can be approximated as the sum of small things.', 'duration': 62.891, 'highlights': ['Understanding how to interpret and compute the area under a graph is a general problem-solving tool, applicable to various disparate problems. It serves as a common language for many problems (quantifiable)', "The velocity-distance duo is not the only concern; the area between a function's graph and the horizontal axis is a key concept, which can be approximated as the sum of small things (quantifiable)", 'The understanding of finding the area under a graph is further enhanced with a background in derivatives, leading to a more complete understanding of the concept (quantifiable)']}], 'duration': 191.705, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/rfG8ce4nNh0/pics/rfG8ce4nNh0381647.jpg', 'highlights': ['The area under a graph is a general problem-solving tool, applicable to various disparate problems. It serves as a common language for many problems (quantifiable)', "The value 'dt' impacts the area and spacing between time steps", 'Smaller choices of dt indicate closer approximations for the original question how far does the car actually go', 'The chapter discusses finding the sum of areas of rectangles over time steps between 0 and 8 seconds', 'This limiting value for the sum, the area under this curve, gives us the precise answer to the question in full, unapproximated precision']}, {'end': 873.903, 'segs': [{'end': 622.078, 'src': 'embed', 'start': 574.393, 'weight': 1, 'content': [{'end': 581.718, 'text': 'For our velocity example, think of this right endpoint as a variable, capital T.', 'start': 574.393, 'duration': 7.325}, {'end': 588.604, 'text': "So we're thinking of this integral of the velocity function between 0 and t, the area under this curve between those inputs,", 'start': 581.718, 'duration': 6.886}, {'end': 591.206, 'text': 'as a function where the upper bound is the variable.', 'start': 588.604, 'duration': 2.602}, {'end': 596.839, 'text': 'That area represents the distance the car has traveled after t seconds, right?', 'start': 592.296, 'duration': 4.543}, {'end': 601.841, 'text': 'So in reality, this is a distance versus time function s.', 'start': 597.319, 'duration': 4.522}, {'end': 604.743, 'text': 'Now ask yourself, what is the derivative of that function?', 'start': 601.841, 'duration': 2.902}, {'end': 611.308, 'text': 'On the one hand, a tiny change in distance over a tiny change in time.', 'start': 607.304, 'duration': 4.004}, {'end': 611.988, 'text': "that's velocity.", 'start': 611.308, 'duration': 0.68}, {'end': 613.95, 'text': 'That is what velocity means.', 'start': 612.329, 'duration': 1.621}, {'end': 622.078, 'text': "But there's another way to see this, purely in terms of this graph and this area, which generalizes a lot better to other integral problems.", 'start': 614.811, 'duration': 7.267}], 'summary': 'The integral of velocity represents distance traveled over time. the derivative is velocity.', 'duration': 47.685, 'max_score': 574.393, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/rfG8ce4nNh0/pics/rfG8ce4nNh0574393.jpg'}, {'end': 688.446, 'src': 'heatmap', 'start': 651.638, 'weight': 4, 'content': [{'end': 658.881, 'text': "And because that's an approximation that gets better and better for smaller dt, the derivative of that area function ds.", 'start': 651.638, 'duration': 7.243}, {'end': 665.944, 'text': 'dt at this point equals vt, the value of the velocity function, at whatever time we started on.', 'start': 658.881, 'duration': 7.063}, {'end': 669.245, 'text': "And that right there, that's a super general argument.", 'start': 666.984, 'duration': 2.261}, {'end': 676.148, 'text': 'The derivative of any function giving the area under a graph like this is equal to the function for the graph itself.', 'start': 669.665, 'duration': 6.483}, {'end': 684.401, 'text': 'So, if our velocity function is t times 8 minus t, what should s be??', 'start': 678.675, 'duration': 5.726}, {'end': 688.446, 'text': 'What function of t has a derivative of t times 8 minus t??', 'start': 685.102, 'duration': 3.344}], 'summary': 'The derivative of the area function is equal to the velocity function. the given velocity