{"version":1,"lectureId":"01M14TZ29ND79HRFMF6W061GNY","attempt":0,"publication":{"slug":"accumulation-and-rate-of-change","title":"Accumulation and the Two Halves of Calculus","subject":"mathematics","summary":"Two questions that look unrelated, the slope of a curve and the area under one, turn out to be the same question asked from opposite ends. Starting from water running into a tank, we build the accumulation function, the area under a rate graph up to time x, and discover that its slope at every point is the height of the rate graph. A thin strip argument shows why that has to happen for any rate at all, the notation of the definite integral turns it into the Fundamental Theorem of Calculus, and we finish by using the theorem to find the exact area under a parabola in three lines of algebra.","metaDescription":"Build the accumulation function from a rate graph, discover its slope is the rate, and reach the Fundamental Theorem of Calculus.","transcript":"Derivatives and integrals. One of them measures how fast something is changing. The other measures how much of something has piled up. Two completely different questions, and this whole lecture is about the fact that they are one question, asked from opposite ends. So let me start with a tank. Somebody has opened a tap, and I know the rate, at every instant, exactly how fast the water is arriving. The question is simple enough. How much water is in there? Let's draw the rate. Time runs along the bottom, and the height of this graph is how fast the water is arriving at that moment. This tap is being opened steadily, so the line climbs. So at time one, the water is arriving at one litre a second. Move out to time three, and it is arriving at three litres a second. Now hold that thought, and imagine a different tap for a moment. Suppose the rate were constant. Say two litres a second, flat, the whole way across. Held there for three seconds, the answer is easy. Two litres a second, for three seconds, is six litres. Rate times time. And on this picture, rate times time is the area of one rectangle. But our tap is not that tap. Our rate climbs the whole way, so there is no single rectangle to draw. So we chop the time up into slivers instead. Over a stretch short enough the rate barely changes, so each sliver is rate times time again. One thin rectangle. Then we add them all up. Now watch what happens as we use more and more of them. The staircase closes on the region underneath the curve, and in the limit the water that has arrived is exactly the area under the rate graph. So here is the object I care about. The amount of water that has arrived by time x is the area under the graph, from the start out to x. Give it a name. A of x, the accumulation function. And notice what kind of object A is. It is a function. You hand it a time, here, and it hands you back a number, the area up to there. Watch what happens when I slide that edge along. More of the region is swept in, and the number gets bigger. Every stopping point gives you one number, so this really is a function of x, built out of nothing but area. Hold on to that picture. The area under a rate graph is itself a function, and everything that follows comes from one question about it. So let me put that question on the board, in as many words. How fast does A grow? Answering that is the whole of the rest of this lecture. A moment ago we ended with the accumulation function. A of x is the area under the rate graph, from zero up to x. For this particular tap we can work that area out exactly, and pleasingly, we need no calculus at all to do it. First, one word about the letters, because there are two of them and they are not the same letter. Along the bottom the clock is called t, since t is what sweeps across the region while we shade it in. The letter x is saved for where we stop. So stop the clock at x. There is the mark, on the axis the clock runs along. And look at what is shaded: it is a triangle. Its base runs from zero out to x, so the base is x. Its height is the rate at time x, and for this tap the rate at time x is x as well. The area of a triangle is a half, base, height. Here that is a half times x times x. Which comes to one half x squared. There it is. The accumulation function for this tap, in closed form. Now let us put the two of them side by side, because this is where it gets good. On the left, the rate. A straight line, climbing steadily. On the right, the accumulation. A parabola, bending upward. As the tap opens further, the water piles up faster and faster. Now a question about the curve on the right. How steep is it? The steepness of A at a point is exactly what the symbol A prime of x will mean, so take the tangent line at x equals two and read off its slope. Rise over run, the slope there is two. Now look across at the rate graph. At time two, the height of the rate line is, two. Coincidence? Let's slide x along and watch both pictures at once. Carry on out to x equals three. The tangent on the accumulation curve now has slope three. And the rate line at time three is sitting at height three. The slope of the accumulation is the height of the rate, every single time. That deserves a line of its own. Two things to write down, and the first one is what the symbol is going to mean. A is a function of x, so A prime of x is its rate of change: how fast the shaded area grows as x moves off to the right. And the second is what we just watched. That rate of change is the height of the rate graph at x. So A prime of x equals f of x. The slope of the accumulation curve is the height of the rate curve, at every single point. We got there from one triangle and one parabola, but it has nothing to do with triangles, and I want to show you why it has to be true for absolutely any rate at all. That was one particular tap, and one particularly friendly shape. But the fact we landed on has nothing to do with triangles. So here is a rate that wanders. No nice formula, and no geometry that will help you. Same set-up as before. A of x is the area underneath it, from zero up to x. We stop the clock here, at x, and everything shaded to the left of that line is A of x. Now nudge it. Push x forward by a small amount, and give that amount a name: h. There is h, on the picture, the little step from x across to x plus h. The area grows by this sliver, the water that arrives between time x and time x plus h. Call that extra area delta A. And here is the whole trick, so let me put it on the screen rather than leave it in the air. That sliver is very nearly a rectangle. There it is, standing on the sliver it is pretending to be. Its width is h, the step we just marked. Its top sits level with the curve at x, at a height of f of x, because over a stretch that short the curve has not had time to change. So delta A is about f of x times h. Now divide both sides by h, and the left-hand side turns into something you have met before. Change in A, divided by change in x. The average rate of change of A. And now let h shrink. Watch the sliver, and watch the rectangle standing on it. The thinner they both get, the better the rectangle fits, and the closer those two sides come to being equal. In the limit, the left-hand side is the derivative of A at