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- 1. Review for Final Exam Integration Math 1a January 13, 2008 Announcements Oﬃce hours on the website (click “Exams”) Email your TF, CA, or me with questions Final: Thursday 9:15am in Hall B
- 2. Outline The Riemann Integral Estimating the integral Properties of the integral Comparison Properties of the Integral The Fundamental Theorem of Calculus Statement Diﬀerentiation of functions deﬁned by integrals Properties of the area function The Second FTC Examples Total Change Indeﬁnite Integrals Integration by Substitution Substitution for Indeﬁnite Integrals Substitution for Deﬁnite Integrals
- 3. The Riemann Integral Learning Objectives Compute the deﬁnite integral using a limit of Riemann sums Estimate the deﬁnite integral using a Riemann sum (e.g., Midpoint Rule) Reason with the deﬁnite integral using its elementary properties.
- 4. The Area Problem Given a function f deﬁned on [a, b], how can one ﬁnd the area between y = 0, y = f (x), x = a, and x = b? We divide and conquer.
- 5. Forming Riemann sums We have many choices of how to approximate the area: Ln = f (x0 )∆x + f (x1 )∆x + · · · + f (xn−1 )∆x Rn = f (x1 )∆x + f (x2 )∆x + · · · + f (xn )∆x x0 + x1 x1 + x2 xn−1 + xn ∆x + · · · + f Mn = f ∆x + f ∆x 2 2 2
- 6. Forming Riemann sums We have many choices of how to approximate the area: Ln = f (x0 )∆x + f (x1 )∆x + · · · + f (xn−1 )∆x Rn = f (x1 )∆x + f (x2 )∆x + · · · + f (xn )∆x x0 + x1 x1 + x2 xn−1 + xn ∆x + · · · + f Mn = f ∆x + f ∆x 2 2 2 In general, choose ci to be a point in the ith interval [xi−1 , xi ]. Form the Riemann sum Sn = f (c1 )∆x + f (c2 )∆x + · · · + f (cn )∆x n = f (ci )∆x i=1
- 7. Theorem Theorem If f is a continuous function on [a, b] or has ﬁnitely many jump discontinuities, then lim Sn = lim {f (c1 )∆x + f (c2 )∆x + · · · + f (cn )∆x} n→∞ n→∞ exists and is the same value no matter what choice of ci we made.
- 8. Theorem Theorem If f is a continuous function on [a, b] or has ﬁnitely many jump discontinuities, then lim Sn = lim {f (c1 )∆x + f (c2 )∆x + · · · + f (cn )∆x} n→∞ n→∞ exists and is the same value no matter what choice of ci we made. Deﬁnition The deﬁnite integral of f from a to b is the number n b f (x) dx = lim f (ci ) ∆x ∆x→0 a i=1
- 9. Example (“Sample Exam”, Problem 6) The rate at which the world’s oil is being consumed is increasing. Suppose that the rate (measured in billions of barrels per year) is given by the function r (t), where t is measured in years and t = 0 represents January 1, 2000. (a) Write a deﬁnite integral that represents the total quantity of oil used between the start of 2000 and the start of 2005. (b) Suppose that r (t) = 32e 0.05t . Find the approximate value for the deﬁnite integral from part (a) using a right-hand sum with n = 5 subintervals. (c) Interpret each of the ﬁve terms in the sum from part (b) in terms of oil consumption.
- 10. Answers 5 (a) r (t) dt 0 (b) 1·32e 0.05(1) +1·32e 0.05(2) +1·32e 0.05(3) +1·32e 0.05(4) +1·32e 0.05(5) (c) Each term stands for the approximate amount of oil used in each year. For instance, the term 1 · 32e 0.05(3) is approximately the amount of oil used between January 1, 2002 and January 1, 2003.
