EXAMPLE 9 Since f(x) = (tan x)/ (1 + x2 + x4) satisfies f(-x) = -f(x), it is odd and so tan x -1+12 + , 4 dx =0 4.5 EXERCISES 1-6 Evaluate the integral by making the given substitution. dt 23. /1+ 23 -dz 24. cost 1 + tant 1. cos 2x dx, u = 2x 2 . x ( 2 x 2 + 3 ) # dx , 1 = 2 x 2 + 3 25. cot x csc'x dx 26. dx tan2x 3. [x' V x3 + 1 dx , 1 = x3 + 1 27 . sec'x tan x dx 28 . x 2 2 + x dx 4. | sin '0 cos0 do, u = sine 29 . * ( 2x + 5 ) 8 dx 30 . [x 3 V x 2 + 1 dx 5 . ( x4 - 5) 2 dx , 1 = 14 - 5 31-34 Evaluate the indefinite integral. Illustrate and check that 6. V 2t + 1 dt , u = 2+ + 1 your answer is reasonable by graphing both the function and its antiderivative (take C = 0). 31 . (x ( x 2 - 1 ) 3 dx 32 . tan '0 sec 2 0 do 7-30 Evaluate the indefinite integral. 7. x V 1 - x2 dx 8. [x2 sin(x 3 ) dx 33 . sin'x cos x dx 34 . sin x cos*x dx 9 . ( 1 - 2x ) dx 10 . sint VI + cost dt 35-51 Evaluate the definite integral. 11. sin ( 20/ 3 ) de 12 . sec 2 20 do 35. COS ( Tr t / 2 ) dt 36 . ( 3t - 1 ) 50 dt 13 . sec 3t tan 3t dt 14 . ) 2 ( 4 - 1 3 ) 213 dy 37. V1 + 7x dx 38 . " x cos ( x ? ) dx 15. cos ( 1 + 5t) dt 16. ( sin Vx dx 39, 7/6 sin t Vx cost dt 40. Ja./3 CSC2 ( 2t) dt 17 . sec -0 tan '0 de 18 . sin x sin(cos x) dx _ma ( x3 + x* tan x) dx 42. [ cos x sin(sin x) dx 19 . ( x 2 + 1 ) ( * 3 + 3x ) + dx 20 . [x x + 2 dx 43 . [13 - dx V ( 1 + 2 x ) 2 44 . " x Vaz - x 2 dx 21. a + bx 2 = dx COS(TT / x) 3ax + bx3 22. -dx x 2 45. " x x2 + a? dx ( a > 0) 46. " x4 sin x dx2 30-32 Sketch the region enclosed by the given curves and find 32 its area. UA W N - O 46 30. y = X X x20 54 V I + x 2 62 31. y = cos x sin x, y = sin x, Ox T 48. The widths (in m 32. y = x\\x2 + 1, y=x2\\x3 + 1 were measured at ure. Use the Mid 33-34 Use calculus to find the area of the triangle with the given vertices. 33. (0, 0), (3, 1), (1, 2) 34. (2, 0), (0, 2), (- 1, 1) 6.2 35-36 Evaluate the integral and interpret it as the area of a region. Sketch the region. 49. A cross-section of 35. | sin x - cos 2x | dx 36. [ | V x + 2 - x / dx ments of the thick 20-centimeter inter 27.3, 23.8, 20.5, 15 37-40 Use a graph to find approximate x-coordinates of the to estimate the area points of intersection of the given curves. Then find (approxi- mately) the area of the region bounded by the curves. 37. y = x sin(x2), y = x4, x20 x 38. y = ( x 2 + 1 ) 2 ' y = x' - x, x20 39. y = 3x2 - 2x, y = x3 - 3x+ 4 50. If the birth rate of a 40. y = x - cos x, y = 2 -x2 b(t) = 2200 + and the death rate is 41-44 Graph the region between the curves and use your d(t) = 146 calculator to compute the area correct to five decimal places. find the area betwee 2 41. y = 1 + x 4 ' y = x2 42. y = x, y = v2 -x4 does this area repres 43. y = tan'x, y = Vx 44. y = cosx, y = x+ 2 sintx 51. In Example 5, we m by a function f. A paNOTE We could have found the area in Example 7 by integrating ( - 1 . - 2 ) instead of y, but the calculation is much more involved. Because the consists of two different curves, it would have meant splitting the y=-V21+ 6 computing the areas labeled A, and Az in Figure 16. The method we FIGURE 16 is much easier. 