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Lesson 25: Evaluating Definite Integrals (handout)
 

Lesson 25: Evaluating Definite Integrals (handout)

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A remarkable theorem about definite integrals is that they can be calculated with antiderivatives.

A remarkable theorem about definite integrals is that they can be calculated with antiderivatives.

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    Lesson 25: Evaluating Definite Integrals (handout) Lesson 25: Evaluating Definite Integrals (handout) Document Transcript

    • . V63.0121.001: Calculus I . Sec on 5.3: Evalua ng Definite Integrals . April 27, 2011 Notes Sec on 5.3 Evalua ng Definite Integrals V63.0121.001: Calculus I Professor Ma hew Leingang New York University April 27, 2011 . . Notes Announcements Today: 5.3 Thursday/Friday: Quiz on 4.1–4.4 Monday 5/2: 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 Use the Evalua on Theorem to evaluate definite integrals. Write an deriva ves as indefinite integrals. Interpret definite integrals as “net change” of a func on over an interval. . . . 1.
    • . V63.0121.001: Calculus I . Sec on 5.3: Evalua ng Definite Integrals . April 27, 2011 Notes Outline Last me: The Definite Integral The definite integral as a limit Proper es of the integral Evalua ng Definite Integrals Examples The Integral as Net Change Indefinite Integrals My first table of integrals Compu ng Area with integrals . . 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.
    • . V63.0121.001: Calculus I . Sec on 5.3: Evalua ng Definite Integrals . April 27, 2011 Notes Notation/Terminology ∫ b f(x) dx a ∫ — integral sign (swoopy S) f(x) — integrand a and b — limits of integra on (a is the lower limit and b the upper limit) dx — ??? (a parenthesis? an infinitesimal? a variable?) The process of compu ng an integral is called integra on . . Example ∫ Notes 1 4 Es mate dx using M4 . 0 1 + x2 Solu on 1 1 3 We have x0 = 0, x1 = , x2 = , x3 = , x4 = 1. 4 2 4 1 3 5 7 So c1 = , c2 = , c3 = , c4 = . 8 8 8 8 . . Example ∫ Notes 1 4 Es mate dx using M4 . 0 1 + x2 Solu on ( ) 1 4 4 4 4 M4 = + + + 4 1 + (1/8)2 1 + (3/8)2 1 + (5/8)2 1 + (7/8)2 ( ) 1 4 4 4 4 = + + + 4 65/64 73/64 89/64 113/64 64 64 64 64 = + + + ≈ 3.1468 65 73 89 113 . . . 3.
    • . V63.0121.001: Calculus I . Sec on 5.3: Evalua ng Definite Integrals . April 27, 2011 Notes Properties of the integral Theorem (Addi ve Proper es of the Integral) Let f and g be integrable func ons on [a, b] and c a constant. Then ∫ b 1. c dx = c(b − a) ∫a b ∫ b ∫ b 2. [f(x) + g(x)] dx = f(x) dx + g(x) dx. ∫a b ∫ b a a 3. cf(x) dx = c f(x) dx. ∫a b a ∫ b ∫ b 4. [f(x) − g(x)] dx = f(x) dx − g(x) dx. a a a . . Notes More Properties of the Integral Conven ons: ∫ ∫ a b f(x) dx = − f(x) dx b a ∫ a f(x) dx = 0 a This allows us to have Theorem ∫ c ∫ b ∫ c 5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c. a a b . . Notes Illustrating Property 5 Theorem ∫ c ∫ b ∫ c 5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c. a a b y ∫ b ∫ c f(x) dx f(x) dx a b . a c x b . . . 4.
    • . V63.0121.001: Calculus I . Sec on 5.3: Evalua ng Definite Integrals . April 27, 2011 Notes Illustrating Property 5 Theorem ∫ c ∫ b ∫ c 5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c. a a b y ∫ ∫ c c f(x) dx f(x) dx = b∫ a b − f(x) dx . c a c x b . . Notes Definite Integrals We Know So Far If the integral computes an area and we know the area, we can use that. For instance, ∫ 1√ y π 1 − x2 dx = 0 4 By brute force we computed . ∫ 1 ∫ 1 x 1 1 x2 dx = x3 dx = 0 3 0 4 . . Notes Comparison Properties of the Integral Theorem Let f and g be integrable func ons on [a, b]. ∫ b 6. If f(x) ≥ 0 for all x in [a, b], then f(x) dx ≥ 0 a ∫ b ∫ b 7. If f(x) ≥ g(x) for all x in [a, b], then f(x) dx ≥ g(x) dx a a 8. If m ≤ f(x) ≤ M for all x in [a, b], then ∫ b m(b − a) ≤ f(x) dx ≤ M(b − a) a . . . 5.
