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Lesson 24: Evaluating Definite Integrals (handout)
 

Lesson 24: Evaluating Definite Integrals (handout)

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Computing integrals with Riemann sums is like computing derivatives with limits. The calculus of integrals turns out to come from antidifferentiation. This startling fact is the Second Fundamental ...

Computing integrals with Riemann sums is like computing derivatives with limits. The calculus of integrals turns out to come from antidifferentiation. This startling fact is the Second Fundamental Theorem of Calculus!

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    Lesson 24: Evaluating Definite Integrals (handout) Lesson 24: Evaluating Definite Integrals (handout) Document Transcript

    • V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010 Section 5.3 Notes Evaluating Definite Integrals V63.0121.006/016, Calculus I New York University April 20, 2010 Announcements April 16: Quiz 4 on §§4.1–4.4 April 29: Movie Day!! April 30: Quiz 5 on §§5.1–5.4 Monday, May 10, 12:00noon (not 10:00am as previously announced) Final Exam Image credit: docman Announcements Notes April 16: Quiz 4 on §§4.1–4.4 April 29: Movie Day!! April 30: Quiz 5 on §§5.1–5.4 Monday, May 10, 12:00noon (not 10:00am as previously announced) Final Exam V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 2 / 44 Objectives Notes Use the Evaluation Theorem to evaluate definite integrals. Write antiderivatives as indefinite integrals. Interpret definite integrals as “net change” of a function over an interval. V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 4 / 44 1
    • V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010 Outline Notes Last time: The Definite Integral The definite integral as a limit Properties of the integral Comparison Properties of the Integral Evaluating Definite Integrals Examples The Integral as Total Change Indefinite Integrals My first table of integrals Computing Area with integrals V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 5 / 44 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 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 ]. Theorem If f is continuous on [a, b] or if f has only finitely many jump discontinuities, then f is integrable on [a, b]; that is, the definite integral b f (x) dx exists and is the same for any choice of ci . a V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 6 / 44 Notation/Terminology Notes b n f (x) dx = lim f (ci ) ∆x a n→∞ i=1 — integral sign (swoopy S) f (x) — integrand a and b — limits of integration (a is the lower limit and b the upper limit) dx — ??? (a parenthesis? an infinitesimal? a variable?) The process of computing an integral is called integration V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 7 / 44 2
    • V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010 Properties of the integral Notes Theorem (Additive Properties of the Integral) Let f and g be integrable functions 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 a a b b 3. cf (x) dx = c f (x) dx. a a b b b 4. [f (x) − g (x)] dx = f (x) dx − g (x) dx. a a a V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 8 / 44 More Properties of the Integral Notes Conventions: a b f (x) dx = − f (x) dx b a a f (x) dx = 0 a This allows us to have c b c 5. f (x) dx = f (x) dx + f (x) dx for all a, b, and c. a a b V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 9 / 44 Definite Integrals We Know So Far Notes If the integral computes an area and we know the area, we can use that. For y instance, 1 π 1 − x 2 dx = 0 2 By brute force we computed x 1 1 1 1 x 2 dx = x 3 dx = 0 3 0 4 V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 10 / 44 3
    • V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010 Example 1 4 Notes Estimate dx using the midpoint rule and four divisions. 0 1 + x2 Solution Dividing up [0, 1] into 4 pieces gives 1 2 3 4 x0 = 0, x1 = , x2 = , x3 = , x4 = 4 4 4 4 So the midpoint rule gives 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 89/64 113/64 150, 166, 784 = ≈ 3.1468 47, 720, 465 V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 11 / 44 Comparison Properties of the Integral Notes Theorem Let f and g be integrable functions 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 V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 12 / 44 The integral of a nonnegative function is nonnegative Notes 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 Since Sn ≥ 0 for all n, the limit of {Sn } is nonnegative, too: b f (x) dx = lim Sn ≥ 0 a n→∞ ≥0 V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 13 / 44 4
