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- 1. Section 5.4 The Fundamental Theorem of Calculus V63.0121.002.2010Su, Calculus I New York University June 21, 2010 Announcements
- 2. Announcements V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 2 / 33
- 3. Objectives State and explain the Fundemental Theorems of Calculus Use the ﬁrst fundamental theorem of calculus to ﬁnd derivatives of functions deﬁned as integrals. Compute the average value of an integrable function over a closed interval. V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 3 / 33
- 4. Outline Recall: The Evaluation Theorem a/k/a 2FTC The First Fundamental Theorem of Calculus The Area Function Statement and proof of 1FTC Biographies Diﬀerentiation of functions deﬁned by integrals “Contrived” examples Erf Other applications V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 4 / 33
- 5. The deﬁnite integral as a limit Deﬁnition If f is a function deﬁned on [a, b], the deﬁnite integral of f from a to b is the number b n f (x) dx = lim f (ci ) ∆x a ∆x→0 i=1 V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 5 / 33
- 6. Big time Theorem Theorem (The Second Fundamental Theorem of Calculus) Suppose f is integrable on [a, b] and f = F for another function F , then b f (x) dx = F (b) − F (a). a V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 6 / 33
- 7. The Integral as Total Change Another way to state this theorem is: b F (x) dx = F (b) − F (a), a or the integral of a derivative along an interval is the total change between the sides of that interval. This has many ramiﬁcations: V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 7 / 33
- 8. The Integral as Total Change Another way to state this theorem is: b F (x) dx = F (b) − F (a), a or the integral of a derivative along an interval is the total change between the sides of that interval. This has many ramiﬁcations: Theorem If v (t) represents the velocity of a particle moving rectilinearly, then t1 v (t) dt = s(t1 ) − s(t0 ). t0 V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 7 / 33
- 9. The Integral as Total Change Another way to state this theorem is: b F (x) dx = F (b) − F (a), a or the integral of a derivative along an interval is the total change between the sides of that interval. This has many ramiﬁcations: Theorem If MC (x) represents the marginal cost of making x units of a product, then x C (x) = C (0) + MC (q) dq. 0 V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 7 / 33
- 10. The Integral as Total Change Another way to state this theorem is: b F (x) dx = F (b) − F (a), a or the integral of a derivative along an interval is the total change between the sides of that interval. This has many ramiﬁcations: Theorem If ρ(x) represents the density of a thin rod at a distance of x from its end, then the mass of the rod up to x is x m(x) = ρ(s) ds. 0 V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 7 / 33
- 11. My ﬁrst table of integrals [f (x) + g (x)] dx = f (x) dx + g (x) dx x n+1 x n dx = + C (n = −1) cf (x) dx = c f (x) dx n+1 1 e x dx = e x + C dx = ln |x| + C x ax sin x dx = − cos x + C ax dx = +C ln a cos x dx = sin x + C csc2 x dx = − cot x + C sec2 x dx = tan x + C csc x cot x dx = − csc x + C 1 sec x tan x dx = sec x + C √ dx = arcsin x + C 1 − x2 1 dx = arctan x + C 1 + x2 V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 8 / 33
- 12. Outline Recall: The Evaluation Theorem a/k/a 2FTC The First Fundamental Theorem of Calculus The Area Function Statement and proof of 1FTC Biographies Diﬀerentiation of functions deﬁned by integrals “Contrived” examples Erf Other applications V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 9 / 33
- 13. An area function x Let f (t) = t 3 and deﬁne g (x) = f (t) dt. Can we evaluate the integral 0 in g (x)? 0 x V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 10 / 33
- 14. An area function x Let f (t) = t 3 and deﬁne g (x) = f (t) dt. Can we evaluate the integral 0 in g (x)? Dividing the interval [0, x] into n pieces x ix gives ∆t = and ti = 0 + i∆t = . So n n x x 3 x (2x)3 x (nx)3 Rn = · 3+ · 3 + ··· + · n n n n n n3 x4 = 4 13 + 23 + 33 + · · · + n3 n x4 1 2 = 4 2 n(n + 1) n 0 x x 4 n2 (n + 1)2 x4 = 4 → 4n 4 as n → ∞. V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 10 / 33