    • . V63.0121.001: Calculus I . Sec on 5.3: Evalua ng Definite Integrals . April 27, 2011 Notes Integral of a nonnegative function is nonnegative Proof. If f(x) ≥ 0 for all x in [a, b], then for any number of divisions n and choice of sample points {ci }: ∑ n ∑ n Sn = f(ci ) ∆x ≥ 0 · ∆x = 0 i=1 ≥0 i=1 . x Since Sn ≥ 0 for all n, the limit of {Sn } is nonnega ve, too: ∫ b f(x) dx = lim Sn ≥ 0 a n→∞ ≥0 . . Notes The integral is “increasing” Proof. Let h(x) = f(x) − g(x). If f(x) ≥ g(x) for all x in [a, b], then h(x) ≥ 0 for all f(x) x in [a, b]. So by the previous h(x) g(x) property ∫ b h(x) dx ≥ 0 . x a This means that ∫ b ∫ b ∫ b ∫ b f(x) dx − g(x) dx = (f(x) − g(x)) dx = h(x) dx ≥ 0 a a a a . . Notes Bounding the integral Proof. If m ≤ f(x) ≤ M on for all x in [a, b], then by y the previous property ∫ b ∫ b ∫ b M m dx ≤ f(x) dx ≤ M dx a a a f(x) By Property 8, the integral of a constant func on is the product of the constant and m the width of the interval. So: ∫ b . x m(b − a) ≤ f(x) dx ≤ M(b − a) a b a . . . 6.
    • . V63.0121.001: Calculus I . Sec on 5.3: Evalua ng Definite Integrals . April 27, 2011 Example Notes ∫ 2 1 Es mate dx using the comparison proper es. 1 x Solu on . . Notes Outline Last me: The Definite Integral The definite integral as a limit Proper es of the integral Evalua ng Definite Integrals Examples The Integral as Net Change Indefinite Integrals My first table of integrals Compu ng Area with integrals . . Notes Socratic proof The definite integral of velocity measures displacement (net distance) The deriva ve of displacement is velocity So we can compute displacement with the definite integral or the an deriva ve of velocity But any func on can be a velocity func on, so . . . . . . 7.
    • . V63.0121.001: Calculus I . Sec on 5.3: Evalua ng Definite Integrals . April 27, 2011 Notes Theorem of the Day 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 Proving the Second FTC Proof. b−a Divide up [a, b] into n pieces of equal width ∆x = as n usual. For each i, F is con nuous on [xi−1 , xi ] and differen able on (xi−1 , xi ). So there is a point ci in (xi−1 , xi ) with F(xi ) − F(xi−1 ) = F′ (ci ) = f(ci ) xi − xi−1 =⇒ f(ci )∆x = F(xi ) − F(xi−1 ) . . Notes Proving the Second FTC Proof. Form the Riemann Sum: ∑ n ∑ n Sn = f(ci )∆x = (F(xi ) − F(xi−1 )) i=1 i=1 = (F(x1 ) − F(x0 )) + (F(x2 ) − F(x1 )) + (F(x3 ) − F(x2 )) + · · · · · · + (F(xn−1 ) − F(xn−2 )) + (F(xn ) − F(xn−1 )) = F(xn ) − F(x0 ) = F(b) − F(a) . . . 8.
    • . V63.0121.001: Calculus I . Sec on 5.3: Evalua ng Definite Integrals . April 27, 2011 Notes Proving the Second FTC Proof. We have shown for each n, Sn = F(b) − F(a) Which does not depend on n. So in the limit ∫ b f(x) dx = lim Sn = lim (F(b) − F(a)) = F(b) − F(a) a n→∞ n→∞ . . Notes Computing area with the 2nd FTC Example Find the area between y = x3 and the x-axis, between x = 0 and x = 1. Solu on ∫ 1 1 x4 1 A= x3 dx = = 0 4 0 4 . Here we use the nota on F(x)|b or [F(x)]b to mean F(b) − F(a). a a . . Notes Computing area with the 2nd FTC Example Find the area enclosed by the parabola y = x2 and the line y = 1. Solu on ∫ 1 [ 3 ]1 x A=2− x2 dx = 2 − 1 −1 3 −1 [ ( )] 1 1 4 =2− − − = . 3 3 3 −1 1 . . . 9.
