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# Lesson 25: The Fundamental Theorem of Calculus (handout)

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### Lesson 25: The Fundamental Theorem of Calculus (handout)

1. 1. V63.0121.006/016, Calculus I 5.4 : The Fundamental Theorem of Calculus Section April 22, 2010 Section 5.4 Notes The Fundamental Theorem of Calculus V63.0121.006/016, Calculus I New York University April 22, 2010 Announcements April 29: Movie Day April 30: Quiz 5 on §§5.1–5.4 Monday, May 10, 12:00noon Final Exam Announcements Notes April 29: Movie Day April 30: Quiz 5 on §§5.1–5.4 Monday, May 10, 12:00noon Final Exam V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem of Calculus April 22, 2010 2 / 31 Resurrection policies Notes 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 If your ﬁnal exam score beats your midterm score, we will re-weight it by 50% and make the midterm 15% V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem of Calculus April 22, 2010 3 / 31 1
2. 2. V63.0121.006/016, Calculus I 5.4 : The Fundamental Theorem of Calculus Section April 22, 2010 Objectives Notes 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. V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem of Calculus April 22, 2010 4 / 31 Outline Notes 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 V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem of Calculus April 22, 2010 5 / 31 The deﬁnite integral as a limit Notes 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 V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem of Calculus April 22, 2010 6 / 31 2
3. 3. V63.0121.006/016, Calculus I 5.4 : The Fundamental Theorem of Calculus Section April 22, 2010 Notes 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 V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem of Calculus April 22, 2010 7 / 31 The Integral as Total Change Notes 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 V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem of Calculus April 22, 2010 8 / 31 The Integral as Total Change Notes 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 V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem of Calculus April 22, 2010 8 / 31 3
4. 4. V63.0121.006/016, Calculus I 5.4 : The Fundamental Theorem of Calculus Section April 22, 2010 The Integral as Total Change Notes 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 V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem of Calculus April 22, 2010 8 / 31 My ﬁrst table of integrals Notes [f (x) + g (x)] dx = f (x) dx + g (x) dx x n+1 x n dx = + C (n = −1) cf (x) dx = c f (x) dx n+1 x x 1 e dx = e + 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 V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem of Calculus April 22, 2010 9 / 31 Outline Notes 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 V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem of Calculus April 22, 2010 10 / 31 4
5. 5. V63.0121.006/016, Calculus I 5.4 : The Fundamental Theorem of Calculus Section April 22, 2010 An area function Notes x Let f (t) = t 3 and deﬁne g (x) = f (t) dt. Can we evaluate the integral 0 in g (x)? Dividing the interval [0, x] into n pieces x ix gives ∆t = and ti = 0 + i∆t = . So n n x x 3 x (2x)3 x (nx)3 Rn = · + · + ··· + · n n3 n n3 n n3 x4 3 = 4 1 + 23 + 33 + · · · + n 3 n x4 1 2 = 4 2 n(n + 1) n 0 x x 4 n2 (n + 1)2 x4 = 4 → 4n 4 as n → ∞. V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem of Calculus April 22, 2010 11 / 31 An area function, continued Notes So x4 g (x) = . 4 This means that g (x) = x 3 . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem of Calculus April 22, 2010 12 / 31 The area function Notes 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 ? V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem of Calculus April 22, 2010 13 / 31 5
6. 6. V63.0121.006/016, Calculus I 5.4 : The Fundamental Theorem of Calculus Section April 22, 2010 Envisioning the area function Notes Example Suppose f (t) is the function graphed below v t0 t1 c t2 t3 t x Let g (x) = f (t) dt. What can you say about g ? t0 V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem of Calculus April 22, 2010 14 / 31 features of g from f Notes Interval sign monotonicity monotonicity concavity of f of g of f of g [t0 , t1 ] + [t1 , c] + [c, t2 ] − [t2 , t3 ] − [t3 , ∞) − → none We see that g is behaving a lot like an antiderivative of f . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem of Calculus April 22, 2010 15 / 31 Notes 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). V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem of Calculus April 22, 2010 16 / 31 6