- 11. Example 1 4 Estimate dx using the midpoint rule and four divisions. 1 + x2 0
- 12. Example 1 4 Estimate dx using the midpoint rule and four divisions. 1 + x2 0 Solution 1 1 3 The partition is 0 < < < < 1, so the estimate is 4 2 4 1 4 4 4 4 M4 = + + + 2 2 2 1 + (7/8)2 4 1 + (1/8) 1 + (3/8) 1 + (5/8)
- 13. Example 1 4 Estimate dx using the midpoint rule and four divisions. 1 + x2 0 Solution 1 1 3 The partition is 0 < < < < 1, so the estimate is 4 2 4 1 4 4 4 4 M4 = + + + 2 2 2 1 + (7/8)2 4 1 + (1/8) 1 + (3/8) 1 + (5/8) 1 4 4 4 4 = + + + 4 65/64 73/64 89/64 113/64
- 14. Example 1 4 Estimate dx using the midpoint rule and four divisions. 1 + x2 0 Solution 1 1 3 The partition is 0 < < < < 1, so the estimate is 4 2 4 1 4 4 4 4 M4 = + + + 2 2 2 1 + (7/8)2 4 1 + (1/8) 1 + (3/8) 1 + (5/8) 1 4 4 4 4 = + + + 4 65/64 73/64 89/64 113/64 150, 166, 784 ≈ 3.1468 = 47, 720, 465
- 15. Properties of the integral Theorem (Additive Properties of the Integral) Let f and g be integrable functions on [a, b] and c a constant. Then b c dx = c(b − a) 1. a
- 16. Properties of the integral Theorem (Additive Properties of the Integral) Let f and g be integrable functions on [a, b] and c a constant. Then b c dx = c(b − a) 1. a b b b 2. [f (x) + g (x)] dx = f (x) dx + g (x) dx. a a a
- 17. Properties of the integral Theorem (Additive Properties of the Integral) Let f and g be integrable functions on [a, b] and c a constant. Then b c dx = c(b − a) 1. 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
- 18. Properties of the integral Theorem (Additive Properties of the Integral) Let f and g be integrable functions on [a, b] and c a constant. Then b c dx = c(b − a) 1. 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 [f (x) − g (x)] dx = f (x) dx − 4. g (x) dx. a a a
- 19. More Properties of the Integral Conventions: a b f (x) dx = − f (x) dx b a
- 20. More Properties of the Integral Conventions: a b f (x) dx = − f (x) dx b a a f (x) dx = 0 a
- 21. More Properties of the Integral 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
- 22. Example Suppose f and g are functions with 4 f (x) dx = 4 0 5 f (x) dx = 7 0 5 g (x) dx = 3. 0 Find 5 [2f (x) − g (x)] dx (a) 0 5 (b) f (x) dx. 4
- 23. Solution We have (a) 5 5 5 [2f (x) − g (x)] dx = 2 f (x) dx − g (x) dx 0 0 0 = 2 · 7 − 3 = 11
- 24. Solution We have (a) 5 5 5 [2f (x) − g (x)] dx = 2 f (x) dx − g (x) dx 0 0 0 = 2 · 7 − 3 = 11 (b) 5 5 4 f (x) dx − f (x) dx = f (x) dx 4 0 0 =7−4=3
- 25. Comparison Properties of the Integral Theorem Let f and g be integrable functions on [a, b].
- 26. Comparison Properties of the Integral Theorem Let f and g be integrable functions on [a, b]. 6. If f (x) ≥ 0 for all x in [a, b], then b f (x) dx ≥ 0 a
- 27. Comparison Properties of the Integral Theorem Let f and g be integrable functions on [a, b]. 6. If f (x) ≥ 0 for all x in [a, b], then b 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
- 28. Comparison Properties of the Integral Theorem Let f and g be integrable functions on [a, b]. 6. If f (x) ≥ 0 for all x in [a, b], then b 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
- 29. Example 4 1 Estimate dx using the comparison properties. x + sin2 πx 1
- 30. Outline The Riemann Integral Estimating the integral Properties of the integral Comparison Properties of the Integral The Fundamental Theorem of Calculus Statement Diﬀerentiation of functions deﬁned by integrals Properties of the area function The Second FTC Examples Total Change Indeﬁnite Integrals Integration by Substitution Substitution for Indeﬁnite Integrals Substitution for Deﬁnite Integrals
- 31. The Fundamental Theorem of Calculus Learning Objectives State and use both fundamental theorems of calculus Understand the relationship between integration and antidiﬀerentiation Use FTC to compute derivatives of integrals with functions in the limits Use FTC to compute areas or other accumulations
- 32. 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).
- 33. Example (Spring 2000 Final, Problem 7c) 100 dy p 2 − p dp Find if y = dx x 3 +x
- 34. Example (Spring 2000 Final, Problem 7c) 100 dy p 2 − p dp Find if y = dx x 3 +x Solution u p 2 − p dp. By the Fundamental Theorem of Let A(u) = 1 u 2 − u. We have Calculus, A (u) = 100 d p 2 − p dp y = dx x 3 +x x 3 +x 100 d p2 p 2 − p dp − p dp − = dx 1 1 d A(100) − A(x 3 + x) = dx = −A (x 3 + x) · (3x 2 + 1) = −(3x 2 + 1) (x 3 + x)2 − (x 3 + x).
