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# Lesson 26: The Fundamental Theorem of Calculus (handout)

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The fundamental theorem shows that differentiation and integration are inverse processes.

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### Lesson 26: The Fundamental Theorem of Calculus (handout)

1. 1. . V63.0121.001: Calculus I . Sec on 5.4: The Fundamental Theorem . May 2, 2011 . Notes Sec on 5.4 The Fundamental Theorem of Calculus V63.0121.001: Calculus I Professor Ma hew Leingang New York University May 2, 2011 . . Notes Announcements Today: 5.4 Wednesday 5/4: 5.5 Monday 5/9: Review and Movie Day! Thursday 5/12: Final Exam, 2:00–3:50pm . . Notes Objectives State and explain the Fundemental Theorems of Calculus Use the ﬁrst fundamental theorem of calculus to ﬁnd deriva ves of func ons deﬁned as integrals. Compute the average value of an integrable func on over a closed interval. . . . 1.
2. 2. . V63.0121.001: Calculus I . Sec on 5.4: The Fundamental Theorem . May 2, 2011 Notes Outline Recall: The Evalua on Theorem a/k/a 2nd FTC The First Fundamental Theorem of Calculus Area as a Func on Statement and proof of 1FTC Biographies Diﬀeren a on of func ons deﬁned by integrals “Contrived” examples Erf Other applica ons . . Notes The deﬁnite integral as a limit Deﬁni on If f is a func on 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 n→∞ i=1 b−a where ∆x = , and for each i, xi = a + i∆x, and ci is a point in n [xi−1 , xi ]. . . Notes The deﬁnite integral as a limit Theorem If f is con nuous on [a, b] or if f has only ﬁnitely many jump discon nui es, then f is integrable on [a, b]; that is, the deﬁnite ∫ b integral f(x) dx exists and is the same for any choice of ci . a . . . 2.
3. 3. . V63.0121.001: Calculus I . Sec on 5.4: The Fundamental Theorem . May 2, 2011 Notes Big time Theorem Theorem (The Second Fundamental Theorem of Calculus) Suppose f is integrable on [a, b] and f = F′ for another func on F, then ∫ b f(x) dx = F(b) − F(a). a . . Notes The Integral as Net Change Corollary If v(t) represents the velocity of a par cle moving rec linearly, then ∫ t1 v(t) dt = s(t1 ) − s(t0 ). t0 . . Notes The Integral as Net Change Corollary If MC(x) represents the marginal cost of making x units of a product, then ∫ x C(x) = C(0) + MC(q) dq. 0 . . . 3.
4. 4. . V63.0121.001: Calculus I . Sec on 5.4: The Fundamental Theorem . May 2, 2011 Notes The Integral as Net Change Corollary 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 . . Notes 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 ∫ ∫ 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 . . Notes Outline Recall: The Evalua on Theorem a/k/a 2nd FTC The First Fundamental Theorem of Calculus Area as a Func on Statement and proof of 1FTC Biographies Diﬀeren a on of func ons deﬁned by integrals “Contrived” examples Erf Other applica ons . . . 4.
5. 5. . V63.0121.001: Calculus I . Sec on 5.4: The Fundamental Theorem . May 2, 2011 Notes Area as a Function Example ∫ x Let f(t) = t3 and deﬁne g(x) = f(t) dt. Find g(x) and g′ (x). 0 Solu on x Dividing the interval [0, x] into n pieces gives ∆t = and n ix ti = 0 + i∆t = . n x x3 x (2x)3 x (nx)3 . Rn = · 3 + · 3 + ··· + · 3 x n n n n n n 0 x4 ( 3 ) x4 [ ]2 . = 4 1 + 2 + 3 + · · · + n = 4 1 n(n + 1) 3 3 3 n n 2 . Notes Area as a Function Example ∫ x Let f(t) = t3 and deﬁne g(x) = f(t) dt. Find g(x) and g′ (x). 0 Solu on x Dividing the interval [0, x] into n pieces gives ∆t = and n ix ti = 0 + i∆t = . n x4 n2 (n + 1)2 . Rn = x 4n4 0 . x4 So g(x) = lim Rn = and g′ (x) = x3 . x→∞ 4 . Notes The area function in general Let f be a func on which is integrable (i.e., con nuous or with ﬁnitely many jump discon nui es) 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. Ques on: What does f tell you about g? . . . 5.
6. 6. . V63.0121.001: Calculus I . Sec on 5.4: The Fundamental Theorem . May 2, 2011 Notes Envisioning the area function Example y Suppose f(t) is the func on graphed to the right. Let ∫ x g(x) = f(t) dt. What can 0 you say about g? . x 2 4 6 8 10f . . Notes 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 6 810 [2, 4.5] + ↗ ↘ ⌢ [4.5, 6] − ↘ ↘ ⌢ [6, 8] − ↘ ↗ ⌣ [8, 10] − ↘ → none . We see that g is behaving a lot like an an deriva ve of f. . Notes Another Big Time Theorem Theorem (The First Fundamental Theorem of Calculus) Let f be an integrable func on on [a, b] and deﬁne ∫ x g(x) = f(t) dt. a If f is con nuous at x in (a, b), then g is diﬀeren able at x and g′ (x) = f(x). . . . 6.
