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- 1. Section 5.4 The Fundamental Theorem of Calculus V63.0121, Calculus I April 23, 2009 Announcements Quiz 6 next week on §§5.1–5.2 . . . . . .
- 2. 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 Worksheet Summary . . . . . .
- 3. 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 ∆x→0 a i=1 . . . . . .
- 4. 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 . . . . . .
- 5. 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: . . . . . .
- 6. 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 . . . . . .
- 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: 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 . . . . . .
- 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 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 . . . . . .
- 9. My ﬁrst table of integrals ∫ ∫ ∫ [f(x) + g(x)] dx = f(x) dx + g(x) dx ∫ ∫ ∫ xn+1 xn dx = cf(x) dx = c f(x) dx + C (n ̸= −1) n+1 ∫ ∫ 1 ex dx = ex + C dx = ln |x| + C x ∫ ∫ ax ax dx = +C sin x dx = − cos x + C ln a ∫ ∫ csc2 x dx = − cot x + C cos x dx = sin x + C ∫ ∫ sec2 x dx = tan x + C csc x cot x dx = − csc x + C ∫ ∫ 1 √ dx = arcsin x + C sec x tan x dx = sec x + C 1 − x2 ∫ 1 dx = arctan x + C 1 + x2 . . . . . .
- 10. 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 Worksheet Summary . . . . . .
- 11. An area function ∫ x 3 Let f(t) = t and deﬁne g(x) = f(t) dt. Can we evaluate the 0 integral in g(x)? . x . 0 . . . . . . .
- 12. An area function ∫ x 3 Let f(t) = t and deﬁne g(x) = f(t) dt. Can we evaluate the 0 integral in g(x)? Dividing the interval [0, x] into n pieces x ix gives ∆x = and xi = 0 + i∆x = . n n So x x3 x (2x)3 x (nx)3 · 3 + · 3 + ··· + · 3 Rn = nn n n n n 4( ) x = 4 1 3 + 2 3 + 3 3 + · · · + n3 n x4 [ 1 ]2 = 4 2 n(n + 1) . n x . 0 . x4 n2 (n + 1)2 x4 → = 4n4 4 as n → ∞. . . . . . .
- 13. An area function, continued So x4 g(x) = . 4 . . . . . .
- 14. An area function, continued So x4 g(x) = . 4 This means that g′ (x) = x3 . . . . . . .
- 15. 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 When is g increasing? . . . . . .
- 16. 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 When is g increasing? When is g decreasing? . . . . . .
- 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 When is g increasing? When is g decreasing? Over a small interval, what’s the average rate of change of g? . . . . . .
- 18. 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). . . . . . .
- 19. Proof. Let h > 0 be given so that x + h < b. We have g(x + h) − g(x) = h . . . . . .
- 20. 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 . . . . . .
- 21. 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 mh the minimum value of f on [x, x + h]. From §5.2 we have ∫ x+h f(t) dt x . . . . . .
- 22. 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 mh the minimum value of f on [x, x + h]. From §5.2 we have ∫ x+h f(t) dt ≤ Mh · h x . . . . . .
- 23. 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 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 . . . . . .
- 24. 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 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 . . . . . .
- 25. 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 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). Zappa-dappa. . . . . . .
- 26. 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 . . . . . .
- 27. Meet the Mathematician: Isaac Barrow English, 1630-1677 Professor of Greek, theology, and mathematics at Cambridge Had a famous student . . . . . .
- 28. Meet the Mathematician: Isaac Newton English, 1643–1727 Professor at Cambridge (England) Philosophiae Naturalis Principia Mathematica published 1687 . . . . . .
- 29. Meet the Mathematician: Gottfried Leibniz German, 1646–1716 Eminent philosopher as well as mathematician Contemporarily disgraced by the calculus priority dispute . . . . . .
- 30. Differentiation and Integration as reverse processes Putting together 1FTC and 2FTC, we get a beautiful relationship between the two fundamental concepts in calculus. ∫ x d f(t) dt = f(x) dx a . . . . . .
- 31. Differentiation and Integration as reverse processes Putting together 1FTC and 2FTC, we get a beautiful relationship between the two fundamental concepts in calculus. ∫ x d f(t) dt = f(x) dx a ∫ b F′ (x) dx = F(b) − F(a). a . . . . . .
- 32. 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 Worksheet Summary . . . . . .
- 33. Differentiation of area functions Example ∫ x t3 dt. We know g′ (x) = x3 . What if instead we had Let g(x) = 0 ∫ 3x t3 dt. h(x) = 0 What is h′ (x)? . . . . . .
- 34. Differentiation of area functions Example ∫ x t3 dt. We know g′ (x) = x3 . What if instead we had Let g(x) = 0 ∫ 3x t3 dt. h(x) = 0 What is h′ (x)? Solution ∫ u t3 dt We can think of h as the composition g k, where g(u) = ◦ 0 and k(x) = 3x. Then h′ (x) = g′ (k(x))k′ (x) = 3(k(x))3 = 3(3x)3 = 81x3 . . . . . . .
- 35. Example ∫ sin2 x (17t2 + 4t − 4) dt. What is h′ (x)? Let h(x) = 0 . . . . . .
- 36. Example ∫ sin2 x (17t2 + 4t − 4) dt. What is h′ (x)? Let 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 . . . . . .
- 37. Erf Here’s a function with a funny name but an important role: ∫x 2 2 e−t dt. √ erf(x) = π0 . . . . . .
- 38. Erf Here’s a function with a funny name but an important role: ∫x 2 2 e−t dt. √ erf(x) = π0 It turns out erf is the shape of the bell curve. . . . . . .
- 39. Erf Here’s a function with a funny name but an important role: ∫x 2 2 e−t dt. √ erf(x) = π0 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) = . . . . . .
- 40. Erf Here’s a function with a funny name but an important role: ∫x 2 2 e−t dt. √ erf(x) = π0 It turns out erf is the shape of the bell curve. We can’t ﬁnd erf(x), explicitly, but we do know its derivative. 2 2 erf′ (x) = √ e−x . π . . . . . .
- 41. Erf Here’s a function with a funny name but an important role: ∫x 2 2 e−t dt. √ erf(x) = π0 It turns out erf is the shape of the bell curve. We can’t ﬁnd erf(x), explicitly, but we do know its derivative. 2 2 erf′ (x) = √ e−x . π Example d erf(x2 ). Find dx . . . . . .
- 42. Erf Here’s a function with a funny name but an important role: ∫x 2 2 e−t dt. √ erf(x) = π0 It turns out erf is the shape of the bell curve. We can’t ﬁnd erf(x), explicitly, but we do know its derivative. 2 2 erf′ (x) = √ e−x . π Example d erf(x2 ). Find dx Solution By the chain rule we have d d 2 4 22 4 erf(x2 ) = erf′ (x2 ) x2 = √ e−(x ) 2x = √ xe−x . dx dx π π . . . . . .
- 43. Other functions deﬁned by integrals The future value of an asset: ∫∞ π(τ )e−rτ dτ FV(t) = 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. . . . . . .
- 44. 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 Worksheet Summary . . . . . .
- 45. Worksheet . . Image: Erick Cifuentes . . . . . .
- 46. 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 Worksheet Summary . . . . . .
- 47. Summary FTC links integration and differentiation When differentiating integral functions, do not forget the chain rule Facts about the integral function can be gleaned from the integrand . . . . . .

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