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- 1. Section 5.4 The Fundamental Theorem of Calculus V63.0121.041, Calculus I New York University December 8, 2010 Announcements Today: Section 5.4 Monday, December 13: Section 5.5 ”Monday,” December 15: Review and Movie Day! Monday, December 20, 12:00–1:50pm: Final Exam (location still TBD). . . . . . .
- 2. Announcements Today: Section 5.4 Monday, December 13: Section 5.5 ”Monday,” December 15: Review and Movie Day! Monday, December 20, 12:00–1:50pm: Final Exam (location still TBD) . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 2 / 32
- 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. . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 3 / 32
- 4. OutlineRecall: The Evaluation Theorem a/k/a 2nd FTCThe First Fundamental Theorem of Calculus Area as a Function Statement and proof of 1FTC BiographiesDifferentiation of functions defined by integrals “Contrived” examples Erf Other applications . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 4 / 32
- 5. The definite integral as a limitDefinitionIf f is a function defined on [a, b], the definite integral of f from a to bis the number ∫ b ∑n f(x) dx = lim f(ci ) ∆x a ∆x→0 i=1 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 5 / 32
- 6. Big time TheoremTheorem (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 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 6 / 32
- 7. The Integral as Total ChangeAnother way to state this theorem is: ∫ b F′ (x) dx = F(b) − F(a), aor the integral of a derivative along an interval is the total changebetween the sides of that interval. This has many ramifications: . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 7 / 32
- 8. The Integral as Total ChangeAnother way to state this theorem is: ∫ b F′ (x) dx = F(b) − F(a), aor the integral of a derivative along an interval is the total changebetween the sides of that interval. This has many ramifications:TheoremIf v(t) represents the velocity of a particle moving rectilinearly, then ∫ t1 v(t) dt = s(t1 ) − s(t0 ). t0 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 7 / 32
- 9. The Integral as Total ChangeAnother way to state this theorem is: ∫ b F′ (x) dx = F(b) − F(a), aor the integral of a derivative along an interval is the total changebetween the sides of that interval. This has many ramifications:TheoremIf MC(x) represents the marginal cost of making x units of a product,then ∫ x C(x) = C(0) + MC(q) dq. 0 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 7 / 32
- 10. The Integral as Total ChangeAnother way to state this theorem is: ∫ b F′ (x) dx = F(b) − F(a), aor the integral of a derivative along an interval is the total changebetween the sides of that interval. This has many ramifications:TheoremIf ρ(x) represents the density of a thin rod at a distance of x from itsend, then the mass of the rod up to x is ∫ x m(x) = ρ(s) ds. 0 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 7 / 32
- 11. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 8 / 32
- 12. OutlineRecall: The Evaluation Theorem a/k/a 2nd FTCThe First Fundamental Theorem of Calculus Area as a Function Statement and proof of 1FTC BiographiesDifferentiation of functions defined by integrals “Contrived” examples Erf Other applications . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 9 / 32
- 13. Area as a FunctionExample ∫ x 3Let f(t) = t and define g(x) = f(t) dt. Find g(x) and g′ (x). 0 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 10 / 32
- 14. Area as a FunctionExample ∫ x 3Let f(t) = t and define g(x) = f(t) dt. Find g(x) and g′ (x). 0Solution Dividing the interval [0, x] into n pieces x ix gives ∆t = and ti = 0 + i∆t = . n n . 0 x . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 10 / 32
- 15. Area as a FunctionExample ∫ x 3Let f(t) = t and define g(x) = f(t) dt. Find g(x) and g′ (x). 0Solution Dividing the interval [0, x] into n pieces x ix gives ∆t = and ti = 0 + i∆t = . So n n . 0 x . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 10 / 32
- 16. Area as a FunctionExample ∫ x 3Let f(t) = t and define g(x) = f(t) dt. Find g(x) and g′ (x). 0Solution 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 . 0 x . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 10 / 32
- 17. Area as a FunctionExample ∫ x 3Let f(t) = t and define g(x) = f(t) dt. Find g(x) and g′ (x). 0Solution 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 . 0 x . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 10 / 32
- 18. Area as a FunctionExample ∫ x 3Let f(t) = t and define g(x) = f(t) dt. Find g(x) and g′ (x). 0Solution 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 . 0 x = 4 1 n(n + 1) n 2 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 10 / 32
- 19. Area as a FunctionExample ∫ x 3Let f(t) = t and define g(x) = f(t) dt. Find g(x) and g′ (x). 0Solution Dividing the interval [0, x] into n pieces x ix gives ∆t = and ti = 0 + i∆t = . So n n x4 n2 (n + 1)2 Rn = 4n4 . 0 x . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 10 / 32