x. That is precisely the definition of a derivative. And the right-hand side never moved. So A prime of x equals f of x, for any rate you like. Said out loud, it is almost a tautology. How fast is the water in the tank going up? At the rate the water is coming in. That is the whole theorem. The surprise is not that it is true, the surprise is what it lets you do. Before we cash this in, we need the notation everybody actually writes. The area under f, from a up to x, has a symbol of its own. A long, stretched S. The function goes in the middle, and dee t closes it off. Out loud: the integral of f of t, dee t, from a to x. And let me say what those two ends are, because nobody ever does. The little a at the bottom is where we start counting. It is the left-hand edge of the region, chosen once and then left exactly where it is. The x on the top is where we stop. So the whole symbol is a function of where you stop, which is exactly what the shaded area was. And the S is for sum, because that is precisely what it is. All those thin rectangles, added up, with dee t as the width of one of them. So in this language, what we proved with the thin sliver reads like this. Differentiate an integral with respect to its upper limit, and you get the integrand straight back. Now the payoff. Here is a rate again, and here is the job in front of us: the area underneath it, between two fixed times. The left-hand end we call a. That is where the counting starts. The right-hand end we call b, and that is where it stops. Everything shaded between those two lines is the area we are trying to find, and we have no formula for a shape like that. Now suppose somebody hands you a function big F whose derivative is little f. Any such function at all will do. There it is, written down: F prime of x equals f of x. And we already have one function like that. Our own area function, A. Last time we proved that A prime of x is f of x too. So big F and our area function A have exactly the same derivative, everywhere, whatever big F happens to be. Which raises a fair question. How different can two functions with the same derivative actually be? Let me draw one of them over here. This is big F, and its slope at every point is the height of the curve beside it. Now take a copy of it, and lift the whole copy upward. Every point goes up by the same amount, so the curve keeps its shape exactly. At every x it is still leaning the way it was leaning. Its slope has not changed anywhere. Push a copy downward and you get the same story again. All three of these are antiderivatives of the same little f. There is a whole family of them stacked up the board, and every one of them is parallel to the others. And here is where the constant comes from. Measure the gap between two of them at x equals a. Now measure it again over at x equals b. It is the same gap. It is the same gap everywhere, and that gap is the number we call C. So two antiderivatives of one function differ by a constant, and only by a constant. Big F of x is our area function A of x, plus some fixed number C. Now watch what that does the moment we subtract. Start from the line we just earned. Big F of x is A of x, plus C. That is true at every single x, so in particular it is true at b. Big F of b is A of b, plus C. And it is just as true at a. Big F of a is A of a, plus C. The very same C both times, because it is the very same big F both times. Now take the second of those two lines away from the first, exactly the way you would take one number off another. So rule a line underneath them both. And there is the whole trick, standing one above the other at the end of those two lines. Plus C on the top, plus C underneath. The same unknown number added on and then taken straight back off. So slash the pair of them out, and whatever C was, it has gone. So what survives on the right is A of b minus A of a. The constant has gone, and it was always going to go, whichever antiderivative you had picked up in the first place. And A of b minus A of a is exactly the area under f between a and b. Which, in the notation we built at the start of this scene, is the integral of f from a to b. So there is the bottom line. Let me write it out properly, because that is the theorem. So there it is, in a single line. The integral of f from a to b equals big F of b, minus big F of a, where big F is any antiderivative of f. That is the Fundamental Theorem of Calculus, and it is the reason integration is something you can actually sit down and do. To find an area, you no longer add up rectangles. You find a function whose derivative is the one you started with, and subtract at the two ends. Let's spend it on a problem you cannot do with geometry. Here is y equals x squared, and here is the piece of it we want: everything under the curve between zero and two. The old route uses thin bars like these, then takes a limit. That is real work, because this is not a triangle, a rectangle, a circle, or any other shape with a school formula. By the new route you need exactly one thing. Some function whose derivative is x squared. So let us go and find one. Start by guessing. Differentiate x cubed, and you get three x squared. Write that down, because it is nearly right. It is three times too big. So divide the whole thing by three before you differentiate it. Now the three on top and the three underneath are the same three, and what comes out is exactly x squared. So there is our big F. F of x is x cubed over three, and its derivative is the x squared we started with. Now the theorem does the rest. The integral from zero to two of x squared, dee x, is F at two minus F at zero. F at two is two cubed over three. F at zero is zero. Take the second one away from the first. And what is left is eight thirds, which is about two point six seven. That is the exact area under the parabola, and we got it by differentiating backwards. No rectangles, no limits, three lines of algebra. Does that number look sensible? The region sits inside this two by four rectangle, which has area eight, and by eye the shaded part is about a third of it. Two point six seven out of eight. That checks out. So here are the two halves of one idea. On the left, differentiate an accumulation, and you get back the rate you were accumulating at. On the right, to total up a rate across an interval, find something that differentiates to it, and subtract at the two ends. Area and slope are not two separate subjects that happen to share a course. They are one machine, run in the two directions. That is the Fundamental Theorem of Calculus.","watch":{"version":1,"scenes":[{"title":"What Piles Up","start":0,"end":165.73575,"objects":{"acc_words":"a Tex [text] that says \"$A(x) =$ the shaded area\"","axes_rate":"an Axes (x_range=(0.0, 4.0), y_range=(0.0, 4.4), x_ticks_every=1.0)","bar_count":"a VariableNumber (initial_value=6.0)","bars":"a RiemannRectangles [blue] drawn in axes_rate (rectangle_count=<VariableNumber bar_count = 60.0>, x_range=(0.0, 2.0), target='rate')","edge_foot":"a Point [red] labelled \"x\" drawn in axes_rate (location=(<VariableNumber x_edge = 3.5>, 0.0))","edge_line":"a Line [red] drawn in axes_rate (start=(<VariableNumber x_edge = 3.5>, 0.0), end=(<VariableNumber x_edge = 3.5>, <VariableNumber x_edge = 3.5>), dashed=True)","flat":"a FunctionPlot [gray] drawn in axes_rate (function=<function>, x_range=(0.0, 3.0))","flat_box":"a RiemannRectangles [gray] drawn in axes_rate (rectangle_count=1, x_range=(0.0, 3.0), stroke_color='text')","flat_law":"a Math [text] that says \"$f(t) = 2$\"","flat_sum":"a Math [text] that says \"$2 times 3 = 6 thin upright(\"L\")$\"","question":"a Panel that says \"Water is running into a tank, and at every instant I know the rate it is arriving at. 