Successfully reported this slideshow.
Upcoming SlideShare
×

1,466 views

Published on

Published in: Education
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

1. 1. 7.5 5.0 2.5Chapter 4 0.0 −2 −1 0 1 2 x −2.5Integration 4. sin x, sin x + 2, sin x − 5 2 x −3 −2 −1 0 1 2 34.1 Antiderivatives 0 −2 x4 x4 x4 1. , + 3, −2 4 4 4 −4 20 −6 15 3 5 3 2 5. (3x4 − 3x)dx = x − x +c 10 5 2 1 4 5 6. (x3 − 2)dx = x − 2x + c 4 √ 1 x−3 −3 −2 −1 1 2 3 7. 3 x− 4 dx = 2x3/2 + +c x 3 1 8. 2x−2 + √ dx x x4 x2 x4 x2 x4 x2 = −2x−1 + 2x1/2 + c 2. − , − − 1, − +4 4 2 4 2 4 2 x1/3 − 3 6 9. dx = (x−1/3 − 3x−2/3 )dx x2/3 5 3 = x2/3 − 9x1/3 + c 4 2 x + 2x3/4 3 10. dx = (x−1/4 + 2x−1/2 )dx x5/4 2 4 = x3/4 + 4x1/2 + c 1 3 0 −2 −1 0 1 2 11. (2 sin x + cos x)dx = −2 cos x + sin x + c x −1 12. (3 cos x − sin x)dx = 3 sin x + cos x + c 3. ex , ex + 1, ex − 3 13. 2 sec x tan xdx = 2 sec x + c 240
2. 2. 4.1. ANTIDERIVATIVES 241 4 d14. √ dx = 4 arcsin x + c 30. ln |sin x · 2| 1 − x2 dx 1 d = (sin x · 2)15. 5 sec2 xdx = 5 tan x + c sin x · 2 dx 2 cos x = = cot x 2 sin x 4 cos x16. dx = −4 csc x + c sin2 x 31. (a) N/A (b) By Power Formula,17. (3ex − 2)dx = 3ex − 2x + c √ 2 ( x3 + 4)dx = x5/2 + 4x + c. 518. (4x − 2ex )dx = 2x2 − 2ex + c 32. (a) By Power Formula, 3x2 − 419. (3 cos x − 1/x)dx = 3 sin x − ln |x| + c dx = (3 − 4x−2 )dx x2 = 3x + 4x−1 + c20. (2x−1 + sin x)dx = 2 ln |x| − cos x + c (b) N/A 33. (a) N/A 4x21. dx = 2 ln |x2 + 4| + c (b) By Reversing derivative formula, x2 + 4 sec2 xdx = tan x + c 3 322. dx = tan−1 x + c 4x2 + 4 4 34. (a) By Power Formula, 1 1 cos x − 1 dx = − − x + c23. dx = ln | sin x| + c x 2 x sin x (b) N/A24. (2 cos x − ex )dx = 2 sin x − ex + c 35. Finding the antiderivative, x2 f (x) = 3ex + + c. ex 225. dx = ln | ex + 3| + c ex + 3 Since f (0) = 4, ex + 3 we have 4 = f (0) = 3 + c.26. dx = (1 + 3e−x )dx Therefore, ex = x − 3e−x + c x2 f (x) = 3ex + + 1. 227. x1/4 (x5/4 − 4)dx = (x3/2 − 4x1/4 )dx 36. Finding the antiderivative, 2 5/2 16 5/4 f (x) = 4 sin x + c. = x − x +c Since f (0) = 3, 5 5 we have 3 = f (0) = c. Therefore,28. x2/3 (x−4/3 − 3)dx = (x−2/3 − 3x2/3 )dx f (x) = 4 sin x + 3. 9 = 3x1/3 − x5/3 + c 37. Finding the antiderivative 5 f (x) = 4x3 + 2ex + c1 . d Since, f (0) = 2.29. ln |sec x + tan x| dx We have 2 = f (0) = 2 + c1 1 d = (sec x + tan x) and therefore sec x + tan x dx f (x) = 4x3 + 2ex . 2 sec x tan x + sec x = Finding the antiderivative, sec x + tan x f (x) = x4 + 2ex + c2 . sec x (tan x + sec x) = Since f (0) = 3, sec x + tan x = sec x We have 3 = f (0) = 2 + c2 Therefore,