    • V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010 Comparison Properties of the Integral Notes Theorem Let f and g be integrable functions 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 V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 14 / 44 The definite integral is “increasing” Notes 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 x in [a, b]. So by the previous property b h(x) dx ≥ 0 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 So b b f (x) dx ≥ g (x) dx a a V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 15 / 44 Comparison Properties of the Integral Notes Theorem Let f and g be integrable functions 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 V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 16 / 44 5
    • V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010 Bounding the integral using bounds of the function Notes Proof. If m ≤ f (x) ≤ M on for all x in [a, b], then by the previous property b b b m dx ≤ f (x) dx ≤ M dx a a a By Property 1, the integral of a constant function is the product of the constant and the width of the interval. So: b m(b − a) ≤ f (x) dx ≤ M(b − a) a V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 17 / 44 Example 2 Notes 1 Estimate dx using Property 8. 1 x Solution Since 1 1 1 1 ≤ x ≤ 2 =⇒ ≤ ≤ 2 x 1 y we have 2 1 1 · (2 − 1) ≤ dx ≤ 1 · (2 − 1) 2 1 x or 1 2 1 x ≤ dx ≤ 1 2 1 x (Not a very good estimate) V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 18 / 44 Outline Notes Last time: The Definite Integral The definite integral as a limit Properties of the integral Comparison Properties of the Integral Evaluating Definite Integrals Examples The Integral as Total Change Indefinite Integrals My first table of integrals Computing Area with integrals V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 19 / 44 6
    • V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010 Socratic dialogue Notes The definite integral of velocity measures displacement (net distance) The derivative of displacement is velocity So we can compute displacement with the definite integral or the antiderivative of velocity But any function can be a velocity function, so . . . V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 20 / 44 Theorem of the Day 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 Note In Section 5.3, this theorem is called “The Evaluation Theorem”. Nobody else in the world calls it that. V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 21 / 44 Proving the Second FTC Notes Proof. b−a Divide up [a, b] into n pieces of equal width ∆x = as usual. For n each i, F is continuous on [xi−1 , xi ] and differentiable 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 Or f (ci )∆x = F (xi ) − F (xi−1 ) V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 22 / 44 7
    • V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010 Proving the Second FTC Notes We have for each i f (ci )∆x = F (xi ) − F (xi−1 ) 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) See if you can spot the invocation of the Mean Value Theorem! V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 23 / 44 Notes We have shown for each n, Sn = F (b) − F (a) so in the limit b f (x) dx = lim Sn = lim (F (b) − F (a)) = F (b) − F (a) a n→∞ n→∞ V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 24 / 44 Verifying earlier computations Notes Example Find the area between y = x 3 the x-axis, x = 0 and x = 1. Solution 1 1 x4 1 A= x 3 dx = = 0 4 0 4 Here we use the notation F (x)|b or [F (x)]b to mean F (b) − F (a). a a V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 25 / 44 8
    • V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010 Verifying Archimedes Notes Example Find the area enclosed by the parabola y = x 2 and y = 1. 1 −1 1 Solution 1 1 x3 1 1 4 A=2− x 2 dx = 2 − =2− − − = −1 3 −1 3 3 3 V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 26 / 44 Computing exactly what we earlier estimated Notes Example 1 4 Evaluate the integral dx. 0 1 + x2 Solution 1 1 4 1 dx = 4 dx 0 1 + x2 0 1 + x2 = 4 arctan(x)|1 0 = 4 (arctan 1 − arctan 0) π =4 −0 =π 4 V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 27 / 44 Example 1 4 Notes Estimate dx using the midpoint rule and four divisions. 0 1 + x2 Solution Dividing up [0, 1] into 4 pieces gives 1 2 3 4 x0 = 0, x1 = , x2 = , x3 = , x4 = 4 4 4 4 So the midpoint rule gives 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 89/64 113/64 150, 166, 784 = ≈ 3.1468 47, 720, 465 V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 28 / 44 9
    • V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010 Computing exactly what we earlier estimated Notes Example 1 4 Evaluate the integral dx. 0 1 + x2 Solution 1 1 4 1 dx = 4 dx 0 1 + x2 0 1 + x2 = 4 arctan(x)|1 0 = 4 (arctan 1 − arctan 0) π =4 −0 =π 4 V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 29 / 44 Computing exactly what we earlier estimated Notes Example 2 1 Evaluate dx. 1 x Solution 2 1 dx = ln x|2 1 1 x = ln 2 − ln 1 = ln 2 V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 30 / 44 Example 2 Notes 1 Estimate dx using Property 8. 1 x Solution Since 1 1 1 1 ≤ x ≤ 2 =⇒ ≤ ≤ 2 x 1 y we have 2 1 1 · (2 − 1) ≤ dx ≤ 1 · (2 − 1) 2 1 x or 1 2 1 x ≤ dx ≤ 1 2 1 x (Not a very good estimate) V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 31 / 44 10