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So let me put that question on the board, in as many words.","live":null,"does":[[156.632,"the_question is shown on the screen, written out."]]},{"start":159.24,"say":"How fast does A grow? Answering that is the whole of the rest of this lecture.","live":["question","rate_law","axes_rate","rate","acc_words","the_question","water","edge_line","edge_foot"],"does":[[159.646,"the_question is indicated — a transient flash."],[164.69408333333334,"acc_words is hidden from the screen — left the board."],[164.69408333333334,"axes_rate is hidden from the screen — left the board."],[164.69408333333334,"rate is hidden from the screen — axes_rate left the board."],[164.69408333333334,"water is hidden from the screen — axes_rate left the board."],[164.69408333333334,"edge_line is hidden from the screen — axes_rate left the board."],[164.69408333333334,"edge_foot is hidden from the screen — axes_rate left the board."],[164.69408333333334,"question is hidden from the screen — left the board."],[164.69408333333334,"rate_law is hidden from the screen — left the board."],[164.69408333333334,"the_question is hidden from the screen — left the board."]]}]},{"title":"The Slope of the Area","start":165.73575,"end":341.06822916666664,"objects":{"acc":"a FunctionPlot [red] drawn in axes_acc (function=<function>, x_range=(0.0, 4.0))","axes_acc":"an Axes (x_range=(0.0, 4.5), y_range=(0.0, 8.6), x_ticks_every=1.0)","axes_rate":"an Axes (x_range=(0.0, 4.5), y_range=(0.0, 4.4), x_ticks_every=1.0)","dot_f":"a PlotPoint [yellow] labelled \"f(x)\" drawn in axes_rate (target='rate', x=<VariableNumber x_t = 3.0>)","heading":"a Heading that says \"The Slope of the Area\"","heading_result":"a Heading that says \"The Big Fact\"","height_line":"a Line [green] labelled \"f(x) = x\" drawn in axes_rate (start=(3.0, 0.0), end=(3.0, 3.0), dashed=True)","lbl_A":"a Tex [text] that says \"The accumulation, $A$\"","lbl_f":"a Tex [text] that says \"The rate, $f$\"","meaning":"a Math [text] that says \"$A'(x) = frac(dif A, dif x)$\"","note":"a Text [text] that says \"The slope of the accumulation curve is the height of the rate curve, at every single point.\"","rate":"a FunctionPlot [blue] drawn in axes_rate (function=<function>, x_range=(0.0, 4.0))","result":"a Math [text] that says \"$A'(x) = f(x)$\"","tangent":"a TangentLine [yellow] drawn in axes_acc (target='acc', x=<VariableNumber x_t = 3.0>, slope_triangle=True)","tri":"an AreaUnder [red] drawn in axes_rate (x_range=(0.0, 3.0), target='rate')","work":"a Derivation [text] that says \"$A(x) &= frac(1, 2) dot upright(\"base\") dot upright(\"height\") \\ &= frac(1, 2) dot x dot x \\ &= frac(1, 2) x^2$\"","x_foot":"a Point [red] labelled \"x\" drawn in axes_rate (location=(3.0, 0.0))","x_t":"a VariableNumber (initial_value=2.0)"},"beats":[{"start":165.73575,"say":"A moment ago we ended with the accumulation function. A of x is the area under the rate graph, from zero up to x. For this particular tap we can work that area out exactly, and pleasingly, we need no calculus at all to do it.","live":[],"does":[[165.73575,"heading is shown on the screen, written out."],[165.73575,"axes_rate is shown on the screen, written out."],[165.73575,"rate is shown on the screen, drawn."]]},{"start":182.27575,"say":"First, one word about the letters, because there are two of them and they are not the same letter. Along the bottom the clock is called t, since t is what sweeps across the region while we shade it in. The letter x is saved for where we stop.","live":["axes_rate","heading","rate"],"does":[[192.72475,"tri is shown on the screen, written out."]]},{"start":197.44875,"say":"So stop the clock at x. There is the mark, on the axis the clock runs along. 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The accumulation function for this tap, in closed form. Now let us put the two of them side by side, because this is where it gets good.","live":null,"does":[[234.56675,"height_line is hidden from the screen."],[234.56675,"tri is hidden from the screen."],[234.56675,"x_foot is hidden from the screen."],[237.28325,"axes_rate moves to a new place on the board."],[237.28325,"work is hidden from the screen — left the board."]]},{"start":238.48325,"say":"On the left, the rate. A straight line, climbing steadily. On the right, the accumulation. A parabola, bending upward. As the tap opens further, the water piles up faster and faster.","live":["axes_rate","heading","rate"],"does":[[238.99475,"lbl_f is shown on the screen, written out."],[243.12775,"lbl_A is shown on the screen, written out."],[243.12775,"axes_acc is shown on the screen, written out."],[245.06675,"acc is shown on the screen, drawn."]]},{"start":252.08675,"say":"Now a question about the curve on the right. How steep is it? The steepness of A at a point is exactly what the symbol A prime of x will mean, so take the tangent line at x equals two and read off its slope.","live":["axes_rate","heading","rate","lbl_f","lbl_A","axes_acc","acc"],"does":[[262.66375,"tangent is shown on the screen, written out."]]},{"start":266.78125,"say":"Rise over run, the slope there is two. Now look across at the rate graph. At time two, the height of the rate line is, two.","live":["axes_rate","heading","rate","lbl_f","lbl_A","axes_acc","acc","tangent"],"does":[[270.70575,"dot_f is shown on the screen, written out."]]},{"start":276.82025,"say":"Coincidence? Let's slide x along and watch both pictures at once.","live":["axes_rate","heading","rate","lbl_f","lbl_A","axes_acc","acc","tangent","dot_f"],"does":[[279.02675,"tangent is redrawn as the numbers it depends on change."],[279.02675,"dot_f is redrawn as the numbers it depends on change."],[279.02675,"x_t ticks to 2.6."]]},{"start":283.14925,"say":"Carry on out to x equals three. The tangent on the accumulation curve now has slope three. And the rate line at time three is sitting at height three. The slope of the accumulation is the height of the rate, every single time. That deserves a line of its own.","live":null,"does":[[283.49775,"tangent is redrawn as the numbers it depends on change."],[283.49775,"dot_f is redrawn as the numbers it depends on change."],[283.49775,"x_t ticks to 3.0."],[299.10175,"axes_acc is hidden from the screen — left the board."],[299.10175,"acc is hidden from the screen — axes_acc left the board."],[299.10175,"tangent is hidden from the screen — axes_acc left the board."],[299.10175,"axes_rate is hidden from the screen — left the board."],[299.10175,"rate is hidden from the screen — axes_rate left the board."],[299.10175,"dot_f is hidden from the screen — axes_rate left the board."],[299.10175,"heading is hidden from the screen — left the board."],[299.10175,"lbl_A is hidden from the screen — left the board."],[299.10175,"lbl_f is hidden from the screen — left the board."]]