Upcoming SlideShare
×

# Lesson 28: The Fundamental Theorem of Calculus

1,144 views

Published on

Published in: Technology, Education
0 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

• Be the first to like this

Views
Total views
1,144
On SlideShare
0
From Embeds
0
Number of Embeds
11
Actions
Shares
0
31
0
Likes
0
Embeds 0
No embeds

No notes for slide

### Lesson 28: The Fundamental Theorem of Calculus

1. 1. Section 5.4 The Fundamental Theorem of Calculus V63.0121.027, Calculus I December 8, 2009 Announcements Final Exam: Friday 12/18, 2:00-3:50pm, Tisch UC50 . . . . . .
2. 2. Redemption 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 [new!] If your ﬁnal exam score beats your midterm score, we will re-weight it by 50% and make the midterm 15% . . . . . .
3. 3. 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 . . . . . .
4. 4. 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 . . . . . .
5. 5. 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 . . . . . .
6. 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: . . . . . .
7. 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 v(t) represents the velocity of a particle moving rectilinearly, then ∫ t1 v(t) dt = s(t1 ) − s(t0 ). t0 . . . . . .
8. 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 MC(x) represents the marginal cost of making x units of a product, then ∫ x C(x) = C(0) + MC(q) dq. 0 . . . . . .
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: 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 . . . . . .
10. 10. My ﬁrst table of integrals ∫ ∫ ∫ [f(x) + g(x)] dx = f(x) dx + g(x) dx ∫ ∫ ∫ x n +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 ∫ ∫ 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 . . . . . .
11. 11. 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 . . . . . .
12. 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)? . 0 . x . . . . . . .
13. 13. 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 ∆t = and ti = 0 + i∆t = . So n n x x3 x (2x)3 x (nx)3 Rn = · 3 + · 3 + ··· + · 3 n n n n n n x4 ( 3 ) = 4 1 + 2 3 + 3 3 + · · · + n3 n x4 [ 1 ]2 = 4 2 n(n + 1) . n 0 . x . x4 n2 (n + 1)2 x4 = → 4n4 4 as n → ∞. . . . . . .
14. 14. An area function, continued So x4 g(x) = . 4 . . . . . .
15. 15. An area function, continued So x4 g(x) = . 4 This means that g ′ (x ) = x 3 . . . . . . .
16. 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 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? . . . . . .
17. 17. Envisioning the area function Example Suppose f(t) is the function graphed below v . . . . . . . t .0 t .1 c . t .2 t t .3 . . ∫ x Let g(x) = f(t) dt. What can you say about g? t0 . . . . . .
18. 18. features of g from f Interval sign monotonicity monotonicity concavity of f of g of f of g [ t0 , t 1 ] + ↗ ↗ ⌣ [t1 , c] + ↗ ↘ ⌢ [c, t2 ] − ↘ ↘ ⌢ [ t2 , t 3 ] − ↘ ↗ ⌣ [t3 , ∞) − ↘ → none . . . . . .
19. 19. features of g from f Interval sign monotonicity monotonicity concavity of f of g of f of g [ t0 , t 1 ] + ↗ ↗ ⌣ [t1 , c] + ↗ ↘ ⌢ [c, t2 ] − ↘ ↘ ⌢ [ t2 , t 3 ] − ↘ ↗ ⌣ [t3 , ∞) − ↘ → none We see that g is behaving a lot like an antiderivative of f. . . . . . .
20. 20. 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). . . . . . .
21. 21. Proof. Let h > 0 be given so that x + h < b. We have g(x + h) − g(x) = h . . . . . .
22. 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 . . . . . .
23. 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 f(t) dt x . . . . . .
24. 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 f(t) dt ≤ Mh · h x . . . . . .
25. 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 . . . . . .
26. 26. 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 . . . . . .
27. 27. 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). . . . . . .
28. 28. 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 . . . . . .
29. 29. Meet the Mathematician: Isaac Barrow English, 1630-1677 Professor of Greek, theology, and mathematics at Cambridge Had a famous student . . . . . .
30. 30. Meet the Mathematician: Isaac Newton English, 1643–1727 Professor at Cambridge (England) Philosophiae Naturalis Principia Mathematica published 1687 . . . . . .
31. 31. Meet the Mathematician: Gottfried Leibniz German, 1646–1716 Eminent philosopher as well as mathematician Contemporarily disgraced by the calculus priority dispute . . . . . .
32. 32. 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 . . . . . .
33. 33. 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 . . . . . .
34. 34. 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 . . . . . .
35. 35. Differentiation of area functions Example ∫ 3x Let h(x) = t3 dt. What is h′ (x)? 0 . . . . . .
36. 36. 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 0 4 . . . . . .
37. 37. 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 0 4 Solution (Using 1FTC) ∫ u We can think of h as the composition g k, where g(u) = ◦ t3 dt 0 and k(x) = 3x. . . . . . .
38. 38. 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 0 4 Solution (Using 1FTC) ∫ u We can think of h as the composition g k, where g(u) = ◦ t3 dt 0 and k(x) = 3x. Then h′ (x) = g′ (k(x))k′ (x) = (k(x))3 · 3 = (3x)3 · 3 = 81x3 . . . . . . .
39. 39. 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) . . . . . .
40. 40. Example ∫ sin2 x Let h(x) = (17t2 + 4t − 4) dt. What is h′ (x)? 0 . . . . . .
41. 41. 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 . . . . . .
42. 42. Example ∫ ex Find the derivative of F(x) = sin4 t dt. x3 . . . . . .
43. 43. 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 . . . . . .
44. 44. 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 . . . . . . .
45. 45. Erf Here’s a function with a funny name but an important role: ∫ x 2 2 erf(x) = √ e−t dt. π 0 . . . . . .
46. 46. 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. . . . . . .
47. 47. 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) = . . . . . .
48. 48. 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. 2 2 erf′ (x) = √ e−x . π . . . . . .
49. 49. 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. 2 2 erf′ (x) = √ e−x . π Example d Find erf(x2 ). dx . . . . . .
50. 50. 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. 2 2 erf′ (x) = √ e−x . π Example d Find erf(x2 ). dx Solution By the chain rule we have d d 2 2 2 4 4 erf(x2 ) = erf′ (x2 ) x2 = √ e−(x ) 2x = √ xe−x . dx dx π π . . . . . .
51. 51. 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. . . . . . .
52. 52. Surplus by picture c . onsumer surplus p . rice (p) s . upply .∗ . p . . quilibrium e . emand f(q) d . . .∗ q q . uantity (q) . . . . . .