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# Lesson 25: The Fundamental Theorem of Calculus

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### Lesson 25: The Fundamental Theorem of Calculus

1. 1. Section 5.4 The Fundamental Theorem of Calculus V63.0121.006/016, Calculus I New York University April 22, 2010 Announcements April 29: Movie Day April 30: Quiz 5 on §§5.1–5.4 Monday, May 10, 12:00noon Final Exam . . . . . . .
2. 2. Announcements April 29: Movie Day April 30: Quiz 5 on §§5.1–5.4 Monday, May 10, 12:00noon Final Exam . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 2 / 34
3. 3. Resurrection policies Current distribution of grade: 40% final, 25% midterm, 15% quizzes, 10% written HW, 10% WebAssign Remember we drop the lowest quiz, lowest written HW, and 5 lowest WebAssign-ments If your final exam score beats your midterm score, we will re-weight it by 50% and make the midterm 15% . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 3 / 34
4. 4. Resurrection policies Current distribution of grade: 40% final, 25% midterm, 15% quizzes, 10% written HW, 10% WebAssign Remember we drop the lowest quiz, lowest written HW, and 5 lowest WebAssign-ments If your final exam score beats your midterm score, we will re-weight it by 50% and make the midterm 15% . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 3 / 34
5. 5. 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. . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 4 / 34
6. 6. 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 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 5 / 34
7. 7. 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 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 6 / 34
8. 8. 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 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 7 / 34
9. 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: . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 8 / 34
10. 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 v(t) represents the velocity of a particle moving rectilinearly, then ∫ t1 v(t) dt = s(t1 ) − s(t0 ). t0 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 8 / 34
11. 11. 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 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 8 / 34
12. 12. 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 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 8 / 34
13. 13. My first table of integrals . ∫ ∫ ∫ [f(x) + g(x)] dx = f(x) dx + g(x) dx ∫ ∫ ∫ xn+1 xn dx = + C (n ̸= −1) cf(x) dx = c f(x) dx n+1 ∫ ∫ 1 ex dx = ex + 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 ∫ ∫ 2 sec 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 . . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 9 / 34
14. 14. 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 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 10 / 34
15. 15. An area function ∫ x Let f(t) = t3 and define g(x) = f(t) dt. Can we evaluate the integral 0 in g(x)? . 0 . x . . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 11 / 34
16. 16. An area function ∫ x Let f(t) = t3 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 x3 x (2x)3 x (nx)3 Rn = · 3+ · + ··· + · n n n n3 n n3 x4 ( ) = 4 13 + 23 + 33 + · · · + n3 n x4 [ ]2 = 4 1 n(n + 1) . n 2 0 . x . x4 n2 (n + 1)2 x4 = → 4n4 4 as n → ∞. . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 11 / 34
17. 17. An area function, continued So x4 g(x) = . 4 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 12 / 34
18. 18. An area function, continued So x4 g(x) = . 4 This means that g′ (x) = x3 . . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 12 / 34
19. 19. 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? . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 13 / 34
20. 20. Envisioning the area function Example Suppose f(t) is the function graphed below: .. y . . . . . . . . . x . 2 . 4 . 6 .. 8 1f . . 0. ∫ x Let g(x) = f(t) dt. What can you say about g? 0 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 14 / 34
21. 21. Envisioning the area function Example Suppose f(t) is the function graphed below: .. y . g . . . . . . . . . x . 2 . 4 . 6 .. 8 1f . . 0. ∫ x Let g(x) = f(t) dt. What can you say about g? 0 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 14 / 34
22. 22. Envisioning the area function Example Suppose f(t) is the function graphed below: .. y . g . . . . . . . . . x . 2 . 4 . 6 .. 8 1f . . 0. ∫ x Let g(x) = f(t) dt. What can you say about g? 0 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 14 / 34
23. 23. Envisioning the area function Example Suppose f(t) is the function graphed below: .. y . g . . . . . . . . . x . 2 . 4 . 6 .. 8 1f . . 0. ∫ x Let g(x) = f(t) dt. What can you say about g? 0 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 14 / 34
24. 24. Envisioning the area function Example Suppose f(t) is the function graphed below: .. y . g . . . . . . . . . x . 2 . 4 . 6 .. 8 1f . . 0. ∫ x Let g(x) = f(t) dt. What can you say about g? 0 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 14 / 34
