20b_01w_2s

# 20b_01w_2s - Math 20B Second Exam Solutions 4:40 PM March 5...

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Math 20B Second Exam Solutions 4:40 PM March 5, 2001 1. (80 pts.) Evaluate the following integrals. Remember to show your work! (a) [7.5 #9] The answer is 1 - 3 / 2. It can be done as an indeﬁnite integral or by carrying the limits along. There are two reasonable substitutions: x = sin t and 1 - x 2 = u . The ﬁrst, without limits, using dx = cos tdt : Z x 1 - x 2 dx = Z sin t cos t (cos tdt )= Z sin tdt = - cos t + C = - p 1 - x 2 + C. Had the limits been carried along, R 1 / 2 0 over x would become R π/ 6 0 over t . The second substitution in the deﬁnite integral form, using - 2 xdx = du : Z 1 / 2 0 x 1 - x 2 dx = Z 3 / 4 1 - u - 1 / 2 2 du = - u 1 / 2 3 / 4 1 =1 - p 3 / 4 . (b) [7.5 #22] Multiply out and integrate: Z t t + 3 t · dt = Z ( t 3 / 2 + t 5 / 6 ) dt =2 t 5 / 2 / 5+6 t 11 / 6 / 11 + C. (c) Replace cos 2 t with 1 - sin 2 t and let sin t = u : Z sin 2 t cos 3 tdt = Z sin 2 t (1 - sin 2 t ) cos tdt = Z u 2 (1 - u 2 ) du = u 3 / 3 - u 5 / 5+ C = sin 3 t 3 - sin 5 t 5 + C.

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20b_01w_2s - Math 20B Second Exam Solutions 4:40 PM March 5...

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