    • V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010 Computing exactly what we earlier estimated Notes Example 2 1 Evaluate dx. 1 x Solution 2 1 dx = ln x|2 1 1 x = ln 2 − ln 1 = ln 2 V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 32 / 44 Outline Notes Last time: The Definite Integral The definite integral as a limit Properties of the integral Comparison Properties of the Integral Evaluating Definite Integrals Examples The Integral as Total Change Indefinite Integrals My first table of integrals Computing Area with integrals V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 33 / 44 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 over 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 5.3 Evaluating Definite Integrals April 20, 2010 34 / 44 11
    • V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 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 over 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 5.3 Evaluating Definite Integrals April 20, 2010 34 / 44 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 over 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 5.3 Evaluating Definite Integrals April 20, 2010 34 / 44 Outline Notes Last time: The Definite Integral The definite integral as a limit Properties of the integral Comparison Properties of the Integral Evaluating Definite Integrals Examples The Integral as Total Change Indefinite Integrals My first table of integrals Computing Area with integrals V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 35 / 44 12
    • V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010 A new notation for antiderivatives Notes To emphasize the relationship between antidifferentiation and integration, we use the indefinite integral notation f (x) dx for any function whose derivative is f (x). Thus x 2 dx = 1 x 3 + C . 3 V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 36 / 44 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 1 e x dx = e x + 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 5.3 Evaluating Definite Integrals April 20, 2010 37 / 44 Outline Notes Last time: The Definite Integral The definite integral as a limit Properties of the integral Comparison Properties of the Integral Evaluating Definite Integrals Examples The Integral as Total Change Indefinite Integrals My first table of integrals Computing Area with integrals V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 38 / 44 13
    • V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010 Example Notes Find the area between the graph of y = (x − 1)(x − 2), the x-axis, and the vertical lines x = 0 and x = 3. Solution 3 Consider (x − 1)(x − 2) dx. Notice the integrand is positive on [0, 1) 0 and (2, 3], and negative on (1, 2). If we want the area of the region, we have to do 1 2 3 A= (x − 1)(x − 2) dx − (x − 1)(x − 2) dx + (x − 1)(x − 2) dx 0 1 2 1 3 3 2 1 1 3 3 2 2 1 3 3 = 3x − 2x + 2x 0 − 3 x − 2x + 2x 1 + 3x − 2 x 2 + 2x 3 2 5 1 5 11 = − − + = . 6 6 6 6 V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 39 / 44 Graph Notes y x 1 2 3 V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 40 / 44 Example Notes Find the area between the graph of y = (x − 1)(x − 2), the x-axis, and the vertical lines x = 0 and x = 3. Solution 3 Consider (x − 1)(x − 2) dx. Notice the integrand is positive on [0, 1) 0 and (2, 3], and negative on (1, 2). If we want the area of the region, we have to do 1 2 3 A= (x − 1)(x − 2) dx − (x − 1)(x − 2) dx + (x − 1)(x − 2) dx 0 1 2 1 3 1 2 3 = 3x − 2 x 2 + 2x 3 0 − 1 3 3 x − 3 x 2 + 2x 2 1 + 1 3 3x − 2 x 2 + 2x 3 2 5 1 5 11 = − − + = . 6 6 6 6 V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 41 / 44 14
    • V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010 Interpretation of “negative area” in motion Notes There is an analog in rectlinear motion: t1 v (t) dt is net distance traveled. t0 t1 |v (t)| dt is total distance traveled. t0 V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 42 / 44 What about the constant? Notes It seems we forgot about the +C when we say for instance 1 1 x4 1 1 x 3 dx = = −0= 0 4 0 4 4 But notice 1 x4 1 1 1 +C = +C − (0 + C ) = +C −C = 4 0 4 4 4 no matter what C is. So in antidifferentiation for definite integrals, the constant is immaterial. V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 43 / 44 Summary Notes b Second FTC: F (x) dx = F (b) − F (a) a Net change theorem The MVT really is the MITC! V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 44 / 44 15