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SalasSV_08_06_ex_ans - ANSWERS TO ODD-NUMBERED EXERCISES 37...

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Unformatted text preview: ANSWERS TO ODD-NUMBERED EXERCISES 37. Let x = a tan u, dx = a sec2 u du; 39. √ x2 + a2 = a sec u. Then (x2 + a2 )n = a2n sec2n u and the result follows by substitution. 41. 12 rθ 2 1 (2x2 4 A-63 3 3x x tan−1 x + +C + 8 8(x2 + 1) 4(x2 + 1)2 12 r sin θ cos θ + 2 r r cos θ − 1) sin−1 x + 8 [10 3 x√ 1 − x2 + C 4 43. π2 π + 8 4 √ √ ( 2 − 1)a 2), xM = √ ln (1 + 2) y = 3a 8 45. A = r 2 − x2 dx = 47. − 9 2 ln 3] 49. M = ln (1 + √ 2a (2 − 2)a , y= √ √ √ √ 3[ 2 − ln ( 2 + 1)] 3[ 2 − ln ( 2 + 1)] √ 55. (a) Let x = a sec u, dx = a sec u tan u du, x2 − a2 = a tan u. √ √ 51. A = 1 a2 [ 2 − ln ( 2 + 1)]; 2 x= √ (b) Let x = a cosh u, dx = a sinh u, x2 − a2 = a sinh u. √ √ √ 2(3 3 − π ) 5 3 57. (b) ln (2 + 3) − (c) x = √ √ √ , y= √ 2 2 ln (2 + 3) − 3 72[2 ln (2 + 3) − 3] 53. Vy = 2 π a3 , 3 SECTION 8.5 1/5 1/5 1. − x+1 x+6 11. x2 − 2x + 17. 3. 1/4 1/4 x /2 + −2 x−1 x+1 x +1 13. 5. 1/2 3/2 1 + − x x+2 x−1 7. 3/2 9 19/2 − + x−1 x−2 x−3 9. ln x−2 +C x+5 3 + 5 ln |x − 1| − 3 ln |x| + C x 19. 14 43 32 x + x + 6x2 + 32x − + 80 ln|x − 2| + C 4 3 x−2 21. 15. 5 ln |x − 2| − 4 ln |x − 1| + C −1 +C 2(x − 1)2 3 1 1 ln |x − 1| − + ln |x + 1| + C 4 2(x − 1) 4 25. 3 10 1 1 x−2 x ln tan−1 + C − 32 x+2 16 2 23. 1 3 5(1 − x) ln (x2 + 1) + tan−1 x + +C 2 2 2(x2 + 1) 3 x + 4 ln +C x x+1 1 6 1 1 x 2 + 2x + 2 1 ln 2 + tan−1 (x + 1) + tan−1 (x − 1) + C 16 x − 2x + 2 8 8 2 15 27. 29. − 1 ln |x| + 6 37. 1 4 ln |x − 2| − 39. ln |x + 3| + C 31. ln ( 125 ) 108 33. ln ( 27 ) − 2 4 35. ln sin θ − 4 +C sin θ + 2 ln ln t − 2 +C ln t + 2 43. u 1 a = 1− a + bu b a + bu 41. u2 ( a 1 (−b/a2 ) (1/a) ( b 2 /a2 ) = + 2+ + bu) u u a + bu v−1 dv, where v = a2 − u2 1 −b/(ad − bc) d /(ad − bc) = + (a + bu)(c + du) a + bu c + du 47. (a) Exercise 44 53. (a) (b) x = a sin u, dx = a cos u du 55. (b) 3 ln 7 − 5 ln 3 57. (b) 11 − ln 12 45. u du = − 1 2 a2 − u 2 π 4 49. (a) ln 7 (b) π (4 − √ 51. x = (2 ln 2)/π , 7) y = (π + 2)/4π 2 5 4 1 − 2+ − x x x+1 (x + 1)3 1 3 4 (b) 2 + − x +4 x+3 x−3 2x − 1 3 (c) 2 − x + 2x + 4 x SECTION 8.6 √ √ 1. −2( x + ln |1 − x|) + C 7. 2 (x 5 √ √ 3. 2 ln ( 1 + ex − 1) − x + 2 1 + ex + C 9. − 17. 2 (x 3 5. 2 (1 5 + x)5/2 − 2 (1 + x)3/2 + C 3 √ √ 13. x + 4 x − 1 + 4 ln | x − 1 − 1| + C 1 (4x 48 − 1)5/2 + 2(x − 1)3/2 + C 1 + 2x2 +C 4(1 + x2 )2 √ − 8) x + 4 + C √ √ 11. x + 2 x + 2 ln | x − 1| + C 19. 1 (4x 16 √ 15. 2 ln ( 1 + ex − 1) − x + C 23. −ln 1 − tan x +C 2 + 1)1/2 + 1 (4x + 1)−1/2 − 8 + 1)−3/2 + C 21. 4b + 2ax +C √ a2 ax + b 2 1 x 25. √ tan−1 √ (2 tan + 1) + C 2 3 3 31. 