- 35. Facts about g from f Let f be the function whose graph is given below. Suppose the the position at time t seconds of a particle moving t along a coordinate axis is s(t) = f (x) dx meters. Use the 0 graph to answer the following questions. 4 3 • (3,3) 2 • • (2,2) (5,2) 1 • (1,1) 1 2 3 4 5 6 7 8 9
- 36. Facts about g from f Let f be the function whose graph is given below. Suppose the the position at time t seconds of a particle moving t along a coordinate axis is s(t) = f (x) dx meters. Use the 0 graph to answer the following questions. 4 What is the particle’s velocity 3 • (3,3) at time t = 5? 2 • • (2,2) (5,2) 1 • (1,1) 1 2 3 4 5 6 7 8 9
- 37. Facts about g from f Let f be the function whose graph is given below. Suppose the the position at time t seconds of a particle moving t along a coordinate axis is s(t) = f (x) dx meters. Use the 0 graph to answer the following questions. 4 What is the particle’s velocity 3 • (3,3) at time t = 5? 2 • • (2,2) (5,2) Solution 1 • (1,1) Recall that by the FTC we have 1 2 3 4 5 6 7 8 9 s (t) = f (t). So s (5) = f (5) = 2.
- 38. Facts about g from f Let f be the function whose graph is given below. Suppose the the position at time t seconds of a particle moving t along a coordinate axis is s(t) = f (x) dx meters. Use the 0 graph to answer the following questions. 4 Is the acceleration of the par- 3 • (3,3) ticle at time t = 5 positive or 2 negative? • • (2,2) (5,2) 1 • (1,1) 1 2 3 4 5 6 7 8 9
- 39. Facts about g from f Let f be the function whose graph is given below. Suppose the the position at time t seconds of a particle moving t along a coordinate axis is s(t) = f (x) dx meters. Use the 0 graph to answer the following questions. 4 Is the acceleration of the par- 3 • (3,3) ticle at time t = 5 positive or 2 negative? • • (2,2) (5,2) 1 • (1,1) Solution We have s (5) = f (5), which 1 2 3 4 5 6 7 8 9 looks negative from the graph.
- 40. Facts about g from f Let f be the function whose graph is given below. Suppose the the position at time t seconds of a particle moving t along a coordinate axis is s(t) = f (x) dx meters. Use the 0 graph to answer the following questions. 4 What is the particle’s position 3 • (3,3) at time t = 3? 2 • • (2,2) (5,2) 1 • (1,1) 1 2 3 4 5 6 7 8 9
- 41. Facts about g from f Let f be the function whose graph is given below. Suppose the the position at time t seconds of a particle moving t along a coordinate axis is s(t) = f (x) dx meters. Use the 0 graph to answer the following questions. 4 What is the particle’s position 3 • (3,3) at time t = 3? 2 • • (2,2) (5,2) Solution 1 • (1,1) Since on [0, 3], f (x) = x, we have 1 2 3 4 5 6 7 8 9 3 9 s(3) = x dx = . 2 0
- 42. Facts about g from f Let f be the function whose graph is given below. Suppose the the position at time t seconds of a particle moving t along a coordinate axis is s(t) = f (x) dx meters. Use the 0 graph to answer the following questions. 4 At what time during the ﬁrst 9 3 • (3,3) seconds does s have its largest 2 value? • • (2,2) (5,2) 1 • (1,1) 1 2 3 4 5 6 7 8 9
- 43. Facts about g from f Let f be the function whose graph is given below. Suppose the the position at time t seconds of a particle moving t along a coordinate axis is s(t) = f (x) dx meters. Use the 0 graph to answer the following questions. 4 At what time during the ﬁrst 9 3 • (3,3) seconds does s have its largest 2 value? • • (2,2) (5,2) 1 • (1,1) Solution 1 2 3 4 5 6 7 8 9
- 44. Facts about g from f Let f be the function whose graph is given below. Suppose the the position at time t seconds of a particle moving t along a coordinate axis is s(t) = f (x) dx meters. Use the 0 graph to answer the following questions. 4 At what time during the ﬁrst 9 3 • (3,3) seconds does s have its largest 2 value? • • (2,2) (5,2) 1 • (1,1) Solution The critical points of s are 1 2 3 4 5 6 7 8 9 the zeros of s = f .