- 15. An area function, continued So x4 g (x) = . 4 V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 11 / 33
- 16. An area function, continued So x4 g (x) = . 4 This means that g (x) = x 3 . V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 11 / 33
- 17. The area function Let f be a function which is integrable (i.e., continuous or with ﬁnitely many jump discontinuities) on [a, b]. Deﬁne x g (x) = f (t) dt. a The variable is x; t is a “dummy” variable that’s integrated over. Picture changing x and taking more of less of the region under the curve. Question: What does f tell you about g ? V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 12 / 33
- 18. Envisioning the area function Example Suppose f (t) is the function graphed below: y x 2 4 6 8 10f x Let g (x) = f (t) dt. What can you say about g ? 0 V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33
- 19. Envisioning the area function Example Suppose f (t) is the function graphed below: y g x 2 4 6 8 10f x Let g (x) = f (t) dt. What can you say about g ? 0 V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33
- 20. Envisioning the area function Example Suppose f (t) is the function graphed below: y g x 2 4 6 8 10f x Let g (x) = f (t) dt. What can you say about g ? 0 V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33
- 21. Envisioning the area function Example Suppose f (t) is the function graphed below: y g x 2 4 6 8 10f x Let g (x) = f (t) dt. What can you say about g ? 0 V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33
- 22. Envisioning the area function Example Suppose f (t) is the function graphed below: y g x 2 4 6 8 10f x Let g (x) = f (t) dt. What can you say about g ? 0 V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33
- 23. Envisioning the area function Example Suppose f (t) is the function graphed below: y g x 2 4 6 8 10f x Let g (x) = f (t) dt. What can you say about g ? 0 V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33
- 24. Envisioning the area function Example Suppose f (t) is the function graphed below: y g x 2 4 6 8 10f x Let g (x) = f (t) dt. What can you say about g ? 0 V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33
- 25. Envisioning the area function Example Suppose f (t) is the function graphed below: y g x 2 4 6 8 10f x Let g (x) = f (t) dt. What can you say about g ? 0 V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33
- 26. Envisioning the area function Example Suppose f (t) is the function graphed below: y g x 2 4 6 8 10f x Let g (x) = f (t) dt. What can you say about g ? 0 V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33
- 27. Envisioning the area function Example Suppose f (t) is the function graphed below: y g x 2 4 6 8 10f x Let g (x) = f (t) dt. What can you say about g ? 0 V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33
- 28. Envisioning the area function Example Suppose f (t) is the function graphed below: y g x 2 4 6 8 10f x Let g (x) = f (t) dt. What can you say about g ? 0 V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33
- 29. Envisioning the area function Example Suppose f (t) is the function graphed below: y g x 2 4 6 8 10f x Let g (x) = f (t) dt. What can you say about g ? 0 V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 13 / 33
- 30. features of g from f y Interval sign monotonicity monotonicity concavity of f of g of f of g g [0, 2] + x 2 4 6 8 10f [2, 4.5] + [4.5, 6] − [6, 8] − [8, 10] − → none V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 14 / 33
- 31. features of g from f y Interval sign monotonicity monotonicity concavity of f of g of f of g g [0, 2] + x 2 4 6 8 10f [2, 4.5] + [4.5, 6] − [6, 8] − [8, 10] − → none We see that g is behaving a lot like an antiderivative of f . V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 14 / 33