function is t times 8 minus t.', 'duration': 24.51, 'max_score': 651.638, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/rfG8ce4nNh0/pics/rfG8ce4nNh0651638.jpg'}, {'end': 754.054, 'src': 'embed', 'start': 728.44, 'weight': 2, 'content': [{'end': 734.244, 'text': 'So if we just scale that down by a third, the derivative of 1 third t cubed is exactly t squared.', 'start': 728.44, 'duration': 5.804}, {'end': 740.949, 'text': "And then making that negative, you'd see that negative 1 third t cubed has a derivative of negative t squared.", 'start': 734.865, 'duration': 6.084}, {'end': 750.696, 'text': 'Therefore, the antiderivative of our function, 8t minus t squared, is 4t squared minus 1 third t cubed.', 'start': 742.233, 'duration': 8.463}, {'end': 754.054, 'text': "But there's a slight issue here.", 'start': 752.452, 'duration': 1.602}], 'summary': 'Derivative and antiderivative example using t-cubed and t-squared.', 'duration': 25.614, 'max_score': 728.44, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/rfG8ce4nNh0/pics/rfG8ce4nNh0728440.jpg'}, {'end': 800.289, 'src': 'embed', 'start': 774.534, 'weight': 5, 'content': [{'end': 783.88, 'text': "So in reality, there's actually infinitely many different possible antiderivative functions, and every one of them looks like 4t squared minus 1,", 'start': 774.534, 'duration': 9.346}, {'end': 788.502, 'text': 'third t cubed plus c for some constant c.', 'start': 783.88, 'duration': 4.622}, {'end': 791.444, 'text': "But there is one piece of information that we haven't used yet,", 'start': 788.502, 'duration': 2.942}, {'end': 797.107, 'text': "that's going to let us zero in on which antiderivative to use the lower bound of the integral.", 'start': 791.444, 'duration': 5.663}, {'end': 800.289, 'text': 'This integral has to be 0.', 'start': 798.308, 'duration': 1.981}], 'summary': 'Infinitely many antiderivative functions; integral has to be 0.', 'duration': 25.755, 'max_score': 774.534, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/rfG8ce4nNh0/pics/rfG8ce4nNh0774534.jpg'}, {'end': 886.161, 'src': 'embed', 'start': 852.25, 'weight': 0, 'content': [{'end': 856.674, 'text': 'is this expression evaluated at t equals eight, which is about 85.33 minus zero.', 'start': 852.25, 'duration': 4.424}, {'end': 863.151, 'text': 'So the answer as a whole is just 85.33.', 'start': 856.734, 'duration': 6.417}, {'end': 868.157, 'text': 'But a more typical example would be something like the integral between 1 and 7.', 'start': 863.151, 'duration': 5.006}, {'end': 873.903, 'text': "That's the area pictured right here, and it represents the distance traveled between 1 second and 7 seconds.", 'start': 868.157, 'duration': 5.746}, {'end': 886.161, 'text': 'What you do is evaluate the antiderivative we found at the top bound, 7, and then subtract off its value at that bottom bound, 1.', 'start': 876.526, 'duration': 9.635}], 'summary': 'Expression at t=8 is 85.33. typical example: integral from 1 to 7, representing distance traveled. antiderivative evaluated at 7, subtracted by value at 1.', 'duration': 33.911, 'max_score': 852.25, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/rfG8ce4nNh0/pics/rfG8ce4nNh0852250.jpg'}], 'start': 574.393, 'title': 'Distance and velocity functions', 'summary': 'Discusses the concept of distance versus time function and its derivative, illustrating the relationship between the area under the velocity graph and the velocity function. it also explains finding the antiderivative of a velocity function using a specific example, ultimately determining the distance traveled as 85.33 units in 8 seconds.', 'chapters': [{'end': 676.148, 'start': 574.393, 'title': 'Velocity and derivative of distance function', 'summary': 'Discusses the concept of distance versus time function and its derivative, illustrating the relationship between the area under the velocity graph and the velocity function, with a general argument that the derivative of any function giving the area under a graph is equal to the function for the graph itself.', 'duration': 101.755, 'highlights': ['The derivative of the distance versus time function is discussed, illustrating the relationship between the area under the velocity graph and the velocity function.', 'A general argument is presented, showing that the derivative of any function giving the area under a graph is equal to the function for the graph itself.', 'The concept of distance versus time function is introduced, emphasizing the area under the curve as a function where the upper bound is the variable, representing the distance traveled by a car after t seconds.']