},{"start":300.30174999999997,"say":"Two things to write down, and the first one is what the symbol is going to mean. A is a function of x, so A prime of x is its rate of change: how fast the shaded area grows as x moves off to the right.","live":[],"does":[[300.30174999999997,"heading_result is shown on the screen, written out."],[309.43875,"meaning is shown on the screen, written out."]]},{"start":315.33325,"say":"And the second is what we just watched. That rate of change is the height of the rate graph at x. So A prime of x equals f of x.","live":["meaning","heading_result"],"does":[[319.12975000000006,"result is shown on the screen, written out."],[322.70575,"A box is drawn around result."]]},{"start":324.89025,"say":"The slope of the accumulation curve is the height of the rate curve, at every single point. We got there from one triangle and one parabola, but it has nothing to do with triangles, and I want to show you why it has to be true for absolutely any rate at all.","live":["meaning","result","heading_result"],"does":[[324.89025,"note is shown on the screen, written out."],[340.02656249999995,"heading_result is hidden from the screen — left the board."],[340.02656249999995,"meaning is hidden from the screen — left the board."],[340.02656249999995,"note is hidden from the screen — left the board."],[340.02656249999995,"result is hidden from the screen — left the board."]]}]},{"title":"Why It Always Works","start":341.06822916666664,"end":478.4562291666666,"objects":{"acc_area":"an AreaUnder [red] labelled \"A(x)\" drawn in axes_gen (x_range=(0.0, 2.0), target='curve')","axes_gen":"an Axes (x_range=(0.0, 4.0), y_range=(0.0, 3.8), x_ticks_every=1.0)","curve":"a FunctionPlot [blue] drawn in axes_gen (function=<function>, x_range=(0.0, 4.0))","drop":"a Line [yellow] drawn in axes_gen (start=(2.0, 0.0), end=(2.0, 2.08), dashed=True)","h_brace":"a Brace [cyan] labelled \"h\" drawn in axes_gen (x_start=2.0, x_end=(h_var + 2.0))","h_var":"a VariableNumber (initial_value=0.7)","heading":"a Heading that says \"Why It Always Works\"","pt":"a PlotPoint [yellow] labelled \"f(x)\" drawn in axes_gen (target='curve', x=2.0)","result":"a Math [text] that says \"$A'(x) = f(x)$\"","strip":"an AreaUnder [yellow] drawn in axes_gen (x_range=(2.0, (h_var + 2.0)), target='curve')","strip_box":"a RiemannRectangles [cyan] drawn in axes_gen (rectangle_count=1, x_range=(2.0, (h_var + 2.0)), sample_type='left')","work":"a Derivation [text] that says \"$Delta A &= A(x + h) - A(x) \\ &approx f(x) dot h \\ frac(Delta A, h) &approx f(x)$\""},"beats":[{"start":341.06822916666664,"say":"That was one particular tap, and one particularly friendly shape. But the fact we landed on has nothing to do with triangles. So here is a rate that wanders. No nice formula, and no geometry that will help you.","live":[],"does":[[341.06822916666664,"heading is shown on the screen, written out."],[341.06822916666664,"axes_gen is shown on the screen, written out."],[341.06822916666664,"curve is shown on the screen, drawn."]]},{"start":355.10372916666665,"say":"Same set-up as before. A of x is the area underneath it, from zero up to x. We stop the clock here, at x, and everything shaded to the left of that line is A of x.","live":["axes_gen","heading","curve"],"does":[[358.7492291666666,"acc_area is shown on the screen, written out."],[362.9632291666666,"drop is shown on the screen, written out."]]},{"start":368.83972916666664,"say":"Now nudge it. Push x forward by a small amount, and give that amount a name: h. There is h, on the picture, the little step from x across to x plus h.","live":["axes_gen","heading","curve","acc_area","drop"],"does":[[374.0642291666666,"h_brace is shown on the screen, written out."]]},{"start":382.10422916666664,"say":"The area grows by this sliver, the water that arrives between time x and time x plus h. Call that extra area delta A.","live":["axes_gen","heading","curve","acc_area","drop","h_brace"],"does":[[383.53222916666664,"strip is shown on the screen, written out."],[389.59222916666664,"axes_gen moves to a new place on the board."],[389.59222916666664,"work is shown on the screen, written out."]]},{"start":391.1097291666666,"say":"And here is the whole trick, so let me put it on the screen rather than leave it in the air. That sliver is very nearly a rectangle.","live":["axes_gen","heading","curve","acc_area","drop","h_brace","strip"],"does":[[398.0062291666666,"strip_box is shown on the screen, written out."]]},{"start":400.33922916666666,"say":"There it is, standing on the sliver it is pretending to be. Its width is h, the step we just marked. Its top sits level with the curve at x, at a height of f of x, because over a stretch that short the curve has not had time to change.","live":["axes_gen","heading","curve","acc_area","drop","h_brace","strip","strip_box"],"does":[[408.95322916666663,"pt is shown on the screen, written out."],[412.65722916666664,"work is shown on the screen, written out."]]},{"start":417.26772916666664,"say":"So delta A is about f of x times h. Now divide both sides by h, and the left-hand side turns into something you have met before. Change in A, divided by change in x. The average rate of change of A.","live":["axes_gen","heading","curve","acc_area","drop","h_brace","strip","strip_box","pt"],"does":[[421.0352291666666,"work is shown on the screen, written out."]]},{"start":432.69972916666666,"say":"And now let h shrink. Watch the sliver, and watch the rectangle standing on it. The thinner they both get, the better the rectangle fits, and the closer those two sides come to being equal.","live":null,"does":[[434.1512291666666,"h_brace is redrawn as the numbers it depends on change."],[434.1512291666666,"strip is redrawn as the numbers it depends on change."],[434.1512291666666,"strip_box is redrawn as the numbers it depends on change."],[434.1512291666666,"h_var ticks to 0.15."]]},{"start":446.4152291666667,"say":"In the limit, the left-hand side is the derivative of A at x. That is precisely the definition of a derivative. And the right-hand side never moved. So A prime of x equals f of x, for any rate you like.","live":null,"does":[[451.94122916666663,"result is shown on the screen, written out."],[459.80122916666664,"A box is drawn around result."]]},{"start":461.87572916666664,"say":"Said out loud, it is almost a tautology. How fast is the water in the tank going up? At the rate the water is coming in. That is the whole theorem. The surprise is not that it is true, the surprise is what it lets you do.","live":["result","axes_gen","heading","curve","acc_area","drop","h_brace","strip","strip_box","pt"],"does":[[464.43022916666666,"strip is hidden from the screen."],[464.43022916666666,"strip_box is hidden from the screen."],[464.43022916666666,"h_brace is hidden from the screen."],[477.4145625,"axes_gen is hidden from the screen — left the board."],[477.4145625,"curve is hidden from the screen — axes_gen left the board."],[477.4145625,"acc_area is hidden from the screen — axes_gen left the board."],[477.4145625,"drop is hidden from the screen — axes_gen left the board."],[477.4145625,"pt is hidden from the screen — axes_gen left the board."],[477.4145625,"heading is hidden from the screen — left the board."],[477.4145625,"result is hidden from the screen — left the board."],[477.4145625,"work is hidden from the screen — left the board."]]