- 45. Facts about g from f Let f be the function whose graph is given below. Suppose the the position at time t seconds of a particle moving t along a coordinate axis is s(t) = f (x) dx meters. Use the 0 graph to answer the following questions. 4 At what time during the ﬁrst 9 3 • (3,3) seconds does s have its largest 2 value? • • (2,2) (5,2) 1 • (1,1) Solution By looking at the graph, we 1 2 3 4 5 6 7 8 9 see that f is positive from t = 0 to t = 6, then negative from t = 6 to t = 9.
- 46. Facts about g from f Let f be the function whose graph is given below. Suppose the the position at time t seconds of a particle moving t along a coordinate axis is s(t) = f (x) dx meters. Use the 0 graph to answer the following questions. 4 At what time during the ﬁrst 9 3 • (3,3) seconds does s have its largest 2 value? • • (2,2) (5,2) 1 • (1,1) Solution Therefore s is increasing on 1 2 3 4 5 6 7 8 9 [0, 6], then decreasing on [6, 9]. So its largest value is at t = 6.
- 47. Facts about g from f Let f be the function whose graph is given below. Suppose the the position at time t seconds of a particle moving t along a coordinate axis is s(t) = f (x) dx meters. Use the 0 graph to answer the following questions. 4 Approximately when is the ac- 3 • (3,3) celeration zero? 2 • • (2,2) (5,2) 1 • (1,1) 1 2 3 4 5 6 7 8 9
- 48. Facts about g from f Let f be the function whose graph is given below. Suppose the the position at time t seconds of a particle moving t along a coordinate axis is s(t) = f (x) dx meters. Use the 0 graph to answer the following questions. 4 Approximately when is the ac- 3 • (3,3) celeration zero? 2 • • (2,2) (5,2) Solution 1 • (1,1) s = 0 when f = 0, which happens at t = 4 and t = 7.5 1 2 3 4 5 6 7 8 9 (approximately)
- 49. Facts about g from f Let f be the function whose graph is given below. Suppose the the position at time t seconds of a particle moving t along a coordinate axis is s(t) = f (x) dx meters. Use the 0 graph to answer the following questions. 4 When is the particle moving 3 • (3,3) toward the origin? Away from 2 the origin? • • (2,2) (5,2) 1 • (1,1) 1 2 3 4 5 6 7 8 9
- 50. Facts about g from f Let f be the function whose graph is given below. Suppose the the position at time t seconds of a particle moving t along a coordinate axis is s(t) = f (x) dx meters. Use the 0 graph to answer the following questions. 4 When is the particle moving 3 • (3,3) toward the origin? Away from 2 the origin? • • (2,2) (5,2) 1 • (1,1) Solution The particle is moving away 1 2 3 4 5 6 7 8 9 from the origin when s > 0 and s > 0.
- 51. Facts about g from f Let f be the function whose graph is given below. Suppose the the position at time t seconds of a particle moving t along a coordinate axis is s(t) = f (x) dx meters. Use the 0 graph to answer the following questions. 4 When is the particle moving 3 • (3,3) toward the origin? Away from 2 the origin? • • (2,2) (5,2) 1 • (1,1) Solution Since s(0) = 0 and s > 0 on 1 2 3 4 5 6 7 8 9 (0, 6), we know the particle is moving away from the origin then.
- 52. Facts about g from f Let f be the function whose graph is given below. Suppose the the position at time t seconds of a particle moving t along a coordinate axis is s(t) = f (x) dx meters. Use the 0 graph to answer the following questions. 4 When is the particle moving 3 • (3,3) toward the origin? Away from 2 the origin? • • (2,2) (5,2) 1 • (1,1) Solution After t = 6, s < 0, so the 1 2 3 4 5 6 7 8 9 particle is moving toward the origin.
- 53. Facts about g from f Let f be the function whose graph is given below. Suppose the the position at time t seconds of a particle moving t along a coordinate axis is s(t) = f (x) dx meters. Use the 0 graph to answer the following questions. 4 On which side (positive or neg- 3 • (3,3) ative) of the origin does the 2 • • particle lie at time t = 9? (2,2) (5,2) 1 • (1,1) 1 2 3 4 5 6 7 8 9
- 54. Facts about g from f Let f be the function whose graph is given below. Suppose the the position at time t seconds of a particle moving t along a coordinate axis is s(t) = f (x) dx meters. Use the 0 graph to answer the following questions. 4 On which side (positive or neg- 3 • (3,3) ative) of the origin does the 2 • • particle lie at time t = 9? (2,2) (5,2) 1 • (1,1) Solution We have s(9) = 1 2 3 4 5 6 7 8 9 6 9 f (x) dx + f (x) dx, 0 6 where the left integral is positive and the right integral is negative.
