lab8_solns_f11 (1)

# lab8_solns_f11 (1) - Math 116 - Lab 8 - Solutions - Fall...

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Unformatted text preview: Math 116 - Lab 8 - Solutions - Fall 2011. 1. a) 7 e t + 1 2 t 2 + C . 1. b) 3 t ln3 + C . 1. c)- 2cos t + C . 1. d) 1 9 sin(9 t ) + C . 1. e)- 9 13 cot(13 t ) + C . 2. a) Z 6- 1- 2 t + 8 dt = (- t 2 + 8 t ) 6- 1 = [- (6) 2 + 8(6)]- [- (- 1) 2 + 8(- 1)] = 21 . 2. b) Z 27 8 t 1 3 dt = 1 1 + 1 3 t 4 3 27 8 = 3 4 27 4 3- 3 4 8 4 3 = 195 4 . 2. c) Z 2 1 9 t 5 + 8 t 3 + 2 tdt = 9 6 t 6 + 8 4 t 4 + 2 2 t 2 2 1 = 9 6 2 6 + 8 4 2 4 + 2 2 2 2- 9 6 1 6 + 8 4 1 4 + 2 2 1 2 = 255 2 3. a) Net area = R 5 (3 t 2 + 1) dt = ( t 3 + t ) | 5 = 130. 3. b) Net area = Z / 4- / 4 sec 2 t- t 2 dt = (tan t- 1 3 t 3 ) | / 4- / 4 = tan 4- 1 3 4 3- tan- 4- 1 3- 4 3 = 2 1- 3 192 . 4. Since cos x sin x for x [0 ,/ 4] and sin x cos x for x [ / 4 ,/ 2] (plot the graphs!!), the area between the two graphs over [0 ,/ 2] is computed as Z 4 cos x- sin xdx + Z 4 sin x- cos xdx = (sin x + cos x ) | 4 + (- cos x- sin x ) | 2 4 = 1 2 + 1 2- [0 + 1] + [-- 1]...
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## lab8_solns_f11 (1) - Math 116 - Lab 8 - Solutions - Fall...

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