Martens_WA5

# Martens_WA5 - 5.1 6. dr =π dθ r = πθ + C Check via...

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Unformatted text preview: 5.1 6. dr =π dθ r = πθ + C Check via differentiation. dr =π dθ 10. ∫x 1 2 dx Rewrite: ∫x −2 dx Integrate: x −1 +C −1 1 − +C x Simplify: 12. ∫ x( x 2 + 3)dx Rewrite: ∫ (x 3 + 3x)dx Integrate: x4 3 2 + x +C 42 Simplify: x 4 3x 2 x4 6x2 x4 + 6x2 + +C = + +C = +C 4 2 4 4 4 32. ∫ (t 2 − sin t )dt t3 = + cos t + C 3 Check via differentiation. t3 + cos t + C 3 3(3t 2 ) − t 3 (0) − sin t dt 9 2 9t − sin t dt 9 (t 2 − sin t )dt 38 ∫ sec y (tan y − sec y)dy = ∫ (sec y tan y − sec y )dy 2 = sec y − tan y + C Check via differentiation. sec y − tan y + C sec y tan y − sec 2 y 54. dy = 2( x − 1) , [3,2] dx ∫ dx dx = ∫ 2( x − 1)dx y = 2 ∫ xdx − ∫ 2dx x2 y = 2 − 2x + C 2 2 y = x − 2x + C 2 = 32 − 2(3) + C 2 = 9−6+C 2 = 3+C C = −1 2 The equation is: 2 = 3 − 2(3) − 1 dy 56. dy 3 =, dx e [e,3], x>0 y = 3ln | x | +C 3 = 3ln | e | +C 3 = 3+C 3−3 = C C =0 The equation is: 3 = 3ln | e | +0 70. f ''( x) = sin x f '(0) = 1 f (0) = 6 f ''( x) = sin x f '( x ) = − cos x + C 1 = − cos 0 + C 1 = −1 + C C=2 f '( x ) = − cos x + 2 f ( x) = − sin x + 2 x + C 6 = 0 + 2(0) + C C =6 f ( x) = − sin x + 2 x + 6 80. s ''(t ) = −322 s '(t ) = −32t + C1 s (t ) = −16t 2 + C1t + C2 Balloon is rising at 8ft/sec s '(0) = −32(0) + C1 s '(0) = −32(0) + 8 Balloon is at 64’. s (0) = −16(0) 2 + 8(0) + C2 s (0) = −16(0) 2 + 8(0) + 64 s(t ) = −16t 2 + 8t + 64 s(t ) = −8(2t 2 − t − 8) The bag hit the ground at s(t)=0 s (t ) = −8(2t 2 − t − 8) = −16t 2 + 8t + 64 = = −8 ± 64 − 4(16)64 2(16) −8 ± −4032 32 −8 ± 63.4980315 = 32 = 71.4980315 32 t ≈ 2.23431348 s '(2.2340625) = −32(2.2340625) + 8 = -63.49 The balls velocity was about -63.49ft/sec when it hit the ground. 5.2 2. ∑ k (k − 2) k =3 6 ∑ (k k =3 6 k =3 6 2 − k 2) 6 ∑ k 2 − 2∑ k k =3 6(6 + 1)(2(6) + 1) 6(6 + 1) −2 6 2 6(7)(13) 6(7) −2 6 2 546 − 42 6 91 − 42 49 16. ∑ (2i − 3) i =1 15 2∑ i − ∑ 3 i =1 i =1 15 15 15(15 + 1) − 15*3 2 15(16) − 45 2 240 − 45 2 120 − 45 75 5.3 28. ∫ 8 0 (8 − x)dx 1 * b * h to calculate it. 2 This is a triangle: I can use the geometric formula of ∫ 8 0 (8 − x)dx = 1 *b * h = 2 1 *8*8 = 2 64 = 2 32 44. Given ∫ f ( x)dx = 0 and ∫ f ( x)dx = 5 evaluate −1 0 0 −1 1 1 a) ∫ f ( x)dx = −5 b) ∫ ∫ 1 0 f ( x )dx − ∫ f ( x) dx = 5 − (−5) = 10 −1 0 1 c) −1 3 f ( x)dx = 3*0 = 0 d) ∫ 3 f ( x)dx = 3(5) = 15 0 1 5.4 10. ∫ 3 1 (3 x 2 + 5 x − 4)dx = 3 3x3 5 x 2 + − 4x = 3 2 1 45 5 − 12) − (1 + − 