}]},{"title":"Running It Backwards","start":478.4562291666666,"end":798.9311666666666,"objects":{"F_dn":"a FunctionPlot [green] drawn in axes_F (function=<function>, x_range=(0.0, 4.0))","F_mid":"a FunctionPlot [blue] labelled \"F\" drawn in axes_F (function=<function>, x_range=(0.0, 4.0))","F_up":"a FunctionPlot [yellow] drawn in axes_F (function=<function>, x_range=(0.0, 4.0))","axes_F":"an Axes (x_range=(0.0, 4.0), y_range=(-2.4, 11.6), x_ticks_every=1.0)","axes_b":"an Axes (x_range=(0.0, 4.0), y_range=(0.0, 3.8), x_ticks_every=1.0)","band":"an AreaUnder [red] drawn in axes_b (x_range=(1.0, 3.0), target='curve_b')","closing":"a Math [text] that says \"$A(b) - A(a) = integral_a^b f(x) dif x$\"","curve_b":"a FunctionPlot [blue] drawn in axes_b (function=<function>, x_range=(0.0, 4.0))","defn":"a Panel that says \"The integral of $f$ from $a$ to $x$ is the area under the graph of $f$ between those two limits, the limit of a sum of thin rectangles.\"","difference":"an Arithmetic [text] that says \"$F(b) = A(b) + C F(a) = A(a) + C F(b) - F(a) = A(b) - A(a)$\" (operator='-', operands=('F(b) = A(b) + C', 'F(a) = A(a) + C'), result='F(b) - F(a) = A(b) - A(a)')","fact":"a Math [text] that says \"$F(x) = A(x) + C$\"","ftc1":"a Math [text] that says \"$frac(dif, dif x) integral_a^x f(t) dif t = f(x)$\"","ftc2":"a Math [text] that says \"$integral_a^b f(x) dif x = F(b) - F(a)$\"","ftc_note":"a Text [text] that says \"where $F$ is any antiderivative of $f$, that is, any function at all whose derivative is $f$.\"","gap_a":"a Line [magenta] labelled \"C\" drawn in axes_F (start=(1.0, 2.623333333333333), end=(1.0, 4.223333333333333))","gap_b":"a Line [magenta] labelled \"C\" drawn in axes_F (start=(3.0, 6.93), end=(3.0, 8.53))","goal":"a Tex [text] that says \"We want the area between $a$ and $b$.\"","h_back":"a Heading that says \"Running It Backwards\"","h_const":"a Heading that says \"Where the Constant Comes From\"","h_ftc":"a Heading that says \"The Fundamental Theorem\"","h_notation":"a Heading that says \"A Symbol for the Area\"","h_sub":"a Heading that says \"Subtract, and Watch C Go\"","lbl_family":"a Tex [text] that says \"Every antiderivative of it\"","lbl_rate":"a Tex [text] that says \"The rate, and the area we want\"","line_a":"a Line [green] labelled \"a\" drawn in axes_b (start=(1.0, 0.0), end=(1.0, 2.3200000000000003), dashed=True)","line_b":"a Line [green] labelled \"b\" drawn in axes_b (start=(3.0, 0.0), end=(3.0, 2.28), dashed=True)","notation":"a Math [text] that says \"$A(x) = integral_a^x f(t) dif t$\"","same":"a Derivation [text] that says \"$F'(x) &= f(x) \\ A'(x) &= f(x) \\ F'(x) &= A'(x)$\""},"beats":[{"start":478.4562291666666,"say":"Before we cash this in, we need the notation everybody actually writes. The area under f, from a up to x, has a symbol of its own.","live":[],"does":[[478.4562291666666,"h_notation is shown on the screen, written out."],[486.3972291666666,"notation is shown on the screen, written out."]]},{"start":488.39072916666663,"say":"A long, stretched S. The function goes in the middle, and dee t closes it off. Out loud: the integral of f of t, dee t, from a to x.","live":["notation","h_notation"],"does":[[491.2232291666666,"notation (the \"f(t)\" part) is emphasized."],[492.7792291666666,"notation (the \"dif t\" part) is emphasized."],[492.7792291666666,"notation (the \"f(t)\" part) is no longer emphasized."],[494.81122916666664,"defn is shown on the screen, written out."],[494.81122916666664,"notation (the \"dif t\" part) is no longer emphasized."]]},{"start":500.6467291666666,"say":"And let me say what those two ends are, because nobody ever does. The little a at the bottom is where we start counting. It is the left-hand edge of the region, chosen once and then left exactly where it is.","live":["notation","defn","h_notation"],"does":[[505.9522291666666,"notation (the \"a\" part) is emphasized."]]},{"start":514.1357291666666,"say":"The x on the top is where we stop. So the whole symbol is a function of where you stop, which is exactly what the shaded area was.","live":null,"does":[[515.1692291666666,"notation (the \"a\" part) is no longer emphasized."],[515.1692291666666,"notation (the \"x#2\" part) is emphasized."]]},{"start":522.7002291666666,"say":"And the S is for sum, because that is precisely what it is. All those thin rectangles, added up, with dee t as the width of one of them.","live":null,"does":[[524.0122291666667,"notation (the \"integral_a^x\" part) is emphasized."],[524.0122291666667,"notation (the \"x#2\" part) is no longer emphasized."],[530.9552291666666,"notation (the \"dif t\" part) is emphasized."],[530.9552291666666,"notation (the \"integral_a^x\" part) is no longer emphasized."]]},{"start":532.9367291666666,"say":"So in this language, what we proved with the thin sliver reads like this. 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Last time we proved that A prime of x is f of x too.","live":["axes_b","h_back","curve_b","line_a","line_b","band"],"does":[[591.3792291666666,"same is shown on the screen, written out."]]},{"start":595.4507291666666,"say":"So big F and our area function A have exactly the same derivative, everywhere, whatever big F happens to be.","live":null,"does":[[599.6892291666667,"same is shown on the screen, written out."],[603.9147291666666,"axes_b moves to a new place on the board."],[603.9147291666666,"h_back is hidden from the screen — left the board."],[603.9147291666666,"same is hidden from the screen — left the board."]]},{"start":604.9147291666666,"say":"Which raises a fair question. How different can two functions with the same derivative actually be? Let me draw one of them over here. 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Big F of x is our area function A of x, plus some fixed number C. Now watch what that does the moment we subtract.","live":["axes_b","curve_b","line_a","line_b","band","lbl_rate","lbl_family","axes_F","h_const","F_mid","F_up","F_dn","gap_a","gap_b"],"does":[[679.1362291666667,"axes_F is hidden from the screen — left the board."],[679.1362291666667,"F_mid is hidden from the screen — axes_F left the board."],[679.1362291666667,"F_up is hidden from the screen — axes_F left the board."],[679.1362291666667,"F_dn is hidden from the screen — axes_F left the board."],[679.1362291666667,"gap_a is hidden from the screen — axes_F left the board."],[679.1362291666667,"gap_b is hidden from the screen — axes_F left the board."],[679.1362291666667,"axes_b is hidden from the screen — left the board."],[679.1362291666667,"curve_b is hidden from the screen — axes_b left the board."],[679.1362291666667,"line_a is hidden from the screen — axes_b left the board."],[679.1362291666667,"line_b is hidden from the screen — axes_b left the board."],[679.1362291666667,"band is hidden from the screen — axes_b left the board."],[679.1362291666667,"h_const is hidden from the screen — left the board."],[679.1362291666667,"lbl_family is hidden from the screen — left the board."],[679.1362291666667,"lbl_rate is hidden from the screen — left the board."]]