}, {'end': 873.903, 'start': 678.675, 'title': 'Finding antiderivative of velocity function', 'summary': 'Explains the process of finding the antiderivative of a velocity function using the example t times 8 minus t, ultimately determining the distance traveled as 85.33 units in 8 seconds and 85.33 - 0 = 85.33.', 'duration': 195.228, 'highlights': ['Antiderivative of the given function t times 8 minus t is 4t squared minus 1 third t cubed, with infinitely many possible antiderivative functions in the form 4t squared minus 1 third t cubed plus c.', 'The integral of the antiderivative function at t equals 8 seconds gives the total distance traveled, which is 85.33 units.', 'The process of determining the constant to add to the antiderivative function involves subtracting off the value of the antiderivative function at the lower bound, ensuring the integral from the lower bound to itself is zero.']}], 'duration': 299.51, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/rfG8ce4nNh0/pics/rfG8ce4nNh0574393.jpg', 'highlights': ['The integral of the antiderivative function at t equals 8 seconds gives the total distance traveled, which is 85.33 units.', 'The concept of distance versus time function is introduced, emphasizing the area under the curve as a function where the upper bound is the variable, representing the distance traveled by a car after t seconds.', 'Antiderivative of the given function t times 8 minus t is 4t squared minus 1 third t cubed, with infinitely many possible antiderivative functions in the form 4t squared minus 1 third t cubed plus c.', 'The derivative of the distance versus time function is discussed, illustrating the relationship between the area under the velocity graph and the velocity function.', 'A general argument is presented, showing that the derivative of any function giving the area under a graph is equal to the function for the graph itself.', 'The process of determining the constant to add to the antiderivative function involves subtracting off the value of the antiderivative function at the lower bound, ensuring the integral from the lower bound to itself is zero.']}, {'end': 1226.054, 'segs': [{'end': 966.533, 'src': 'embed', 'start': 924.742, 'weight': 0, 'content': [{'end': 931.765, 'text': 'Then the integral equals this antiderivative evaluated at the top bound minus its value at the bottom bound.', 'start': 924.742, 'duration': 7.023}, {'end': 937.527, 'text': "And this fact right here that you're staring at is the fundamental theorem of calculus.", 'start': 932.845, 'duration': 4.682}, {'end': 941.169, 'text': 'And I want you to appreciate something kind of crazy about this fact.', 'start': 938.308, 'duration': 2.861}, {'end': 942.846, 'text': 'The integral,', 'start': 941.966, 'duration': 0.88}, {'end': 951.809, 'text': 'the limiting value for the sum of all of these thin rectangles takes into account every single input on the continuum from the lower bound to the upper bound.', 'start': 942.846, 'duration': 8.963}, {'end': 954.009, 'text': "That's why we use the word integrate.", 'start': 952.229, 'duration': 1.78}, {'end': 955.79, 'text': 'It brings them all together.', 'start': 954.309, 'duration': 1.481}, {'end': 964.472, 'text': 'And yet, to actually compute it using an antiderivative, you only look at two inputs, the top bound and the bottom bound.', 'start': 956.81, 'duration': 7.662}, {'end': 966.533, 'text': 'It almost feels like cheating.', 'start': 965.453, 'duration': 1.08}], 'summary': 'The fundamental theorem of calculus states that the integral equals the antiderivative evaluated at the top bound minus its value at the bottom bound, bringing together all inputs on the continuum from the lower bound to the upper bound, yet computed using only two inputs.', 'duration': 41.791, 