- 32. Another Big Time Theorem Theorem (The First Fundamental Theorem of Calculus) Let f be an integrable function on [a, b] and deﬁne x g (x) = f (t) dt. a If f is continuous at x in (a, b), then g is diﬀerentiable at x and g (x) = f (x). V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 15 / 33
- 33. Proving the Fundamental Theorem Proof. Let h > 0 be given so that x + h < b. We have g (x + h) − g (x) = h V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 16 / 33
- 34. Proving the Fundamental Theorem Proof. Let h > 0 be given so that x + h < b. We have x+h g (x + h) − g (x) 1 = f (t) dt. h h x V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 16 / 33
- 35. Proving the Fundamental Theorem Proof. Let h > 0 be given so that x + h < b. We have x+h g (x + h) − g (x) 1 = f (t) dt. h h x Let Mh be the maximum value of f on [x, x + h], and let mh the minimum value of f on [x, x + h]. From §5.2 we have x+h f (t) dt x V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 16 / 33
- 36. Proving the Fundamental Theorem Proof. Let h > 0 be given so that x + h < b. We have x+h g (x + h) − g (x) 1 = f (t) dt. h h x Let Mh be the maximum value of f on [x, x + h], and let mh the minimum value of f on [x, x + h]. From §5.2 we have x+h f (t) dt ≤ Mh · h x V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 16 / 33
- 37. Proving the Fundamental Theorem Proof. Let h > 0 be given so that x + h < b. We have x+h g (x + h) − g (x) 1 = f (t) dt. h h x Let Mh be the maximum value of f on [x, x + h], and let mh the minimum value of f on [x, x + h]. From §5.2 we have x+h mh · h ≤ f (t) dt ≤ Mh · h x V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 16 / 33
- 38. Proving the Fundamental Theorem Proof. Let h > 0 be given so that x + h < b. We have x+h g (x + h) − g (x) 1 = f (t) dt. h h x Let Mh be the maximum value of f on [x, x + h], and let mh the minimum value of f on [x, x + h]. From §5.2 we have x+h mh · h ≤ f (t) dt ≤ Mh · h x So g (x + h) − g (x) mh ≤ ≤ Mh . h V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 16 / 33
- 39. Proving the Fundamental Theorem Proof. Let h > 0 be given so that x + h < b. We have x+h g (x + h) − g (x) 1 = f (t) dt. h h x Let Mh be the maximum value of f on [x, x + h], and let mh the minimum value of f on [x, x + h]. From §5.2 we have x+h mh · h ≤ f (t) dt ≤ Mh · h x So g (x + h) − g (x) mh ≤ ≤ Mh . h As h → 0, both mh and Mh tend to f (x). V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 16 / 33
- 40. Meet the Mathematician: James Gregory Scottish, 1638-1675 Astronomer and Geometer Conceived transcendental numbers and found evidence that π was transcendental Proved a geometric version of 1FTC as a lemma but didn’t take it further V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 17 / 33
- 41. Meet the Mathematician: Isaac Barrow English, 1630-1677 Professor of Greek, theology, and mathematics at Cambridge Had a famous student V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 18 / 33
- 42. Meet the Mathematician: Isaac Newton English, 1643–1727 Professor at Cambridge (England) Philosophiae Naturalis Principia Mathematica published 1687 V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 19 / 33
- 43. Meet the Mathematician: Gottfried Leibniz German, 1646–1716 Eminent philosopher as well as mathematician Contemporarily disgraced by the calculus priority dispute V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 20 / 33
- 44. Diﬀerentiation and Integration as reverse processes Putting together 1FTC and 2FTC, we get a beautiful relationship between the two fundamental concepts in calculus. Theorem (The Fundamental Theorem(s) of Calculus) I. If f is a continuous function, then x d f (t) dt = f (x) dx a So the derivative of the integral is the original function. II. If f is a diﬀerentiable function, then b f (x) dx = f (b) − f (a). a So the integral of the derivative of is (an evaluation of) the original function. V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 21 / 33
- 45. Outline Recall: The Evaluation Theorem a/k/a 2FTC The First Fundamental Theorem of Calculus The Area Function Statement and proof of 1FTC Biographies Diﬀerentiation of functions deﬁned by integrals “Contrived” examples Erf Other applications V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 22 / 33