7. 7. V63.0121.006/016, Calculus I 5.4 : The Fundamental Theorem of Calculus Section April 22, 2010 Proof. Notes 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). V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem of Calculus April 22, 2010 17 / 31 Meet the Mathematician: James Gregory Notes 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 V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem of Calculus April 22, 2010 18 / 31 Meet the Mathematician: Isaac Barrow Notes English, 1630-1677 Professor of Greek, theology, and mathematics at Cambridge Had a famous student V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem of Calculus April 22, 2010 19 / 31 7
8. 8. V63.0121.006/016, Calculus I 5.4 : The Fundamental Theorem of Calculus Section April 22, 2010 Meet the Mathematician: Isaac Newton Notes English, 1643–1727 Professor at Cambridge (England) Philosophiae Naturalis Principia Mathematica published 1687 V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem of Calculus April 22, 2010 20 / 31 Meet the Mathematician: Gottfried Leibniz Notes German, 1646–1716 Eminent philosopher as well as mathematician Contemporarily disgraced by the calculus priority dispute V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem of Calculus April 22, 2010 21 / 31 Diﬀerentiation and Integration as reverse processes Notes 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 V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem of Calculus April 22, 2010 22 / 31 8
9. 9. V63.0121.006/016, Calculus I 5.4 : The Fundamental Theorem of Calculus Section April 22, 2010 Outline Notes 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 V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem of Calculus April 22, 2010 23 / 31 Diﬀerentiation of area functions Notes Example 3x Let h(x) = t 3 dt. What is h (x)? 0 Solution (Using 2FTC) 3x t4 1 h(x) = = (3x)4 = 1 4 · 81x 4 , so h (x) = 81x 3 . 4 0 4 Solution (Using 1FTC) u We can think of h as the composition g ◦ k, where g (u) = t 3 dt and 0 k(x) = 3x. Then h (x) = g (k(x))k (x) = (k(x))3 · 3 = (3x)3 · 3 = 81x 3 . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem of Calculus April 22, 2010 24 / 31 Diﬀerentiation of area functions, in general Notes 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) V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem of Calculus April 22, 2010 25 / 31 9
10. 10. V63.0121.006/016, Calculus I 5.4 : The Fundamental Theorem of Calculus Section April 22, 2010 Notes Example sin2 x Let h(x) = (17t 2 + 4t − 4) dt. What is h (x)? 0 Solution We have sin2 x d (17t 2 + 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 V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem of Calculus April 22, 2010 26 / 31 Notes Example ex Find the derivative of F (x) = sin4 t dt. x3 Solution ex d sin4 t dt = sin4 (e x ) · e x − sin4 (x 3 ) · 3x 2 dx x3 Notice here it’s much easier than ﬁnding an antiderivative for sin4 . V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem of Calculus April 22, 2010 27 / 31 Erf Notes 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), 2 2 explicitly, but we do know its derivative: erf (x) = √ e −x . π Example d Find erf(x 2 ). dx Solution By the chain rule we have d d 2 2 2 4 4 erf(x 2 ) = erf (x 2 ) x 2 = √ e −(x ) 2x = √ xe −x . dx dx π π V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem of Calculus April 22, 2010 28 / 31 10
11. 11. V63.0121.006/016, Calculus I 5.4 : The Fundamental Theorem of Calculus Section April 22, 2010 Other functions deﬁned by integrals Notes The future value of an asset: ∞ FV (t) = π(τ )e −r τ 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. V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem of Calculus April 22, 2010 29 / 31 Surplus by picture Notes consumer surplus price (p) supply p∗ equilibrium demand f (q) q∗ quantity (q) V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem of Calculus April 22, 2010 30 / 31 Summary Notes 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 V63.0121.006/016, Calculus I (NYU) Section 5.4 The Fundamental Theorem of Calculus April 22, 2010 31 / 31 11