MAT137Y1 Midterm: 2007 Test 3 solution
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MAT137Y1 Full Course Notes
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Mat 137y, 2007 2008 winter session, solutions to term test 3: evaluate the following integrals, and simplify your answer. (cid:90) dx = x + 3 + 2 x2 + 3x + 3ln|x| 1 x. +c. x2 dx. (cid:90) (x + 1)3 (cid:90) x3 + 3x2 + 3x + 1 (cid:90) x2 e2x. 1 + e4x dx. (8%) (i) (8%) (ii) (cid:90) (10%) (iii) x2 (x2 + 64)3/2 dx. Then du = 2e2x dx and we have (cid:90) By trigonometric substitution, let x = 8tan , so dx = 8sec2 d . 8sec2 d = d sec d = sec sec cos d (cid:90) sec2 1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:90) tan2 (cid:90) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . 8 x2 + 64 + x| x2 + 64. By parts, we let u = ln(x2 9), dv = dx, so du = 2x v = x. I = xln(x2 9) dx = xln(x2 9) 2 (cid:90) (cid:18) (cid:19) 9 x2 9 dx and x2 9 (cid:90)
