Lesson 25: Evaluating Definite Integrals (Section 021 handout)
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Lesson 25: Evaluating Definite Integrals (Section 021 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 (Section 021 handout) Lesson 25: Evaluating Definite Integrals (Section 021 handout) Document Transcript

    • V63.0121.021, Calculus I Section 5.3 : Evaluating Definite Integrals December 7, 2010 Section 5.3 Notes Evaluating Definite Integrals V63.0121.021, Calculus I New York University December 7, 2010 Announcements Today: Section 5.3 Thursday: Section 5.4 ”Thursday,” December 14: Section 5.5 ”Monday,” December 15: (WWH 109, 12:30–1:45pm) Review and Movie Day! Monday, December 20, 12:00–1:50pm: Final Exam Announcements Notes Today: Section 5.3 Thursday: Section 5.4 ”Thursday,” December 14: Section 5.5 ”Monday,” December 15: (WWH 109, 12:30–1:45pm) Review and Movie Day! Monday, December 20, 12:00–1:50pm: Final Exam V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 2 / 41 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. V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 3 / 41 1
    • V63.0121.021, Calculus I Section 5.3 : Evaluating Definite Integrals December 7, 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 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 4 / 41 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 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 5 / 41 Notation/Terminology Notes b f (x) dx a — 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 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 6 / 41 2
    • V63.0121.021, Calculus I Section 5.3 : Evaluating Definite Integrals December 7, 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 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 7 / 41 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 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 8 / 41 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 Theorem c b c 5. f (x) dx = f (x) dx + f (x) dx for all a, b, and c. a a b V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 9 / 41 3
    • V63.0121.021, Calculus I Section 5.3 : Evaluating Definite Integrals December 7, 2010 Illustrating Property 5 Notes 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 b c x V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 10 / 41 Illustrating Property 5 Notes 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 = a b b − f (x) dx c a c x b V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 10 / 41 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 4 By brute force we computed x 1 1 1 1 x 2 dx = x 3 dx = 0 3 0 4 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 11 / 41 4
    • V63.0121.021, Calculus I Section 5.3 : Evaluating Definite Integrals December 7, 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 7. If f (x) ≥ g (x) for all x in [a, b], then b b 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 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 12 / 41 Estimating an integral with inequalities Notes Example 2 1 Estimate dx using Property 8. 1 x Solution Since 1 1 1 1 ≤ x ≤ 2 =⇒ ≤ ≤ 2 x 1 we have 2 1 1 · (2 − 1) ≤ dx ≤ 1 · (2 − 1) 2 1 x or 2 1 1 ≤ dx ≤ 1 2 1 x V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 13 / 41 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 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 14 / 41 5
    • V63.0121.021, Calculus I Section 5.3 : Evaluating Definite Integrals December 7, 2010 Socratic proof 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 . . . V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 15 / 41 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 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 16 / 41 Proving the Second FTC Notes b−a Divide up [a, b] into n pieces of equal width ∆x = as usual. n For 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 ) V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 17 / 41 6
    • V63.0121.021, Calculus I Section 5.3 : Evaluating Definite Integrals December 7, 2010 Proof continued 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) V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 18 / 41 Proof Completed Notes 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→∞ V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 19 / 41 Computing area with the Second FTC Notes Example Find the area between y = x 3 and the x-axis, between 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 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 20 / 41 7
    • V63.0121.021, Calculus I Section 5.3 : Evaluating Definite Integrals December 7, 2010 Computing area with the Second FTC Notes Example Find the area enclosed by the parabola y = x 2 and the line 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 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 21 / 41 Computing an integral we estimated before Notes Example 1 4 Evaluate the integral dx. 0 1 + x2 Solution V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 22 / 41 Computing an integral we estimated before Notes Example 2 1 Evaluate dx. 1 x Solution V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 25 / 41 8
    • V63.0121.021, Calculus I Section 5.3 : Evaluating Definite Integrals December 7, 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 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 28 / 41 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 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 29 / 41 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 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 29 / 41 9
    • V63.0121.021, Calculus I Section 5.3 : Evaluating Definite Integrals December 7, 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 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 29 / 41 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 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 30 / 41 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 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 31 / 41 10
    • V63.0121.021, Calculus I Section 5.3 : Evaluating Definite Integrals December 7, 2010 My first table of integrals Notes [f (x) + g (x)] dx = f (x) dx + g (x) dx n x n+1 x 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 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 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 32 / 41 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 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 33 / 41 Computing Area with integrals Notes Example Find the area of the region bounded by the lines x = 1, x = 4, the x-axis, and the curve y = e x . Solution The answer is 4 e x dx = e x |4 = e 4 − e. 1 1 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 34 / 41 11
    • V63.0121.021, Calculus I Section 5.3 : Evaluating Definite Integrals December 7, 2010 Computing Area with integrals Notes Example Find the area of the region bounded by the curve y = arcsin x, the x-axis, and the line x = 1. Solution V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 35 / 41 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 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 36 / 41 Graph Notes V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 37 / 41 12
    • V63.0121.021, Calculus I Section 5.3 : Evaluating Definite Integrals December 7, 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 V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 39 / 41 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. V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 40 / 41 Summary Notes 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 function over an interval. We write antiderivatives as indefinite integrals f (x) dx V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 41 / 41 13