},{"start":680.3362291666666,"say":"Start from the line we just earned. 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So rule a line underneath them both.","live":null,"does":[[713.8762291666667,"difference is shown on the screen, drawn."],[714.6762291666666,"difference is shown on the screen, drawn."]]},{"start":717.3662291666666,"say":"And there is the whole trick, standing one above the other at the end of those two lines. Plus C on the top, plus C underneath. The same unknown number added on and then taken straight back off. So slash the pair of them out, and whatever C was, it has gone.","live":null,"does":[[731.0892291666667,"difference (the \"C\" part) is slashed through — it cancels."],[731.0892291666667,"difference (the \"C\" part) is slashed through — it cancels."]]},{"start":736.9727291666666,"say":"So what survives on the right is A of b minus A of a. 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The integral of f from a to b equals big F of b, minus big F of a, where big F is any antiderivative of f.","live":[],"does":[[768.2952291666666,"h_ftc is shown on the screen, written out."],[771.8942291666667,"ftc2 is shown on the screen, written out."],[778.7092291666667,"ftc_note is shown on the screen, written out."]]},{"start":780.8417291666666,"say":"That is the Fundamental Theorem of Calculus, and it is the reason integration is something you can actually sit down and do. To find an area, you no longer add up rectangles. You find a function whose derivative is the one you started with, and subtract at the two ends.","live":["ftc2","ftc_note","h_ftc"],"does":[[781.7012291666666,"A box is drawn around ftc2."],[797.8895,"ftc2 is hidden from the screen — left the board."],[797.8895,"ftc_note is hidden from the screen — left the board."],[797.8895,"h_ftc is hidden from the screen — left the board."]]}]},{"title":"Area Under a Parabola","start":798.9311666666666,"end":941.7526041666666,"objects":{"axes_p":"an Axes (x_range=(0.0, 2.4), y_range=(0.0, 4.4), x_ticks_every=1.0)","box_check":"a Polygon [gray] drawn in axes_p (vertices=((0.0, 0.0), (2.0, 0.0), (2.0, 4.0), (0.0, 4.0)), filled=False, dashed=True)","ftc1_final":"a Math [text] that says \"$frac(dif, dif x) integral_a^x f(t) dif t = f(x)$\"","ftc2_final":"a Math [text] that says \"$integral_a^b F'(x) dif x = F(b) - F(a)$\"","guess":"a Math [text] that says \"$F(x) = frac(x^3, 3)$\"","guesswork":"a Derivation [text] that says \"$frac(dif, dif x) x^3 &= 3 x^2 \\ frac(dif, dif x) frac(x^3, 3) &= frac(3 x^2, 3) = x^2$\"","h_sum":"a Heading that says \"Two Halves of One Idea\"","ladder":"a Derivation [text] that says \"$integral_0^2 x^2 dif x &= F(2) - F(0) \\ &= frac(2^3, 3) - frac(0^3, 3) \\ &= frac(8, 3) approx 2.67$\"","lbl_left":"a Tex [text] that says \"Differentiate an accumulation\"","lbl_right":"a Tex [text] that says \"Add up a rate\"","old_bars":"a RiemannRectangles [blue] drawn in axes_p (rectangle_count=8, x_range=(0.0, 2.0), opacity=0.5)","para":"a FunctionPlot [blue] labelled \"y = x^2\" drawn in axes_p (function=<function>, x_range=(0.0, 2.0))","problem":"a Tex [text] that says \"Find the area under $y = x^2$, from $x = 0$ to $x = 2$.\"","shade":"an AreaUnder [red] drawn in axes_p (x_range=(0.0, 2.0), target='para')"},"beats":[{"start":798.9311666666666,"say":"Let's spend it on a problem you cannot do with geometry. Here is y equals x squared, and here is the piece of it we want: everything under the curve between zero and two.","live":[],"does":[[798.9311666666666,"problem is shown on the screen, written out."],[798.9311666666666,"axes_p is shown on the screen, written out."],[798.9311666666666,"para is shown on the screen, drawn."],[806.6401666666666,"shade is shown on the screen, written out."]]},{"start":809.8926666666666,"say":"The old route uses thin bars like these, then takes a limit. That is real work, because this is not a triangle, a rectangle, a circle, or any other shape with a school formula.","live":["axes_p","problem","para","shade"],"does":[[811.3501666666666,"old_bars is shown on the screen, written out."]]},{"start":821.9576666666667,"say":"By the new route you need exactly one thing. Some function whose derivative is x squared. So let us go and find one.","live":["axes_p","problem","para","shade","old_bars"],"does":[[821.9576666666667,"old_bars is hidden from the screen."]]},{"start":830.9191666666666,"say":"Start by guessing. Differentiate x cubed, and you get three x squared. Write that down, because it is nearly right.","live":["axes_p","problem","para","shade"],"does":[[837.1541666666666,"axes_p moves to a new place on the board."],[837.1541666666666,"guesswork is shown on the screen, written out."]]},{"start":840.7956666666666,"say":"It is three times too big. So divide the whole thing by three before you differentiate it. Now the three on top and the three underneath are the same three, and what comes out is exactly x squared.","live":null,"does":[[843.6861666666666,"guesswork is shown on the screen, written out."]]},{"start":853.8296666666666,"say":"So there is our big F. F of x is x cubed over three, and its derivative is the x squared we started with.","live":null,"does":[[854.3171666666666,"guess is shown on the screen, written out."]]},{"start":863.2906666666667,"say":"Now the theorem does the rest. The integral from zero to two of x squared, dee x, is F at two minus F at zero.","live":["guess","axes_p","problem","para","shade"],"does":[[864.0101666666667,"ladder is shown on the screen, written out."]]},{"start":873.2716666666666,"say":"F at two is two cubed over three. F at zero is zero. Take the second one away from the first.","live":null,"does":[[875.0591666666667,"ladder is shown on the screen, written out."]]},{"start":882.0796666666666,"say":"And what is left is eight thirds, which is about two point six seven. That is the exact area under the parabola, and we got it by differentiating backwards. No rectangles, no limits, three lines of algebra.","live":null,"does":[[882.8461666666666,"ladder is shown on the screen, written out."],[887.0251666666666,"A box is drawn around ladder."]]},{"start":896.4836666666666,"say":"Does that number look sensible? The region sits inside this two by four rectangle, which has area eight, and by eye the shaded part is about a third of it. Two point six seven out of eight. That checks out.","live":null,"does":[[901.2791666666666,"box_check is shown on the screen, written out."],[910.2536666666666,"axes_p is hidden from the screen — left the board."],[910.2536666666666,"para is hidden from the screen — axes_p left the board."],[910.2536666666666,"shade is hidden from the screen — axes_p left the board."],[910.2536666666666,"box_check is hidden from the screen — axes_p left the board."],[910.2536666666666,"guess is hidden from the screen — left the board."],[910.2536666666666,"guesswork is hidden from the screen — left the board."],[910.2536666666666,"ladder is hidden from the screen — left the board."],[910.2536666666666,"problem is hidden from the screen — left the board."]]