'max_score': 924.742, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/rfG8ce4nNh0/pics/rfG8ce4nNh0924742.jpg'}, {'end': 1057.93, 'src': 'heatmap', 'start': 1034.054, 'weight': 0.774, 'content': [{'end': 1040.576, 'text': 'you can deduce that the derivative of that area function must equal the height of the graph at every point.', 'start': 1034.054, 'duration': 6.522}, {'end': 1042.577, 'text': "And that's really the key right there.", 'start': 1041.316, 'duration': 1.261}, {'end': 1049.419, 'text': 'It means that to find a function giving this area, you ask what function has v as a derivative?', 'start': 1043.196, 'duration': 6.223}, {'end': 1057.93, 'text': 'There are actually infinitely many antiderivatives of a given function, since you can always just add some constant without affecting the derivative.', 'start': 1050.6, 'duration': 7.33}], 'summary': "The derivative of the area function equals the graph's height at every point, leading to infinitely many antiderivatives.", 'duration': 23.876, 'max_score': 1034.054, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/rfG8ce4nNh0/pics/rfG8ce4nNh01034054.jpg'}, {'end': 1161.328, 'src': 'embed', 'start': 1117.322, 'weight': 1, 'content': [{'end': 1125.048, 'text': 'Whenever a graph dips below the horizontal axis, the area between that portion of the graph and the horizontal axis is counted as negative.', 'start': 1117.322, 'duration': 7.726}, {'end': 1134.135, 'text': "And what you'll commonly hear is that integrals don't measure area per se, they measure the signed area between the graph and the horizontal axis.", 'start': 1125.928, 'duration': 8.207}, {'end': 1141.24, 'text': "Next up, I'm going to bring up more context where this idea of an integral and area under curves comes up,", 'start': 1135.856, 'duration': 5.384}, {'end': 1144.683, 'text': 'along with some other intuitions for this fundamental theorem of calculus.', 'start': 1141.24, 'duration': 3.443}, {'end': 1152.861, 'text': 'Maybe you remember, Chapter 2 of this series, Introducing the Derivative, was sponsored by the Art of Problem Solving.', 'start': 1146.837, 'duration': 6.024}, {'end': 1158.505, 'text': "So I think there's something elegant to the fact that this video, which is kind of a duel to that one,", 'start': 1153.542, 'duration': 4.963}, {'end': 1161.328, 'text': 'was also supported in part by the Art of Problem Solving.', 'start': 1158.505, 'duration': 2.823}], 'summary': 'Integrals measure signed area under curves, related to fundamental theorem of calculus.', 'duration': 44.006, 'max_score': 1117.322, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/rfG8ce4nNh0/pics/rfG8ce4nNh01117322.jpg'}, {'end': 1202.862, 'src': 'embed', 'start': 1178.795, 'weight': 4, 'content': [{'end': 1185.978, 'text': "or if you're a student who wants to see what math has to offer beyond rote schoolwork, I cannot recommend The Art of Problem Solving enough.", 'start': 1178.795, 'duration': 7.183}, {'end': 1196.88, 'text': "Whether that's their newest development to build the right intuitions in elementary school kids called Beast Academy or their courses in higher level topics and contest preparation.", 'start': 1186.698, 'duration': 10.182}, {'end': 1202.862, 'text': 'going to aops.com, slash 3blue1brown, or clicking on the link in the description.', 'start': 1196.88, 'duration': 5.982}], 'summary': 'Art of problem solving offers math resources for students, including beast academy for elementary school kids and higher-level courses.', 'duration': 24.067, 'max_score': 1178.795, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/rfG8ce4nNh0/pics/rfG8ce4nNh01178795.jpg'}], 'start': 876.526, 'title': 'Fundamental theorem of calculus and sponsorship elegance', 'summary': "Explains the fundamental theorem of calculus, focusing on evaluating integrals and signed area, and discusses the elegant symmetry in sponsorship, recommending the art of problem solving and highlighting the company's influence.", 'chapters': [{'end': 1141.24, 'start': 876.526, 'title': 'Fundamental theorem of calculus', 'summary': 'Explains the fundamental theorem of calculus, emphasizing the process of evaluating integrals using antiderivatives and the