    • . V63.0121.001: Calculus I . Sec on 5.3: Evalua ng Definite Integrals . April 27, 2011 Computing an integral we Notes estimated before Example ∫ 1 4 Evaluate the integral dx. 0 1 + x2 Solu on . . Computing an integral we Notes estimated before Example ∫ 2 1 Evaluate dx. 1 x Solu on . . Notes Outline Last me: The Definite Integral The definite integral as a limit Proper es of the integral Evalua ng Definite Integrals Examples The Integral as Net Change Indefinite Integrals My first table of integrals Compu ng Area with integrals . . . 10.
    • . V63.0121.001: Calculus I . Sec on 5.3: Evalua ng Definite Integrals . April 27, 2011 Notes The Integral as Net Change Another way to state this theorem is: ∫ b F′ (x) dx = F(b) − F(a), a or the integral of a deriva ve along an interval is the net change over that interval. This has many interpreta ons. . . 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 . . . 11.
    • . V63.0121.001: Calculus I . Sec on 5.3: Evalua ng Definite Integrals . April 27, 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 Outline Last me: The Definite Integral The definite integral as a limit Proper es of the integral Evalua ng Definite Integrals Examples The Integral as Net Change Indefinite Integrals My first table of integrals Compu ng Area with integrals . . Notes A new notation for antiderivatives To emphasize the rela onship between an differen a on and integra on, we use the indefinite integral nota on ∫ f(x) dx for any func on whose deriva ve is f(x). Thus ∫ x2 dx = 1 x3 + C. 3 . . . 12.
    • . V63.0121.001: Calculus I . Sec on 5.3: Evalua ng Definite Integrals . April 27, 2011 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 x ax sin x dx = − cos x + C a 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 Last me: The Definite Integral The definite integral as a limit Proper es of the integral Evalua ng Definite Integrals Examples The Integral as Net Change Indefinite Integrals My first table of integrals Compu ng Area with integrals . . Notes Computing Area with integrals ,label=area-exp Example Find the area of the region bounded by the lines x = 1, x = 4, the x-axis, and the curve y = ex . ,label=area-exp Solu on The answer is ∫ 4 ex dx = ex |4 = e4 − e. 1 1 . . . 13.
    • . V63.0121.001: Calculus I . Sec on 5.3: Evalua ng Definite Integrals . April 27, 2011 Notes Computing Area with integrals Example Find the area of the region bounded by the curve y = arcsin x, the x-axis, and the line x = 1. Solu on . . Example Notes Find the area between the graph of y = (x − 1)(x − 2), the x-axis, and the ver cal lines x = 0 and x = 3. Solu on . . Interpretation of “negative area” Notes in motion There is an analog in rectlinear mo on: ∫ t1 v(t) dt is net distance traveled. t ∫ 0t1 |v(t)| dt is total distance traveled. t0 . . . 14.
    • . V63.0121.001: Calculus I . Sec on 5.3: Evalua ng Definite Integrals . April 27, 2011 Notes What about the constant? It seems we forgot about the +C when we say for instance ∫ 1 1 x4 1 1 x3 dx = = −0= 0 4 0 4 4 But no ce [ 4 ]1 ( ) x 1 1 1 +C = + C − (0 + C) = + C − C = 4 0 4 4 4 no ma er what C is. So in an differen a on for definite integrals, the constant is immaterial. . . Notes Summary The second Fundamental Theorem of Calculus: ∫ b f(x) dx = F(b) − F(a) a ′ where F = f. Definite integrals represent net change of a func on over an interval. ∫ We write an deriva ves as indefinite integrals f(x) dx . . Notes . . . 15.