- 46. Diﬀerentiation of area functions Example 3x Let h(x) = t 3 dt. What is h (x)? 0 V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 23 / 33
- 47. Diﬀerentiation of area functions Example 3x Let h(x) = t 3 dt. What is h (x)? 0 Solution (Using 2FTC) 3x t4 1 h(x) = = (3x)4 = 1 4 · 81x 4 , so h (x) = 81x 3 . 4 0 4 V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 23 / 33
- 48. Diﬀerentiation of area functions Example 3x Let h(x) = t 3 dt. What is h (x)? 0 Solution (Using 2FTC) 3x t4 1 h(x) = = (3x)4 = 1 4 · 81x 4 , so h (x) = 81x 3 . 4 0 4 Solution (Using 1FTC) u We can think of h as the composition g ◦ k, where g (u) = t 3 dt and 0 k(x) = 3x. V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 23 / 33
- 49. Diﬀerentiation of area functions Example 3x Let h(x) = t 3 dt. What is h (x)? 0 Solution (Using 2FTC) 3x t4 1 h(x) = = (3x)4 = 1 4 · 81x 4 , so h (x) = 81x 3 . 4 0 4 Solution (Using 1FTC) u We can think of h as the composition g ◦ k, where g (u) = t 3 dt and 0 k(x) = 3x. Then h (x) = g (u) · k (x), or h (x) = g (k(x)) · k (x) = (k(x))3 · 3 = (3x)3 · 3 = 81x 3 . V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 23 / 33
- 50. Diﬀerentiation of area functions, in general by 1FTC k(x) d f (t) dt = f (k(x))k (x) dx a by reversing the order of integration: b h(x) d d f (t) dt = − f (t) dt = −f (h(x))h (x) dx h(x) dx b by combining the two above: k(x) k(x) 0 d d f (t) dt = f (t) dt + f (t) dt dx h(x) dx 0 h(x) = f (k(x))k (x) − f (h(x))h (x) V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 24 / 33
- 51. Another Example Example sin2 x Let h(x) = (17t 2 + 4t − 4) dt. What is h (x)? 0 V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 25 / 33
- 52. Another Example Example sin2 x Let h(x) = (17t 2 + 4t − 4) dt. What is h (x)? 0 Solution We have sin2 x d (17t 2 + 4t − 4) dt dx 0 d = 17(sin2 x)2 + 4(sin2 x) − 4 · sin2 x dx = 17 sin4 x + 4 sin2 x − 4 · 2 sin x cos x V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 25 / 33
- 53. A Similar Example Example sin2 x Let h(x) = (17t 2 + 4t − 4) dt. What is h (x)? 3 V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 26 / 33
- 54. A Similar Example Example sin2 x Let h(x) = (17t 2 + 4t − 4) dt. What is h (x)? 3 Solution We have sin2 x d (17t 2 + 4t − 4) dt dx 0 d = 17(sin2 x)2 + 4(sin2 x) − 4 · sin2 x dx = 17 sin4 x + 4 sin2 x − 4 · 2 sin x cos x V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 26 / 33
- 55. Compare Question Why is sin2 x sin2 x d 2 d (17t + 4t − 4) dt = (17t 2 + 4t − 4) dt? dx 0 dx 3 Or, why doesn’t the lower limit appear in the derivative? V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 27 / 33
- 56. Compare Question Why is sin2 x sin2 x d 2 d (17t + 4t − 4) dt = (17t 2 + 4t − 4) dt? dx 0 dx 3 Or, why doesn’t the lower limit appear in the derivative? Answer Because sin2 x 3 sin2 x 2 2 (17t + 4t − 4) dt = (17t + 4t − 4) dt + (17t 2 + 4t − 4) dt 0 0 3 So the two functions diﬀer by a constant. V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 27 / 33
- 57. The Full Nasty Example ex Find the derivative of F (x) = sin4 t dt. x3 V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 28 / 33
- 58. The Full Nasty Example ex Find the derivative of F (x) = sin4 t dt. x3 Solution ex d sin4 t dt = sin4 (e x ) · e x − sin4 (x 3 ) · 3x 2 dx x3 V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 28 / 33
- 59. The Full Nasty Example ex Find the derivative of F (x) = sin4 t dt. x3 Solution ex d sin4 t dt = sin4 (e x ) · e x − sin4 (x 3 ) · 3x 2 dx x3 Notice here it’s much easier than ﬁnding an antiderivative for sin4 . V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 28 / 33
- 60. Why use 1FTC? Question Why would we use 1FTC to ﬁnd the derivative of an integral? It seems like confusion for its own sake. V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 29 / 33