},{"start":911.4536666666665,"say":"So here are the two halves of one idea. On the left, differentiate an accumulation, and you get back the rate you were accumulating at. On the right, to total up a rate across an interval, find something that differentiates to it, and subtract at the two ends.","live":[],"does":[[911.4536666666665,"h_sum is shown on the screen, written out."],[915.1801666666667,"lbl_left is shown on the screen, written out."],[915.1801666666667,"ftc1_final is shown on the screen, written out."],[921.3681666666666,"lbl_right is shown on the screen, written out."],[921.3681666666666,"ftc2_final is shown on the screen, written out."]]},{"start":929.7121666666666,"say":"Area and slope are not two separate subjects that happen to share a course. They are one machine, run in the two directions. That is the Fundamental Theorem of Calculus.","live":["lbl_left","ftc1_final","lbl_right","ftc2_final","h_sum"],"does":[[935.4361666666666,"ftc1_final is indicated — a transient flash."],[935.4361666666666,"ftc2_final is indicated — a transient flash."],[940.7109375,"ftc1_final is hidden from the screen — left the board."],[940.7109375,"ftc2_final is hidden from the screen — left the board."],[940.7109375,"h_sum is hidden from the screen — left the board."],[940.7109375,"lbl_left is hidden from the screen — left the board."],[940.7109375,"lbl_right is hidden from the screen — left the board."]]}]}]},"durationSeconds":942,"chapters":[{"title":"What Piles Up","startSeconds":0,"narration":"Derivatives and integrals. One of them measures how fast something is changing. The other measures how much of something has piled up. Two completely different questions, and this whole lecture is about the fact that they are one question, asked from opposite ends. So let me start with a tank. Somebody has opened a tap, and I know the rate, at every instant, exactly how fast the water is arriving. The question is simple enough. How much water is in there? Let's draw the rate. Time runs along the bottom, and the height of this graph is how fast the water is arriving at that moment. This tap is being opened steadily, so the line climbs. So at time one, the water is arriving at one litre a second. Move out to time three, and it is arriving at three litres a second. Now hold that thought, and imagine a different tap for a moment. Suppose the rate were constant. Say two litres a second, flat, the whole way across. Held there for three seconds, the answer is easy. Two litres a second, for three seconds, is six litres. Rate times time. And on this picture, rate times time is the area of one rectangle. But our tap is not that tap. Our rate climbs the whole way, so there is no single rectangle to draw. So we chop the time up into slivers instead. Over a stretch short enough the rate barely changes, so each sliver is rate times time again. One thin rectangle. Then we add them all up. Now watch what happens as we use more and more of them. The staircase closes on the region underneath the curve, and in the limit the water that has arrived is exactly the area under the rate graph. So here is the object I care about. The amount of water that has arrived by time x is the area under the graph, from the start out to x. Give it a name. A of x, the accumulation function. And notice what kind of object A is. It is a function. You hand it a time, here, and it hands you back a number, the area up to there. Watch what happens when I slide that edge along. More of the region is swept in, and the number gets bigger. Every stopping point gives you one number, so this really is a function of x, built out of nothing but area. Hold on to that picture. The area under a rate graph is itself a function, and everything that follows comes from one question about it. So let me put that question on the board, in as many words. How fast does A grow? Answering that is the whole of the rest of this lecture."},{"title":"The Slope of the Area","startSeconds":165.73575,"narration":"A moment ago we ended with the accumulation function. A of x is the area under the rate graph, from zero up to x. For this particular tap we can work that area out exactly, and pleasingly, we need no calculus at all to do it. First, one word about the letters, because there are two of them and they are not the same letter. Along the bottom the clock is called t, since t is what sweeps across the region while we shade it in. The letter x is saved for where we stop. So stop the clock at x. There is the mark, on the axis the clock runs along. And look at what is shaded: it is a triangle. Its base runs from zero out to x, so the base is x. Its height is the rate at time x, and for this tap the rate at time x is x as well. The area of a triangle is a half, base, height. Here that is a half times x times x. Which comes to one half x squared. There it is. The accumulation function for this tap, in closed form. Now let us put the two of them side by side, because this is where it gets good. On the left, the rate. A straight line, climbing steadily. On the right, the accumulation. A parabola, bending upward. As the tap opens further, the water piles up faster and faster. Now a question about the curve on the right. How steep is it? The steepness of A at a point is exactly what the symbol A prime of x will mean, so take the tangent line at x equals two and read off its slope. Rise over run, the slope there is two. Now look across at the rate graph. At time two, the height of the rate line is, two. Coincidence? Let's slide x along and watch both pictures at once. Carry on out to x equals three. The tangent on the accumulation curve now has slope three. And the rate line at time three is sitting at height three. The slope of the accumulation is the height of the rate, every single time. That deserves a line of its own. Two things to write down, and the first one is what the symbol is going to mean. A is a function of x, so A prime of x is its rate of change: how fast the shaded area grows as x moves off to the right. And the second is what we just watched. That rate of change is the height of the rate graph at x. So A prime of x equals f of x. The slope of the accumulation curve is the height of the rate curve, at every single point. We got there from one triangle and one parabola, but it has nothing to do with triangles, and I want to show you why it has to be true for absolutely any rate at all."},{"title":"Why It Always Works","startSeconds":341.06822916666664,"narration":"That was one particular tap, and one particularly friendly shape. But the fact we landed on has nothing to do with triangles. So here is a rate that wanders. No nice formula, and no geometry that will help you. Same set-up as before. A of x is the area underneath it, from zero up to x. We stop the clock here, at x, and everything shaded to the left of that line is A of x. Now nudge it. Push x forward by a small amount, and give that amount a name: h. There is h, on the picture, the little step from x across to x plus h. The area grows by this sliver, the water that arrives between time x and time x plus h. Call that extra area delta A. And here is the whole trick, so let me put it on the screen rather than leave it in the air. That sliver is very nearly a rectangle. There it is, standing on the sliver it is pretending to be. Its width is h, the step we just marked. Its top sits level with the curve at x, at a height of f of x, because over a stretch that short the curve has not had time to change. So delta A is about f of x times h. Now divide both sides by h, and the left-hand side turns into something you have met before. Change in A, divided by change in x. The average rate of change of A. And now let h shrink. Watch the sliver, and watch the rectangle standing on it. The thinner they both get, the better the rectangle fits, and the closer those two sides come to being equal. In the limit, the left-hand side is the derivative of A at x. That is precisely the definition of a derivative. And the right-hand side never moved. So A prime of x equals f of x, for any rate you like. Said out loud, it is almost a tautology. How fast is the water in the tank going up? At the rate the water is coming in. That is the whole theorem. The surprise is not that it is true, the surprise is what it lets you do."