- 20. Area as a FunctionExample ∫ x 3Let f(t) = t and define g(x) = f(t) dt. Find g(x) and g′ (x). 0Solution Dividing the interval [0, x] into n pieces x ix gives ∆t = and ti = 0 + i∆t = . So n n x4 n2 (n + 1)2 Rn = 4n4 x4 So g(x) = lim Rn = . x→∞ 4 0 x . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 10 / 32
- 21. Area as a FunctionExample ∫ x 3Let f(t) = t and define g(x) = f(t) dt. Find g(x) and g′ (x). 0Solution Dividing the interval [0, x] into n pieces x ix gives ∆t = and ti = 0 + i∆t = . So n n x4 n2 (n + 1)2 Rn = 4n4 x4 So g(x) = lim Rn = and g′ (x) = x3 . . x→∞ 4 0 x . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 10 / 32
- 22. The area function in generalLet f be a function which is integrable (i.e., continuous or with finitelymany 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? . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 11 / 32
- 23. Envisioning the area functionExampleSuppose f(t) is the function graphed below: y . x 2 4 6 8 10f ∫ xLet g(x) = f(t) dt. What can you say about g? 0 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 12 / 32
- 24. Envisioning the area functionExampleSuppose f(t) is the function graphed below: y g . x 2 4 6 8 10f ∫ xLet g(x) = f(t) dt. What can you say about g? 0 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 12 / 32
- 25. Envisioning the area functionExampleSuppose f(t) is the function graphed below: y g . x 2 4 6 8 10f ∫ xLet g(x) = f(t) dt. What can you say about g? 0 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 12 / 32
- 26. Envisioning the area functionExampleSuppose f(t) is the function graphed below: y g . x 2 4 6 8 10f ∫ xLet g(x) = f(t) dt. What can you say about g? 0 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 12 / 32
- 27. Envisioning the area functionExampleSuppose f(t) is the function graphed below: y g . x 2 4 6 8 10f ∫ xLet g(x) = f(t) dt. What can you say about g? 0 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 12 / 32
- 28. Envisioning the area functionExampleSuppose f(t) is the function graphed below: y g . x 2 4 6 8 10f ∫ xLet g(x) = f(t) dt. What can you say about g? 0 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 12 / 32
- 29. Envisioning the area functionExampleSuppose f(t) is the function graphed below: y g . x 2 4 6 8 10f ∫ xLet g(x) = f(t) dt. What can you say about g? 0 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 12 / 32
- 30. Envisioning the area functionExampleSuppose f(t) is the function graphed below: y g . x 2 4 6 8 10f ∫ xLet g(x) = f(t) dt. What can you say about g? 0 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 12 / 32
- 31. Envisioning the area functionExampleSuppose f(t) is the function graphed below: y g . x 2 4 6 8 10f ∫ xLet g(x) = f(t) dt. What can you say about g? 0 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 12 / 32
- 32. Envisioning the area functionExampleSuppose f(t) is the function graphed below: y g . x 2 4 6 8 10f ∫ xLet g(x) = f(t) dt. What can you say about g? 0 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 12 / 32
- 33. Envisioning the area functionExampleSuppose f(t) is the function graphed below: y g . x 2 4 6 8 10f ∫ xLet g(x) = f(t) dt. What can you say about g? 0 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 12 / 32
- 34. Envisioning the area functionExampleSuppose f(t) is the function graphed below: y g . x 2 4 6 8 10f ∫ xLet g(x) = f(t) dt. What can you say about g? 0 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 12 / 32
- 35. features of g from f y Interval sign monotonicity monotonicity concavity of f of g of f of g g . [0, 2] + ↗ ↗ ⌣ fx 2 4 6 8 10 [2, 4.5] + ↗ ↘ ⌢ [4.5, 6] − ↘ ↘ ⌢ [6, 8] − ↘ ↗ ⌣ [8, 10] − ↘ → none . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 13 / 32
- 36. features of g from f y Interval sign monotonicity monotonicity concavity of f of g of f of g g . [0, 2] + ↗ ↗ ⌣ fx 2 4 6 8 10 [2, 4.5] + ↗ ↘ ⌢ [4.5, 6] − ↘ ↘ ⌢ [6, 8] − ↘ ↗ ⌣ [8, 10] − ↘ → noneWe see that g is behaving a lot like an antiderivative of f. . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 13 / 32
- 37. Another Big Time TheoremTheorem (The First Fundamental Theorem of Calculus)Let f be an integrable function on [a, b] and define ∫ x g(x) = f(t) dt. aIf f is continuous at x in (a, b), then g is differentiable at x and g′ (x) = f(x). . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 14 / 32
- 38. Proving the Fundamental TheoremProof.Let h > 0 be given so that x + h < b. We have g(x + h) − g(x) = h . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 15 / 32
- 39. Proving the Fundamental TheoremProof.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 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 15 / 32