significance of signed area in integrals, with a focus on the context of distance traveled and velocity functions.', 'duration': 264.714, 'highlights': ['The fundamental theorem of calculus states that the integral of a function equals the antiderivative evaluated at the top bound minus its value at the bottom bound, with the constant added to the antiderivative canceling out.', 'Evaluating integrals using antiderivatives involves considering the entire continuum of inputs between the lower and upper bounds, despite only directly using the top and bottom bounds in the computation.', 'The concept of negative area in integrals is crucial, as the area below the horizontal axis is counted as negative, particularly relevant when dealing with velocity functions and distance traveled, where negative velocity results in backward motion.']}, {'end': 1226.054, 'start': 1141.24, 'title': 'Sponsorship elegance and influence', 'summary': "Discusses the elegant symmetry in the sponsorship, recommends the art of problem solving for fostering a love for math, and emphasizes the impact of the company's influence on the speaker's own development.", 'duration': 84.814, 'highlights': ['The chapter discusses the elegant symmetry in the sponsorship, emphasizing the support by the Art of Problem Solving for both the current video and Chapter 2 of the series.', "The speaker highly recommends The Art of Problem Solving for parents looking to foster their child's love for math and for students seeking to explore math beyond rote schoolwork.", "The Art of Problem Solving had a significant influence on the speaker's development as a student, contributing to their love for creative math."]}], 'duration': 349.528, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/rfG8ce4nNh0/pics/rfG8ce4nNh0876526.jpg', 'highlights': ['The fundamental theorem of calculus states that the integral of a function equals the antiderivative evaluated at the top bound minus its value at the bottom bound, with the constant added to the antiderivative canceling out.', 'The concept of negative area in integrals is crucial, as the area below the horizontal axis is counted as negative, particularly relevant when dealing with velocity functions and distance traveled, where negative velocity results in backward motion.', 'Evaluating integrals using antiderivatives involves considering the entire continuum of inputs between the lower and upper bounds, despite only directly using the top and bottom bounds in the computation.', 'The chapter discusses the elegant symmetry in the sponsorship, emphasizing the support by the Art of Problem Solving for both the current video and Chapter 2 of the series.', "The speaker highly recommends The Art of Problem Solving for parents looking to foster their child's love for math and for students seeking to explore math beyond rote schoolwork.", "The Art of Problem Solving had a significant influence on the speaker's development as a student, contributing to their love for creative math."]}], 'highlights': ['The integral of the antiderivative function at t equals 8 seconds gives the total distance traveled, which is 85.33 units.', 'The fundamental theorem of calculus states that the integral of a function equals the antiderivative evaluated at the top bound minus its value at the bottom bound, with the constant added to the antiderivative canceling out.', 'The area under a graph is a general problem-solving tool, applicable to various disparate problems. It serves as a common language for many problems (quantifiable)', "The concept of finding the area bounded by the velocity graph is related to integral problems in math and science, with the example of a car's changing velocity illustrating the concept (relevance: 5)", 'Illustrates how, in the case of constant velocity, multiplying velocity by time gives the distance traveled, visualized as an area, and explains how this visualization corresponds to meters (relevance: 4)', 'The concept of negative area in integrals is crucial, as the area below the horizontal axis is counted as negative, particularly relevant when dealing with velocity functions and distance traveled, where negative velocity results in backward motion.']}