- 61. Why use 1FTC? Question Why would we use 1FTC to ﬁnd the derivative of an integral? It seems like confusion for its own sake. Answer Some functions are diﬃcult or impossible to integrate in elementary terms. V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 29 / 33
- 62. Why use 1FTC? Question Why would we use 1FTC to ﬁnd the derivative of an integral? It seems like confusion for its own sake. Answer Some functions are diﬃcult or impossible to integrate in elementary terms. Some functions are naturally deﬁned in terms of other integrals. V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 29 / 33
- 63. Erf Here’s a function with a funny name but an important role: x 2 2 erf(x) = √ e −t dt. π 0 V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 30 / 33
- 64. Erf Here’s a function with a funny name but an important role: x 2 2 erf(x) = √ e −t dt. π 0 It turns out erf is the shape of the bell curve. V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 30 / 33
- 65. Erf Here’s a function with a funny name but an important role: x 2 2 erf(x) = √ e −t dt. π 0 It turns out erf is the shape of the bell curve. We can’t ﬁnd erf(x), explicitly, but we do know its derivative: erf (x) = V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 30 / 33
- 66. Erf Here’s a function with a funny name but an important role: x 2 2 erf(x) = √ e −t dt. π 0 It turns out erf is the shape of the bell curve. We can’t ﬁnd erf(x), 2 2 explicitly, but we do know its derivative: erf (x) = √ e −x . π V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 30 / 33
- 67. Erf Here’s a function with a funny name but an important role: x 2 2 erf(x) = √ e −t dt. π 0 It turns out erf is the shape of the bell curve. We can’t ﬁnd erf(x), 2 2 explicitly, but we do know its derivative: erf (x) = √ e −x . π Example d Find erf(x 2 ). dx V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 30 / 33
- 68. Erf Here’s a function with a funny name but an important role: x 2 2 erf(x) = √ e −t dt. π 0 It turns out erf is the shape of the bell curve. We can’t ﬁnd erf(x), 2 2 explicitly, but we do know its derivative: erf (x) = √ e −x . π Example d Find erf(x 2 ). dx Solution By the chain rule we have d d 2 2 2 4 4 erf(x 2 ) = erf (x 2 ) x 2 = √ e −(x ) 2x = √ xe −x . dx dx π π V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 30 / 33
- 69. Other functions deﬁned by integrals The future value of an asset: ∞ FV (t) = π(s)e −rs ds t where π(s) is the proﬁtability at time s and r is the discount rate. The consumer surplus of a good: q∗ CS(q ∗ ) = (f (q) − p ∗ ) dq 0 where f (q) is the demand function and p ∗ and q ∗ the equilibrium price and quantity. V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 31 / 33
- 70. Surplus by picture price (p) quantity (q) V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 32 / 33
- 71. Surplus by picture price (p) demand f (q) quantity (q) V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 32 / 33
- 72. Surplus by picture price (p) supply demand f (q) quantity (q) V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 32 / 33
- 73. Surplus by picture price (p) supply p∗ equilibrium demand f (q) q∗ quantity (q) V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 32 / 33
- 74. Surplus by picture price (p) supply p∗ equilibrium market revenue demand f (q) q∗ quantity (q) V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 32 / 33
- 75. Surplus by picture consumer surplus price (p) supply p∗ equilibrium market revenue demand f (q) q∗ quantity (q) V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 32 / 33
- 76. Surplus by picture consumer surplus price (p) producer surplus supply p∗ equilibrium demand f (q) q∗ quantity (q) V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 32 / 33
- 77. Summary Functions deﬁned as integrals can be diﬀerentiated using the ﬁrst FTC: x d f (t) dt = f (x) dx a The two FTCs link the two major processes in calculus: diﬀerentiation and integration F (x) dx = F (x) + C Follow the calculus wars on twitter: #calcwars V63.0121.002.2010Su, Calculus I (NYU) Section 5.4 The Fundamental Theorem June 21, 2010 33 / 33

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