4 5 27. 1 x x 1 ln tan − tan2 + C 2 2 4 2 2 3 29. ln 2 1 +C − 1 + sin x 1 + tan (x/2) + 2 tan−1 2 33. 2 + 4 ln 35. ln √ 3−1 √ 3 A-64 ANSWERS TO ODD-NUMBERED EXERCISES 1 dx = 2 cos x du and the result follows. 1 − u2 du where u = cos x. The result follows. 1 − u2 x 37. Let u = tan . Then 2 39. csc x dx = sin x dx = sin2 x x +C 2 43. sin x dx = − 1 − cos2 x −2 +C 1 + tanh (x/2) 41. 2 tan−1 tanh SECTION 8.7 1. (a) 506 (b) 650 5. (a) π ∼ 3. 1312 = (c) 572 (d) 578 (e) 576 3. (a) 1.394 (b) 1.7915 (b) 0.9122 (c) 1.8090 (c) 1.1776 (d) 1.1533 (e) 1.1614 (b) π ∼ 3. 1416 = 7. (a) 1.8440 9. (a) 0.8818 (b) 0.8821 11. Such a curve passes through the three points (a1 , b1 ), (a2 , b2 ), (a3 , b3 ) iff b1 = a2 A + a1 B + C , 1 which happens iff A= b1 (a2 − a3 ) − b2 (a1 − a3 ) + b3 (a1 − a2 ) , (a1 − a3 )(a1 − a2 )(a2 − a3 ) B=− b1 (a2 − a2 ) − b2 (a2 − a2 ) + b3 (a2 − a2 ) 2 3 1 3 1 2 , (a1 − a3 )(a1 − a2 )(a2 − a3 ) b2 = a2 A + a2 B + C , 2 b3 = a2 A + a3 B + C , 3 C= a2 (a2 b3 − a3 b2 ) − a2 (a1 b3 − a3 b1 ) + a2 (a1 b2 − a2 b1 ) 1 2 3 . (a1 − a3 )(a1 − a2 )(a2 − a3 ) (b) n ≥ 2 15. (a) n ≥ 238 (b) n ≥ 10 17. (a) n ≥ 51 (b) n ≥ 4 19. (a) n ≥ 37 (b) n ≥ 3 13. (a) n ≥ 8 21. (a) 78 (b) 7 1 0 23. f (4) (x) = 0 for all x; therefore by (8.7.3) the theoretical error is zero 1 1 3 T −= = E2 8 3 24 (b) S1 − 0 b a 1 25. (a) T2 − x2 dx = x4 dx = 1 1 5 S −= = E1 24 5 120 29. (a) 49. 4578 (b) 1280.56 31. error ≤ 4. 01 × 10−7 27. Using the hint, Mn = area ABCD = area AEFD ≤ 1 f (x) dx ≤ Tn . 33. 0 4 dx = 4 tan−1 x 1 + x2 1 0 =4 π 4 −0 =π 3. 14159 (a) 3.14141 (b) 3.14159 SECTION 8.8 1. y1 is; y2 is not 11. y = x + C e2x 23. y = 2 e−x + x − 1 3. y1 and y2 are solutions 13. y = 2 3 5. y1 and y2 are solutions 15. y = C ee x 7. y = − 1 + C e2x 2 9. y = 2 2 5 + C e−(5/2)x 21. y = C (x + 1)−2 nx + Cx4 17. y = 1 + C (e−x + 1) 27. y = x2 (ex − e) 19. y = e−x 12 x 2 +C 25. y = e−x ln(1 + ex ) + e − ln 2 29. y = C1 ex + C2 x ex 41. (a) 200 ( 4 )t /5 5 (b) 200 2 ( 4 )t /25 5 35. T (1) ∼ 40. 10◦ ; 1. 62 min = dP = k (M − P ) dt (b) P (t ) = M (1 − e−0.0357t ) (c) 65 days 37. (a) v(t ) = 32 (1 − e−kt ) k (b) 1 − e−kt < 1; e−kt → 0 as t → ∞ 39. (a) i(t ) = E [1 − e−(R/L)t ] R E (b) i(t ) → (amps) as t → ∞ R L (c) t = ln 10 seconds R 43. (a) liters 45. (a) P (t ) = 1000 e( sin 2π t )/π (b) P (t ) = 2000 e( sin 2π t )/π − 1000 SECTION 8.9 1. y = C e−(1/2) cos (2x+3) 9. ln | y + 1| + 3. x4 + 2 =C y2 5. y sin y + cos y = − cos 1 x +C √ 7. e−y = ex − xex + C 1 − x2 15. y + ln | y| = x3 −x−5 3 1 = ln | ln x| + C y+1 11. y2 = C ( ln x)2 − 1 13. sin−1 y = 1 − ...
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