5.1 EXERCISES 1-4 Find the area of the shaded region. 9. y = Vx + 3, y = (x+ 3)/2 2. 10. y = sin x, y = 2x/7, x20 11. x = 1- y?, x= y? -1 y = VX 12. 4x + y 2 = 12, * = y (1, 1) x=4 13-28 Sketch the region enclosed by the given c 0 y = 1/x2 area. 13. y = 12 - x2, y = x2 - 6 14. y = x2, y = 4x -x2 3. 4. 15. y = sec2x, y = 8 cos x, - T/3 5 x TT/ *= 12 - 4y 16. y = cos x, y = 2 - cos x, Ox - 2TT (- 3, 3) - 17. x = 2y?, x=4+y2 x = Vy 18. y = Vx - 1, x- y =1 19. y = COS TIX, y = 4x2 - 1 * = 2y -yz 20. x = y4, y = v2 - x, y = 0 21. y = cos x, y = 1 - 2x/T 5-12 Sketch the region enclosed by the given curves. Decide 22. y = x3, y = x whether to integrate with respect to x or y. Draw a typical approxi- 23. y = V2x, y = 3x2 05 x56 mating rectangle and label its height and width. Then find the area of the region. 24 . y = Cos x, y = 1 - cosx, OS X - TT 5. y = x + 1, y = 9- x, x = -1, x=2 25. y = x4, y = 2- |x) 6. y = sin x, y = x, x = 1/2, X = T 26. ) y = 3x - x, y = x, x=3 7. y = (x - 2)2, y = x 27. y = 1/x2, y = x, y = 3x 8. y = x2 - 4x, y = 2x 28. y = 4x2, y = 2x2, x + y = 3, x20this equation geomet 50. sin(271 / T - a) dt 63. If a and b are positiv dx 57. Jo ( 1 + Vx )' 64. If f is continuous on 52. Verify that f(x) = sin Vx is an odd function and use that to show that fact to show that J." xf (sin o, sin Vidx 1 65. If f is continuous, pro 53-54 Use a graph to give a rough estimate of the area of the region that lies under the given curve. Then find the exact area. f(cos x 53. y = V2x + 1, Ox=1 54. y = 2 sin x - sin 2x, OS X T 66. Use Exercise 65 to eva 55. Evaluate 2, (x + 3)v4 - x2 dx by writing it as a sum of The following exercises are two integrals and interpreting one of those integrals in terms already covered Chapter 6. of an area. 67-84 Evaluate the integral 56. Evaluate foxv1 - x4 dx by making a substitution and interpreting the resulting integral in terms of an area. dx 67. 5-3x 57. Breathing is cyclic and a full respiratory cycle from the beginning of inhalation to the end of exhalation takes 69 . . (In x ) 2 about 5 s. The maximum rate of air flow into the lungs x " dx is about 0.5 L/s. This explains, in part, why the function f(t) = 2 sin(27r t/5) has often been used to model the rate 71 . e VItex dx of air flow into the lungs. Use this model to find the volume of inhaled air in the lungs at time t. 73 . (arctan x)2 - dx x2+ 1 58. A model for the basal metabolism rate, in kcal/h, of a young man is R(t) = 85 - 0.18 cos( 7r t/12), where t is the 1+x time in hours measured from 5:00 AM. What is this man's 2 dx total basal metabolism, Jo R(t) dt, over a 24-hour time period? 77. sin 2x 1 + cos x - dx 59. If f is continuous and f(x) dx - 10, find Sof(2x) dx. 79. cot x dx 60. If f is continuous and J. f(x) dx - 4, find f3 xf(x?) dx. 81. fe dx x VIn x 61. If f is continuous on R, prove that 1 e+ dz [ ." f ( - x ) dx = [" f(x)dx Joetz For the case where f(x) = 0 and 0