7. 7. . V63.0121.001: Calculus I . Sec on 5.4: The Fundamental Theorem . May 2, 2011 Notes 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 . g(x + h) − g(x) =⇒ mh ≤ ≤ Mh . . h As h → 0, both mh and Mh tend to f(x). Notes Proving the Fundamental Theorem Proof. From §5.2 we have ∫ x+h mh · h ≤ f(t) dt ≤ Mh · h x g(x + h) − g(x) =⇒ mh ≤ ≤ Mh . h As h → 0, both mh and Mh tend to f(x). . . Meet the Mathematician: James Notes Gregory Sco sh, 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 . . . 7.
8. 8. . V63.0121.001: Calculus I . Sec on 5.4: The Fundamental Theorem . May 2, 2011 Meet the Mathematician: Isaac Notes Barrow English, 1630-1677 Professor of Greek, theology, and mathema cs at Cambridge Had a famous student . . Meet the Mathematician: Isaac Notes Newton English, 1643–1727 Professor at Cambridge (England) Philosophiae Naturalis Principia Mathema ca published 1687 . . Meet the Mathematician: Gottfried Notes Leibniz German, 1646–1716 Eminent philosopher as well as mathema cian Contemporarily disgraced by the calculus priority dispute . . . 8.
9. 9. . V63.0121.001: Calculus I . Sec on 5.4: The Fundamental Theorem . May 2, 2011 Diﬀerentiation and Integration as Notes reverse processes Pu ng together 1FTC and 2FTC, we get a beau ful rela onship between the two fundamental concepts in calculus. Theorem (The Fundamental Theorem(s) of Calculus) I. If f is a con nuous func on, then ∫ d x f(t) dt = f(x) dx a So the deriva ve of the integral is the original func on. . . Diﬀerentiation and Integration as Notes reverse processes Pu ng together 1FTC and 2FTC, we get a beau ful rela onship between the two fundamental concepts in calculus. Theorem (The Fundamental Theorem(s) of Calculus) II. If f is a diﬀeren able func on, then ∫ b f′ (x) dx = f(b) − f(a). a So the integral of the deriva ve of is (an evalua on of) the original func on. . . Notes Outline Recall: The Evalua on Theorem a/k/a 2nd FTC The First Fundamental Theorem of Calculus Area as a Func on Statement and proof of 1FTC Biographies Diﬀeren a on of func ons deﬁned by integrals “Contrived” examples Erf Other applica ons . . . 9.
10. 10. . V63.0121.001: Calculus I . Sec on 5.4: The Fundamental Theorem . May 2, 2011 Notes Diﬀerentiation of area functions Example ∫ 3x Let h(x) = t3 dt. What is h′ (x)? 0 Solu on (Using 2FTC) 3x t4 1 h(x) = = (3x)4 = 1 4 · 81x4 , so h′ (x) = 81x3 . 4 0 4 . . Notes Diﬀerentiation of area functions Example ∫ 3x Let h(x) = t3 dt. What is h′ (x)? 0 Solu on (Using 1FTC) ∫ u We can think of h as the composi on g ◦ k, where g(u) = t3 dt 0 and 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 . . . Notes Diﬀerentiation of area functions, in general by 1FTC ∫ d k(x) f(t) dt = f(k(x))k′ (x) dx a by reversing the order of integra on: ∫ ∫ d b d h(x) f(t) dt = − f(t) dt = −f(h(x))h′ (x) dx h(x) dx b by combining the two above: ∫ (∫ ∫ 0 ) d k(x) d k(x) 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) . . . 10.
11. 11. . V63.0121.001: Calculus I . Sec on 5.4: The Fundamental Theorem . May 2, 2011 Notes Another Example Example ∫ sin2 x Let h(x) = (17t2 + 4t − 4) dt. What is h′ (x)? 0 Solu on . . Notes A Similar Example Example ∫ sin2 x Let h(x) = (17t2 + 4t − 4) dt. What is h′ (x)? 3 Solu on . . Notes Compare Ques on ∫ 2 ∫ 2 d sin x d sin x Why is (17t2 + 4t − 4) dt = (17t2 + 4t − 4) dt? dx 0 dx 3 Or, why doesn’t the lower limit appear in the deriva ve? Answer . . . 11.
12. 12. . V63.0121.001: Calculus I . Sec on 5.4: The Fundamental Theorem . May 2, 2011 Notes The Full Nasty Example ∫ ex Find the deriva ve of F(x) = sin4 t dt. x3 Solu on No ce here it’s much easier than ﬁnding an an deriva ve for sin4 . . . Notes Why use 1FTC? Ques on Why would we use 1FTC to ﬁnd the deriva ve of an integral? It seems like confusion for its own sake. Answer Some func ons are diﬃcult or impossible to integrate in elementary terms. Some func ons are naturally deﬁned in terms of other integrals. . . Notes Erf ∫ x 2 e−t dt 2 erf(x) = √ π 0 erf measures area the bell curve. erf(x) We can’t ﬁnd erf(x), explicitly, but we do know its deriva ve: . ′ 2 x erf (x) = √ e−x . 2 π . . . 12.
13. 13. . V63.0121.001: Calculus I . Sec on 5.4: The Fundamental Theorem . May 2, 2011 Notes Example of erf Example d Find erf(x2 ). dx Solu on . . Notes 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 me 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 func on and p∗ and q∗ the equilibrium price and quan ty. . . Notes Surplus by picture consumer surplus price (p) producer surplus supply p∗ equilibrium market revenue demand f(q) . q∗ quan ty (q) . . . 13.
14. 14. . V63.0121.001: Calculus I . Sec on 5.4: The Fundamental Theorem . May 2, 2011 Notes Summary Func ons deﬁned as integrals can be diﬀeren ated using the ﬁrst FTC: ∫ d x f(t) dt = f(x) dx a The two FTCs link the two major processes in calculus: diﬀeren a on and integra on ∫ F′ (x) dx = F(x) + C . . Notes . . Notes . . . 14.