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# Lesson 26: The Fundamental Theorem of Calculus (Section 4 version)

## by Matthew Leingang, Clinical Associate Professor of Mathematics at New York University on Apr 23, 2009

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The First Fundamental Theorem of Calculus looks at the area function and its derivative. It so happens that the derivative of the area function is the original integrand.

The First Fundamental Theorem of Calculus looks at the area function and its derivative. It so happens that the derivative of the area function is the original integrand.

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## Lesson 26: The Fundamental Theorem of Calculus (Section 4 version)Presentation Transcript

• 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 . . . . . .
• 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 . . . . . .
• 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 . . . . . .
• 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 . . . . . .
• 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: . . . . . .
• 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 . . . . . .
• 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 . . . . . .
• 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 . . . . . .
• 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 . . . . . .
• 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 . . . . . .
• 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 . . . . . . .
• 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 → ∞. . . . . . .
• An area function, continued So x4 g(x) = . 4 . . . . . .
• An area function, continued So x4 g(x) = . 4 This means that g′ (x) = x3 . . . . . . .
• 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? . . . . . .
• 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? . . . . . .
• 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? . . . . . .
• 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). . . . . . .
• Proof. Let h > 0 be given so that x + h < b. We have g(x + h) − g(x) = h . . . . . .
• 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 . . . . . .
• 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 . . . . . .
• 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 . . . . . .
• 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 . . . . . .
• 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 . . . . . .
• 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. . . . . . .
• 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 . . . . . .
• Meet the Mathematician: Isaac Barrow English, 1630-1677 Professor of Greek, theology, and mathematics at Cambridge Had a famous student . . . . . .
• Meet the Mathematician: Isaac Newton English, 1643–1727 Professor at Cambridge (England) Philosophiae Naturalis Principia Mathematica published 1687 . . . . . .
• Meet the Mathematician: Gottfried Leibniz German, 1646–1716 Eminent philosopher as well as mathematician Contemporarily disgraced by the calculus priority dispute . . . . . .
• 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 . . . . . .
• 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 . . . . . .
• 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 . . . . . .
• 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)? . . . . . .
• 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 . . . . . . .
• Example ∫ sin2 x (17t2 + 4t − 4) dt. What is h′ (x)? Let h(x) = 0 . . . . . .
• 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 . . . . . .
• Erf Here’s a function with a funny name but an important role: ∫x 2 2 e−t dt. √ erf(x) = π0 . . . . . .
• 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. . . . . . .
• 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) = . . . . . .
• 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 . . . . . .
• 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 . . . . . .
• 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 . . . . . .
• 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. . . . . . .
• 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 . . . . . .
• Worksheet . . Image: Erick Cifuentes . . . . . .
• 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 . . . . . .
• 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 . . . . . .