• Share
  • Email
  • Embed
  • Like
  • Save
  • Private Content
Lesson 28: The Fundamental Theorem of Calculus
 

Lesson 28: The Fundamental Theorem of Calculus

on

  • 1,191 views

 

Statistics

Views

Total Views
1,191
Views on SlideShare
1,189
Embed Views
2

Actions

Likes
0
Downloads
19
Comments
0

1 Embed 2

http://www.slideshare.net 2

Accessibility

Categories

Upload Details

Uploaded via as Adobe PDF

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Processing…
Post Comment
Edit your comment

    Lesson 28: The Fundamental Theorem of Calculus Lesson 28: The Fundamental Theorem of Calculus Presentation Transcript

    • Section 5.4 The Fundamental Theorem of Calculus V63.0121.034, Calculus I December 7, 2009 Announcements Final Exam: Friday 12/18, 2:00-3:50pm, Tisch UC50 . . . . . .
    • 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% . . . . . .
    • 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 . . . . . .
    • 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 . . . . . .
    • 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 ∫ ∫ ∫ 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 . . . . . .
    • 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 . . . . . .
    • 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 . . . . . . .
    • 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 → ∞. . . . . . .
    • An area function, continued So x4 g(x) = . 4 . . . . . .
    • An area function, continued So x4 g(x) = . 4 This means that g ′ (x ) = x 3 . . . . . . .
    • 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? . . . . . .
    • 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 . . . . . .
    • 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 . . . . . .
    • 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. . . . . . .
    • 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). . . . . . .
    • 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 . . . . . .
    • Differentiation of area functions Example ∫ 3x Let h(x) = t3 dt. What is h′ (x)? 0 . . . . . .
    • 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 . . . . . .
    • 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. . . . . . .
    • 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 . . . . . . .
    • 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) . . . . . .
    • Example ∫ sin2 x Let h(x) = (17t2 + 4t − 4) dt. What is h′ (x)? 0 . . . . . .
    • 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 . . . . . .
    • Example ∫ ex Find the derivative of F(x) = sin4 t dt. x3 . . . . . .
    • 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 . . . . . .
    • 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 . . . . . . .
    • Erf Here’s a function with a funny name but an important role: ∫ x 2 2 erf(x) = √ e−t dt. π 0 . . . . . .
    • 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. . . . . . .
    • 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) = . . . . . .
    • 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 . . . . . .
    • 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 . . . . . .
    • 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 . . . . . .
    • 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. . . . . . .
    • Surplus by picture c . onsumer surplus p . rice (p) s . upply .∗ . p . . quilibrium e . emand f(q) d . . .∗ q q . uantity (q) . . . . . .