3. 3. 242 CHAPTER 4. INTEGRATION f (x) = x4 + 2ex + 1. 1 f (x) = −3 sin x + x4 + c1 x + c2 . 3 42. Taking antiderivatives, 38. Finding the antiderivative, f (x) = x1/2 − 2 cos x f (x) = 5x4 + e2x + c1 . 2 Since f (0) = −3, f (x) = x3/2 − 2 sin x + c1 3 we have −3 = f (0) = 1 + c1 4 5/2 f (x) = x + 2 cos x + c1 x + c2 . Therefore, 15 f (x) = 5x4 + e2x − 4. 43. Taking antiderivatives, Finding the antiderivative, f (x) = 4 − 2/x3 e2x f (x) = x5 + − 4x + c2 . f (x) = 4x + x−2 + c1 2 f (x) = 2x2 − x−1 + c1 x + c2 Since f (0) = 2, 1 2 c1 We have 2 = f (0) = + c2 f (x) = x3 − ln |x| + x2 + c2 x + c3 2 3 2 Therefore, 44. Taking antiderivatives, e2x 3 f (x) = sin x − ex f (x) = x5 + − 4x + . 2 2 f (x) = − cos x − ex + c1 f (x) = − sin x − ex + c1 x + c2 c1 39. Taking antiderivatives, f (x) = cos x − ex + x2 + c2 x + c3 f (t) = 2t + t2 + c1 2 t3 45. Position is the antiderivative of velocity, f (t) = t2 + + c1 t + c2 3 s(t) = 3t − 6t2 + c. Since f (0) = 2, Since s(0) = 3, we have c = 3. Thus, we have 2 = f (0) = c2 s(t) = 3t − 6t2 + 3. Therefore, t3 46. Position is the antiderivative of velocity, f (t) = t2 + + c1 t + 2. 3 s(t) = −3e−t − 2t + c. Since f (3) = 2, Since s(0) = 0, we have −3 + c = 0 and there- we have fore c = 3. Thus, 2 = f (3) = 9 + 9 + 3c1 + 2 s(t) = −3e−t − 2t + 3. − 6 = c1 Therefore, 47. First we ﬁnd velocity, which is the antideriva- t3 tive of acceleration, f (t) = + t2 − 6t + 2. 3 v(t) = −3 cos t + c1 . Since v(0) = 0 we have −3 + c1 = 0, c1 = 3 and 40. Taking antiderivatives, v(t) = −3 cos t + 3. f (t) = 4t + 3t2 + c1 Position is the antiderivative of velocity, f (t) = 2t2 + t3 + c1 t + c2 s(t) = −3 sin t + 3t + c2 . Since f (1) = 3, Since s(0) = 4, we have c2 = 4. Thus, we have 3 = f (1) = 2 + 1 + c1 + c2 s(t) = −3 sin t + 3t + 4. Therefore, c1 + c2 = 0 48. First we ﬁnd velocity, which is the antideriva- Since f (−1) = −2, tive of acceleration, we have −2 = f (−1) = 2 − 1 − c1 + c2 1 v(t) = t3 + t + c1 . Therefore, −c1 + c2 = −3. 3 So, c1 = 2 and c2 = − 3 3 2 Since v(0) = 4 we have c1 = 4 and Hence, 1 3 3 v(t) = t3 + t + 4. f (t) = t3 + 2t2 + t − . 3 2 2 Position is the antiderivative of velocity, 1 4 1 2 41. Taking antiderivatives, s(t) = t + t + 4t + c2 . 12 2 f (x) = 3 sin x + 4x2 Since s(0) = 0, we have c2 = 0. Thus, 4 1 4 1 2 f (x) = −3 cos x + x3 + c1 s(t) = t + t + 4t. 3 12 2