4) = 2 2 45 5 (15 + ) − ( − 3) = 2 2 45 5 15 + + 3 − = 2 2 40 18 + = 2 38 (27 + 14. ∫ −1 −2 (u − 1 )du = u2 −1 u 2 −1 − = 2 x −2 −12 −1 −22 −1 − − − = −2 2 −1 2 −1 −1 − 1 − 2 − = −2 2 −1 −1 −1+ −2 = 2 −2 −1 − 2 = =3 18. ∫ 8 1 2 dx = x 8 23/2 x1/2 = 1 (2 8 ) − (23/211/2 ) = 3/2 1/2 ( 8 * 8) − ( 8 *1) = 8− 8 = ≈ 5.17157288 28. ∫ ∫ ∫ π 4 0 1 − sin 2 θ dθ = cos 2 θ cos 2 θ dθ = cos 2 θ 1dθ = π 4 0 π 4 0 x ] 04 = π π −0 = 4 π 4 32. x +1 dx = x 5x 1 ∫1 x + x dx = ∫ 5 1 x ] 1 + ln | x |] 1 = 5 5 (5 − 1) + (ln 5 − ln1) = 4 + ln 5 ≈ 5.60943791 38. ∫ 2e e 1 cos x − dx = x 2e 2e sin x ] e − ln x ] e = (sin 2e − sin e) − (ln | 2e | − ln | e |) = (sin 2e − sin e) − (ln | 2e | −1) sin 2e − sin e + ln | 2e | +1 ≈ -0.749046468 - .410781291 - 1.69314718 + 1 ≈ -1.85297494 5.5 14. ∫ x(4 x 2 + 3) 2 dx = 1 2 2 ∫ 8 x(4 x + 3) dx 8 u = 4x2 + 3 du = 8 xdx 12 u du = 8∫ 1 u3 * +C = 83 1 (4 x 2 + 3)3 * +C 8 3 52. ∫ x sin x dx 2 u = x2 du = 2 xdx 1 2 ∫ 2 x sin x dx = 2 1 sin udu = 2∫ − cos u + C = − cos x 2 + C = 56. ∫ ( x + 1)e x2 + 2 x dx u = x2 + 2x du = 2 x + 2dx = 1 du = ( x + 1)dx 2 1u e du = 2∫ 1 x2 + 2 x e +C 2 62. ∫ cos sin x dx = 3 x −3 ∫ (cos x) (sin x)dx u = cos x du = − sin x (dx ) = −du = sin x(dx ) ∫ (cos x) (sin x)dx = − ∫ u du = −3 −3 − u −2 +C −2 88. ∫x 2 x + 3dx u = 2x + 3 du = 2dx du dx = 2 u −3 x= 2 ∫ u − 3 1/2 du u = 2 2 1 1/2 ∫ (u − 3) * u du = 4 1 3/2 1/2 ∫ (u − 3u )du = 4 1 3/2 1/2 ∫ (u − 3u )du = 4 1 (2 x + 3)5/2 3(2 x + 3)3/2 − +C = 4 5/ 2 3/ 2 1 1 (2 x + 3)5/2 − (2 x + 3)3/2 + C = 10 2 102. ∫ 2 1 e1− x dx u = 1− x du = −1 a2 = 1 − 1 = 0 b2 = 1 − 2 = −1 ∫ ∫ 2 1 e1− x dx = eu (− du ) = −1 0 − ∫ eu (−du ) = −1 0 −(e0 − e1 ) = − 1 − e −1 ≈ −1 − 0.367879441 ≈ -1.36787944 106. ∫ 2 0 x 3 4 + x 2 dx u = 4 + x2 du = 2 xdx du dx = 2 a2 = 4 + 0 2 = 4 b2 = 4 + 22 = 8 ∫ 8 4 u − 4 * u1/3 du = 2 18 u − 4 * u1/3du = 4 ∫4 1 1 1/3 2(u − 4)1/2 u = 4 4 1 [ 8 − 0] = 4 2 8 5.7 6. ∫ 3x + 2 dx u = 3x + 2 du = 3dx 1 1 ∫ 3x + 2 3dx = 3 11 du = 3∫u 1 ln u + C = 3 1 ln 3x + 2 + C 3 1 8. x2 ∫ 3 − x3 dx u = 3 − x3 du = −3 x 2 − du = 3x 2 1 3x 2 dx 3 ∫ 3 − x3 1 ln | 3 − x3 | 3 20. ∫ x ln( x )dx 3 1 1 1 ∫ x ln( x)dx 3 u = ln x 1 du = dx x 1 udu 3∫ 1 u2 3 2 1 (ln x) 2 3 2 30. ∫ tan 5θ dθ = 1 −5sin 5θ −∫ dθ = 5 cos 5θ 1 − | cos 5θ | 5 ...
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## This note was uploaded on 12/30/2009 for the course MAT 231 taught by Professor Thurber during the Spring '09 term at Thomas Edison State.

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