},{"title":"Running It Backwards","startSeconds":478.4562291666666,"narration":"Before we cash this in, we need the notation everybody actually writes. The area under f, from a up to x, has a symbol of its own. A long, stretched S. The function goes in the middle, and dee t closes it off. Out loud: the integral of f of t, dee t, from a to x. And let me say what those two ends are, because nobody ever does. The little a at the bottom is where we start counting. It is the left-hand edge of the region, chosen once and then left exactly where it is. The x on the top is where we stop. So the whole symbol is a function of where you stop, which is exactly what the shaded area was. And the S is for sum, because that is precisely what it is. All those thin rectangles, added up, with dee t as the width of one of them. So in this language, what we proved with the thin sliver reads like this. Differentiate an integral with respect to its upper limit, and you get the integrand straight back. Now the payoff. Here is a rate again, and here is the job in front of us: the area underneath it, between two fixed times. The left-hand end we call a. That is where the counting starts. The right-hand end we call b, and that is where it stops. Everything shaded between those two lines is the area we are trying to find, and we have no formula for a shape like that. Now suppose somebody hands you a function big F whose derivative is little f. Any such function at all will do. There it is, written down: F prime of x equals f of x. And we already have one function like that. Our own area function, A. Last time we proved that A prime of x is f of x too. So big F and our area function A have exactly the same derivative, everywhere, whatever big F happens to be. Which raises a fair question. How different can two functions with the same derivative actually be? Let me draw one of them over here. This is big F, and its slope at every point is the height of the curve beside it. Now take a copy of it, and lift the whole copy upward. Every point goes up by the same amount, so the curve keeps its shape exactly. At every x it is still leaning the way it was leaning. Its slope has not changed anywhere. Push a copy downward and you get the same story again. All three of these are antiderivatives of the same little f. There is a whole family of them stacked up the board, and every one of them is parallel to the others. And here is where the constant comes from. Measure the gap between two of them at x equals a. Now measure it again over at x equals b. It is the same gap. It is the same gap everywhere, and that gap is the number we call C. So two antiderivatives of one function differ by a constant, and only by a constant. Big F of x is our area function A of x, plus some fixed number C. Now watch what that does the moment we subtract. Start from the line we just earned. Big F of x is A of x, plus C. That is true at every single x, so in particular it is true at b. Big F of b is A of b, plus C. And it is just as true at a. Big F of a is A of a, plus C. The very same C both times, because it is the very same big F both times. Now take the second of those two lines away from the first, exactly the way you would take one number off another. So rule a line underneath them both. And there is the whole trick, standing one above the other at the end of those two lines. Plus C on the top, plus C underneath. The same unknown number added on and then taken straight back off. So slash the pair of them out, and whatever C was, it has gone. So what survives on the right is A of b minus A of a. The constant has gone, and it was always going to go, whichever antiderivative you had picked up in the first place. And A of b minus A of a is exactly the area under f between a and b. Which, in the notation we built at the start of this scene, is the integral of f from a to b. So there is the bottom line. Let me write it out properly, because that is the theorem. So there it is, in a single line. The integral of f from a to b equals big F of b, minus big F of a, where big F is any antiderivative of f. That is the Fundamental Theorem of Calculus, and it is the reason integration is something you can actually sit down and do. To find an area, you no longer add up rectangles. You find a function whose derivative is the one you started with, and subtract at the two ends."},{"title":"Area Under a Parabola","startSeconds":798.9311666666666,"narration":"Let's spend it on a problem you cannot do with geometry. Here is y equals x squared, and here is the piece of it we want: everything under the curve between zero and two. The old route uses thin bars like these, then takes a limit. That is real work, because this is not a triangle, a rectangle, a circle, or any other shape with a school formula. By the new route you need exactly one thing. Some function whose derivative is x squared. So let us go and find one. Start by guessing. Differentiate x cubed, and you get three x squared. Write that down, because it is nearly right. It is three times too big. So divide the whole thing by three before you differentiate it. Now the three on top and the three underneath are the same three, and what comes out is exactly x squared. So there is our big F. F of x is x cubed over three, and its derivative is the x squared we started with. Now the theorem does the rest. The integral from zero to two of x squared, dee x, is F at two minus F at zero. F at two is two cubed over three. F at zero is zero. Take the second one away from the first. And what is left is eight thirds, which is about two point six seven. That is the exact area under the parabola, and we got it by differentiating backwards. No rectangles, no limits, three lines of algebra. Does that number look sensible? The region sits inside this two by four rectangle, which has area eight, and by eye the shaded part is about a third of it. Two point six seven out of eight. That checks out. So here are the two halves of one idea. On the left, differentiate an accumulation, and you get back the rate you were accumulating at. On the right, to total up a rate across an interval, find something that differentiates to it, and subtract at the two ends. Area and slope are not two separate subjects that happen to share a course. They are one machine, run in the two directions. That is the Fundamental Theorem of Calculus."}]}}