MATH 172 Midterm: MATH 172 TAMU 172-Spring 18 Exam 1 Solutions
Document Summary
0 (a) ln(3) (b) ln(6) (c) (d) (e) Solution: e we make the substitution u = 2 + tan(2x), du = 2 sec2(2x) dx: X2 + 1 dx (b) 5 (c) 10 (d) 15 (e) 20. Solution: b we make the substitution u = x2 + 1, du = 2x dx: Evaluate z 2 (a) 5e4 1 (b) 4e4 + 1 (c) 3e4 2 (d) e4 4 (e) e4 + e2. Solution: a we use integration by parts, starting with u = 4x2, dv = e2x dx. From this we see that du = 8x dx, v = 1. 2 e2x 8x dx = 2x2e2x z 4xe2x dx. From here we use integration by parts again, with u = 4x, dv = e2x dx. Finding du = 4 dx and v = 1. Z 4x2e2x dx = 2x2e2x (cid:18)2xe2x z 2e2x dx(cid:19) = 2x2e2x 2xe2x + e2x + c.