- 40. Proving the Fundamental TheoremProof.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 xLet Mh be the maximum value of f on [x, x + h], and let mh the minimumvalue of f on [x, x + h]. From §5.2 we have ∫ x+h f(t) dt x . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 15 / 32
- 41. Proving the Fundamental TheoremProof.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 xLet Mh be the maximum value of f on [x, x + h], and let mh the minimumvalue of f on [x, x + h]. From §5.2 we have ∫ x+h f(t) dt ≤ Mh · h x . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 15 / 32
- 42. Proving the Fundamental TheoremProof.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 xLet Mh be the maximum value of f on [x, x + h], and let mh the minimumvalue of f on [x, x + h]. From §5.2 we have ∫ x+h mh · h ≤ f(t) dt ≤ Mh · h x . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 15 / 32
- 43. Proving the Fundamental TheoremProof.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 xLet Mh be the maximum value of f on [x, x + h], and let mh the minimumvalue of f on [x, x + h]. From §5.2 we have ∫ x+h mh · h ≤ f(t) dt ≤ Mh · h xSo g(x + h) − g(x) mh ≤ ≤ Mh . h . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 15 / 32
- 44. Proving the Fundamental TheoremProof.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 xLet Mh be the maximum value of f on [x, x + h], and let mh the minimumvalue of f on [x, x + h]. From §5.2 we have ∫ x+h mh · h ≤ f(t) dt ≤ Mh · h xSo g(x + h) − g(x) mh ≤ ≤ Mh . hAs h → 0, both mh and Mh tend to f(x). . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 15 / 32
- 45. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 16 / 32
- 46. Meet the Mathematician: Isaac Barrow English, 1630-1677 Professor of Greek, theology, and mathematics at Cambridge Had a famous student . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 17 / 32
- 47. Meet the Mathematician: Isaac Newton English, 1643–1727 Professor at Cambridge (England) Philosophiae Naturalis Principia Mathematica published 1687 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 18 / 32
- 48. Meet the Mathematician: Gottfried Leibniz German, 1646–1716 Eminent philosopher as well as mathematician Contemporarily disgraced by the calculus priority dispute . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 19 / 32
- 49. Differentiation and Integration as reverse processesPutting together 1FTC and 2FTC, we get a beautiful relationshipbetween 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. . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 20 / 32
- 50. OutlineRecall: The Evaluation Theorem a/k/a 2nd FTCThe First Fundamental Theorem of Calculus Area as a Function Statement and proof of 1FTC BiographiesDifferentiation of functions defined by integrals “Contrived” examples Erf Other applications . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 21 / 32
- 51. Differentiation of area functionsExample ∫ 3xLet h(x) = t3 dt. What is h′ (x)? 0 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 22 / 32
- 52. Differentiation of area functionsExample ∫ 3xLet h(x) = t3 dt. What is h′ (x)? 0Solution (Using 2FTC) 3x t4 1h(x) = = (3x)4 = 1 4 · 81x4 , so h′ (x) = 81x3 . 4 4 0 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 22 / 32
- 53. Differentiation of area functionsExample ∫ 3xLet h(x) = t3 dt. What is h′ (x)? 0Solution (Using 2FTC) 3x t4 1h(x) = = (3x)4 = 1 4 · 81x4 , so h′ (x) = 81x3 . 4 4 0Solution (Using 1FTC) ∫ uWe can think of h as the composition g k, where g(u) = ◦ t3 dt and 0k(x) = 3x. . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 22 / 32
- 54. Differentiation of area functionsExample ∫ 3xLet h(x) = t3 dt. What is h′ (x)? 0Solution (Using 2FTC) 3x t4 1h(x) = = (3x)4 = 1 4 · 81x4 , so h′ (x) = 81x3 . 4 4 0Solution (Using 1FTC) ∫ uWe can think of h as the composition g k, where g(u) = ◦ t3 dt and 0k(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 . . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 22 / 32
- 55. 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) . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 23 / 32
- 56. Another ExampleExample ∫ sin2 xLet h(x) = (17t2 + 4t − 4) dt. What is h′ (x)? 0 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 24 / 32
- 57. Another ExampleExample ∫ sin2 xLet h(x) = (17t2 + 4t − 4) dt. What is h′ (x)? 0SolutionWe have ∫ sin2 x d (17t2 + 4t − 4) dt dx 0 ( d ) 2 2 = 17(sin x) + 4(sin x) − 4 · 2 sin2 x ( ) dx = 17 sin4 x + 4 sin2 x − 4 · 2 sin x cos x . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 24 / 32
- 58. A Similar ExampleExample ∫ sin2 xLet h(x) = (17t2 + 4t − 4) dt. What is h′ (x)? 3 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 25 / 32
- 59. A Similar ExampleExample ∫ sin2 xLet h(x) = (17t2 + 4t − 4) dt. What is h′ (x)? 3SolutionWe have ∫ sin2 x d (17t2 + 4t − 4) dt dx 0 ( d ) 2 2 = 17(sin x) + 4(sin x) − 4 · 2 sin2 x ( ) dx = 17 sin4 x + 4 sin2 x − 4 · 2 sin x cos x . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 25 / 32