- 55. Facts about g from f Let f be the function whose graph is given below. Suppose the the position at time t seconds of a particle moving t along a coordinate axis is s(t) = f (x) dx meters. Use the 0 graph to answer the following questions. 4 On which side (positive or neg- 3 • (3,3) ative) of the origin does the 2 • • particle lie at time t = 9? (2,2) (5,2) 1 • (1,1) Solution In order to decide whether 1 2 3 4 5 6 7 8 9 s(9) is positive or negative, we need to decide if the ﬁrst area is more positive than the second area is negative.
- 56. Facts about g from f Let f be the function whose graph is given below. Suppose the the position at time t seconds of a particle moving t along a coordinate axis is s(t) = f (x) dx meters. Use the 0 graph to answer the following questions. 4 On which side (positive or neg- 3 • (3,3) ative) of the origin does the 2 • • particle lie at time t = 9? (2,2) (5,2) 1 • (1,1) Solution This appears to be the case, 1 2 3 4 5 6 7 8 9 so s(9) is positive.
- 57. 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
- 58. Examples Find the following integrals: 1 1 2 2 1 x 2 dx, x 3 dx, x n dx (n = −1), dx x 0 0 1 1 π 1 e x dx sin θ dθ, 0 0
- 59. 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:
- 60. 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
- 61. 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
- 62. 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
- 63. A new notation for antiderivatives 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).
- 64. A new notation for antiderivatives 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 = 3 x 3 + C . 1
- 65. My ﬁrst table of integrals [f (x) + g (x)] dx = f (x) dx + g (x) dx x n+1 x n dx = cf (x) dx = c f (x) dx + C (n = −1) n+1 1 e x dx = e x + C dx = ln |x| + C x ax ax dx = +C sin x dx = − cos x + C ln a csc2 x dx = − cot x + C cos x dx = sin x + C sec2 x dx = tan x + C csc x cot x dx = − csc x + C 1 √ dx = arcsin x + C sec x tan x dx = sec x + C 1 − x2 1 dx = arctan x + C 1 + x2
- 66. Outline The Riemann Integral Estimating the integral Properties of the integral Comparison Properties of the Integral The Fundamental Theorem of Calculus Statement Diﬀerentiation of functions deﬁned by integrals Properties of the area function The Second FTC Examples Total Change Indeﬁnite Integrals Integration by Substitution Substitution for Indeﬁnite Integrals Substitution for Deﬁnite Integrals
- 67. Integration by Substitution Learning Objectives Given an integral and a speciﬁc substitution, perform that substitution Use the substitution method to evaluate deﬁnite and indeﬁnite integrals
- 68. Theorem (The Substitution Rule) If u = g (x) is a diﬀerentiable function whose range is an interval I and f is continuous on I , then f (g (x))g (x) dx = f (u) du or du f (u) dx = f (u) du dx This is the “anti” version of the chain rule.
- 69. Example 2 xe x dx Find
- 70. Example 2 xe x dx Find Solution Let u = x 2 . Then du = 2x dx and x dx = 1 du. So 2 2 xe x dx = e u du 1 2 = 1 eu + C 2 2 = 1 ex + C 2
- 71. Theorem (The Substitution Rule for Deﬁnite Integrals) If g is continuous and f is continuous on the range of u = g (x), then b g (b) f (g (x))g (x) dx = f (u) du. a g (a)
- 72. Example π cos2 x sin x dx. Compute 0
- 73. Example π cos2 x sin x dx. Compute 0 Solution (Slow Way) cos2 x sin x dx and then First compute the indeﬁnite integral evaluate.
- 74. Example π cos2 x sin x dx. Compute 0 Solution (Slow Way) cos2 x sin x dx and then First compute the indeﬁnite integral evaluate. Let u = cos x. Then du = − sin x dx and cos2 x sin x dx = − u 2 du = − 1 u 3 + C = − 1 cos3 x + C . 3 3 Therefore π π cos2 x sin x dx = − 1 cos3 x = 2. 3 3 0 0
- 75. Solution (Fast Way) Do both the substitution and the evaluation at the same time.
- 76. Solution (Fast Way) Do both the substitution and the evaluation at the same time. Let u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1.
- 77. Solution (Fast Way) Do both the substitution and the evaluation at the same time. Let u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1. So −1 π cos2 x sin x dx = −u 2 du 0 1 1 u 2 du = −1 2 1 = 3 u3 1 =. −1 3

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