25. 25. Envisioning the area function Example Suppose f(t) is the function graphed below: .. y . g . . . . . . . . . x . 2 . 4 . 6 .. 8 1f . . 0. ∫ x Let g(x) = f(t) dt. What can you say about g? 0 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 14 / 34
26. 26. Envisioning the area function Example Suppose f(t) is the function graphed below: .. y . g . . . . . . . . . x . 2 . 4 . 6 .. 8 1f . . 0. ∫ x Let g(x) = f(t) dt. What can you say about g? 0 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 14 / 34
27. 27. Envisioning the area function Example Suppose f(t) is the function graphed below: .. y . g . . . . . . . . . x . 2 . 4 . 6 .. 8 1f . . 0. ∫ x Let g(x) = f(t) dt. What can you say about g? 0 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 14 / 34
28. 28. Envisioning the area function Example Suppose f(t) is the function graphed below: .. y . g . . . . . . . . . x . 2 . 4 . 6 .. 8 1f . . 0. ∫ x Let g(x) = f(t) dt. What can you say about g? 0 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 14 / 34
29. 29. Envisioning the area function Example Suppose f(t) is the function graphed below: .. y . g . . . . . . . . . x . 2 . 4 . 6 .. 8 1f . . 0. ∫ x Let g(x) = f(t) dt. What can you say about g? 0 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 14 / 34
30. 30. Envisioning the area function Example Suppose f(t) is the function graphed below: .. y . g . . . . . . . . . x . 2 . 4 . 6 .. 8 1f . . 0. ∫ x Let g(x) = f(t) dt. What can you say about g? 0 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 14 / 34
31. 31. Envisioning the area function Example Suppose f(t) is the function graphed below: .. y . g . . . . . . . . . x . 2 . 4 . 6 .. 8 1f . . 0. ∫ x Let g(x) = f(t) dt. What can you say about g? 0 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 14 / 34
32. 32. features of g from f .. y Interval sign monotonicity monotonicity concavity . of f of g of f of g g . . . . . . . . [0, 2] + ↗ ↗ ⌣ . ..fx . . .. . . 0 2 4 6 81 [2, 4.5] + ↗ ↘ ⌢ [4.5, 6] − ↘ ↘ ⌢ [6, 8] − ↘ ↗ ⌣ [8, 10] − ↘ → none . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 15 / 34
33. 33. features of g from f .. y Interval sign monotonicity monotonicity concavity . of f of g of f of g g . . . . . . . . [0, 2] + ↗ ↗ ⌣ . ..fx . . .. . . 0 2 4 6 81 [2, 4.5] + ↗ ↘ ⌢ [4.5, 6] − ↘ ↘ ⌢ [6, 8] − ↘ ↗ ⌣ [8, 10] − ↘ → none We see that g is behaving a lot like an antiderivative of f. . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 15 / 34
34. 34. 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). . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 16 / 34
35. 35. Proving the Fundamental Theorem Proof. Let h > 0 be given so that x + h < b. We have g(x + h) − g(x) = h . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 17 / 34
36. 36. Proving the Fundamental Theorem Proof. Let h > 0 be given so that x + h < b. We have ∫ g(x + h) − g(x) 1 x+h = f(t) dt. h h x . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 17 / 34
37. 37. Proving the Fundamental Theorem Proof. Let h > 0 be given so that x + h < b. We have ∫ g(x + h) − g(x) 1 x+h = 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 So g(x + h) − g(x) mh ≤ ≤ Mh . h . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 17 / 34
38. 38. Proving the Fundamental Theorem Proof. Let h > 0 be given so that x + h < b. We have ∫ g(x + h) − g(x) 1 x+h = 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 So g(x + h) − g(x) mh ≤ ≤ Mh . h As h → 0, both mh and Mh tend to f(x). . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 17 / 34
39. 39. Proving the Fundamental Theorem Proof. Let h > 0 be given so that x + h < b. We have ∫ g(x + h) − g(x) 1 x+h = 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). . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 17 / 34
40. 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 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 18 / 34
41. 41. Meet the Mathematician: Isaac Barrow English, 1630-1677 Professor of Greek, theology, and mathematics at Cambridge Had a famous student . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 19 / 34
42. 42. Meet the Mathematician: Isaac Newton English, 1643–1727 Professor at Cambridge (England) Philosophiae Naturalis Principia Mathematica published 1687 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 20 / 34
43. 43. Meet the Mathematician: Gottfried Leibniz German, 1646–1716 Eminent philosopher as well as mathematician Contemporarily disgraced by the calculus priority dispute . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 21 / 34
44. 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. . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 22 / 34
45. 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 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 23 / 34
46. 46. Differentiation of area functions Example ∫ 3x Let h(x) = t3 dt. What is h′ (x)? 0 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 24 / 34
47. 47. Differentiation of area functions Example ∫ 3x Let h(x) = t3 dt. What is h′ (x)? 0 Solution (Using 2FTC) 3x t4 1 h(x) = = (3x)4 = 1 4 · 81x4 , so h′ (x) = 81x3 . 4 4 0 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 24 / 34