- 60. CompareQuestionWhy is ∫ sin2 x ∫ sin2 x d d (17t + 4t − 4) dt = 2 (17t2 + 4t − 4) dt? dx 0 dx 3Or, why doesn’t the lower limit appear in the derivative? . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 26 / 32
- 61. CompareQuestionWhy is ∫ sin2 x ∫ sin2 x d d (17t + 4t − 4) dt = 2 (17t2 + 4t − 4) dt? dx 0 dx 3Or, why doesn’t the lower limit appear in the derivative?AnswerBecause∫ sin2 x ∫ 3 ∫ sin2 x (17t2 + 4t − 4) dt = (17t2 + 4t − 4) dt + (17t2 + 4t − 4) dt 0 0 3So the two functions differ by a constant. . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 26 / 32
- 62. The Full NastyExample ∫ exFind the derivative of F(x) = sin4 t dt. x3 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 27 / 32
- 63. The Full NastyExample ∫ exFind the derivative of F(x) = sin4 t dt. x3Solution ∫ ex d sin4 t dt = sin4 (ex ) · ex − sin4 (x3 ) · 3x2 dx x3 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 27 / 32
- 64. The Full NastyExample ∫ exFind the derivative of F(x) = sin4 t dt. x3Solution ∫ ex d sin4 t dt = sin4 (ex ) · ex − sin4 (x3 ) · 3x2 dx x3Notice here it’s much easier than finding an antiderivative for sin4 . . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 27 / 32
- 65. Why use 1FTC?QuestionWhy would we use 1FTC to find the derivative of an integral? It seemslike confusion for its own sake. . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 28 / 32
- 66. Why use 1FTC?QuestionWhy would we use 1FTC to find the derivative of an integral? It seemslike confusion for its own sake.Answer Some functions are difficult or impossible to integrate in elementary terms. . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 28 / 32
- 67. Why use 1FTC?QuestionWhy would we use 1FTC to find the derivative of an integral? It seemslike 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. . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 28 / 32
- 68. ErfHere’s a function with a funny name but an important role: ∫ x 2 e−t dt. 2 erf(x) = √ π 0 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 29 / 32
- 69. ErfHere’s a function with a funny name but an important role: ∫ x 2 e−t dt. 2 erf(x) = √ π 0It turns out erf is the shape of the bell curve. . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 29 / 32
- 70. ErfHere’s a function with a funny name but an important role: ∫ x 2 e−t dt. 2 erf(x) = √ π 0It 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) = . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 29 / 32
- 71. ErfHere’s a function with a funny name but an important role: ∫ x 2 e−t dt. 2 erf(x) = √ π 0It turns out erf is the shape of the bell curve. We can’t find erf(x), 2explicitly, but we do know its derivative: erf′ (x) = √ e−x . 2 π . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 29 / 32
- 72. ErfHere’s a function with a funny name but an important role: ∫ x 2 e−t dt. 2 erf(x) = √ π 0It turns out erf is the shape of the bell curve. We can’t find erf(x), 2explicitly, but we do know its derivative: erf′ (x) = √ e−x . 2 πExample dFind erf(x2 ). dx . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 29 / 32
- 73. ErfHere’s a function with a funny name but an important role: ∫ x 2 e−t dt. 2 erf(x) = √ π 0It turns out erf is the shape of the bell curve. We can’t find erf(x), 2explicitly, but we do know its derivative: erf′ (x) = √ e−x . 2 πExample dFind erf(x2 ). dxSolutionBy 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 29 / 32
- 74. 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. . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 30 / 32
- 75. Surplus by picture price (p) . quantity (q) . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 31 / 32
- 76. Surplus by picture price (p) demand f(q) . quantity (q) . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 31 / 32
- 77. Surplus by picture price (p) supply demand f(q) . quantity (q) . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 31 / 32
- 78. Surplus by picture price (p) supply p∗ equilibrium demand f(q) . q∗ quantity (q) . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 31 / 32
- 79. Surplus by picture price (p) supply p∗ equilibrium market revenue demand f(q) . q∗ quantity (q) . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 31 / 32
- 80. Surplus by picture consumer surplus price (p) supply p∗ equilibrium market revenue demand f(q) . q∗ quantity (q) . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 31 / 32
- 81. Surplus by picture consumer surplus price (p) producer surplus supply p∗ equilibrium demand f(q) . q∗ quantity (q) . . . . . . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 31 / 32
- 82. 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 . . . . . .V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 32 / 32

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