MATH 1205 Midterm: MATH 1205 VT 1205c Test f f05
Document Summary
Math 106: review for final exam, part i - solutions: find the following. [see review for exam ii for integration tips and strategies. ] (a) let u = x3, so du = 3x2dx and du/3 = x2dx. Z 12x2 cos(x3) dx =12z cos(x3)x2 dx du. = 4 sin(x3) + c (b) we"ll use integration by parts: u = x du = dx and dv = e 3x v = e 3x. 3 xe 3x dx xe 3x dx xe 3x dx = lim. So, the integral converges (to this value). (c) this integral is improper at x = 4 because the integrand has a vertical asymptote there, so we split into two integrals. 0 dx dx dx (x 4)2 =z 4 (x 4)2 (x 4)2. = lim a 4 (x 4)2 a 4 z a b 4+z 6 (x 4)2 + lim. 1 (x 4)(cid:12)(cid:12)(cid:12) (x 4)(cid:12)(cid:12)(cid:12) a 4 (cid:20) 1 (0 4)(cid:21) + lim (a 4) 1.