48. 48. Differentiation of area functions Example ∫ 3x Let h(x) = t3 dt. What is h′ (x)? 0 Solution (Using 2FTC) 3x t4 1 h(x) = = (3x)4 = 1 4 · 81x4 , so h′ (x) = 81x3 . 4 4 0 Solution (Using 1FTC) ∫ u We can think of h as the composition g k, where g(u) = ◦ t3 dt and 0 k(x) = 3x. . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 24 / 34
49. 49. Differentiation of area functions Example ∫ 3x Let h(x) = t3 dt. What is h′ (x)? 0 Solution (Using 2FTC) 3x t4 1 h(x) = = (3x)4 = 1 4 · 81x4 , so h′ (x) = 81x3 . 4 4 0 Solution (Using 1FTC) ∫ u We can think of h as the composition g k, where g(u) = ◦ t3 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 = 81x3 . . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 24 / 34
50. 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) . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 25 / 34
51. 51. Example ∫ sin2 x Let h(x) = (17t2 + 4t − 4) dt. What is h′ (x)? 0 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 26 / 34
52. 52. Example ∫ sin2 x Let h(x) = (17t2 + 4t − 4) dt. What is h′ (x)? 0 Solution We have ∫ sin2 x d (17t2 + 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 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 26 / 34
53. 53. Example ∫ sin2 x Let h(x) = (17t2 + 4t − 4) dt. What is h′ (x)? 3 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 27 / 34
54. 54. Example ∫ sin2 x Let h(x) = (17t2 + 4t − 4) dt. What is h′ (x)? 3 Solution We have ∫ sin2 x d (17t2 + 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 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 27 / 34
55. 55. Question Why is ∫ sin2 x ∫ sin2 x d d (17t2 + 4t − 4) dt = (17t2 + 4t − 4) dt? dx 0 dx 3 Or, why doesn’t the lower limit appear in the derivative? . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 28 / 34
56. 56. Question Why is ∫ sin2 x ∫ sin2 x d d (17t2 + 4t − 4) dt = (17t2 + 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 (17t + 4t − 4) dt = 2 (17t + 4t − 4) dt + (17t2 + 4t − 4) dt 0 0 3 So the two functions differ by a constant. . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 28 / 34
57. 57. Example ∫ ex Find the derivative of F(x) = sin4 t dt. x3 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 29 / 34
58. 58. Example ∫ ex Find the derivative of F(x) = sin4 t dt. x3 Solution ∫ ex d sin4 t dt = sin4 (ex ) · ex − sin4 (x3 ) · 3x2 dx x3 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 29 / 34
59. 59. Example ∫ ex Find the derivative of F(x) = sin4 t dt. x3 Solution ∫ ex d sin4 t dt = sin4 (ex ) · ex − sin4 (x3 ) · 3x2 dx x3 Notice here it’s much easier than finding an antiderivative for sin4 . . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 29 / 34
60. 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. . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 30 / 34
61. 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. . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 30 / 34
62. 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. . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 30 / 34
63. 63. Erf Here’s a function with a funny name but an important role: ∫ x 2 e−t dt. 2 erf(x) = √ π 0 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 31 / 34
64. 64. Erf Here’s a function with a funny name but an important role: ∫ x 2 e−t dt. 2 erf(x) = √ π 0 It turns out erf is the shape of the bell curve. . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 31 / 34
65. 65. Erf Here’s a function with a funny name but an important role: ∫ x 2 e−t dt. 2 erf(x) = √ 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) = . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 31 / 34
66. 66. Erf Here’s a function with a funny name but an important role: ∫ x 2 e−t dt. 2 erf(x) = √ π 0 It turns out erf is the shape of the bell curve. We can’t find erf(x), 2 explicitly, but we do know its derivative: erf′ (x) = √ e−x . 2 π . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 31 / 34
67. 67. Erf Here’s a function with a funny name but an important role: ∫ x 2 e−t dt. 2 erf(x) = √ π 0 It turns out erf is the shape of the bell curve. We can’t find erf(x), 2 explicitly, but we do know its derivative: erf′ (x) = √ e−x . 2 π Example d Find erf(x2 ). dx . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 31 / 34
68. 68. Erf Here’s a function with a funny name but an important role: ∫ x 2 e−t dt. 2 erf(x) = √ π 0 It turns out erf is the shape of the bell curve. We can’t find erf(x), 2 explicitly, but we do know its derivative: erf′ (x) = √ e−x . 2 π Example d Find erf(x2 ). dx Solution By the chain rule we have d d 2 4 erf(x2 ) = erf′ (x2 ) x2 = √ e−(x ) 2x = √ xe−x . 2 2 4 dx dx π π . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 31 / 34
69. 69. Other functions defined by integrals The future value of an asset: ∫ ∞ FV(t) = π(τ )e−rτ dτ t where π(τ ) is the profitability at time τ 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. . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 32 / 34
70. 70. Surplus by picture c . onsumer surplus p . rice (p) s . upply .∗ . p . . quilibrium e d . emand f(q) . . .∗ q q . uantity (q) . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 33 / 34
71. 71. 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 . . . . . . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem April 22, 2010 34 / 34