APPM 1345 Midterm: appm1345spring2018exam2_sol
Document Summary
Spring 2018: [20 pts] evaluate the following integrals. (a)z (t 1)2. 0 sec2 (1 + 26 tan )4/3 d (d)z x (2x + 1)3 dx. 5 t4 dt =z t2 2t + 1 t4/5 t11/5 . 11 dt =z (cid:16)t6/5 2t1/5 + t 4/5(cid:17) dt t6/5 + 5t1/5 + c = 5 5 t(cid:18) 1 t2 . 11 t + 1(cid:19) + c (b) note that (cid:12)(cid:12)(cid:12) = (4 2) ( 8 8) + (8 8) (2 4) = 20. +(cid:18) x2 (c) let u = 1 + 26 tan = du = 26 sec2 d = d = du. Also, = 0 = u = 1 + 26 tan 0 = 1 and. = u = 1 + 26 tan(cid:16) . 4(cid:17) = 27. (1 + 26 tan )4/3 d =z 27 sec2 . 1 (d) let u = 2x + 1 = du = 2 dx = dx = du.
