MATH 132 Midterm: MATH132 Exam 1 2009 Spring Solution
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= (area under y = 6 for 2 x 3) 8 (b) [10%] by the fundamental theorem of calculus: G (3) = f (3) = 6 , G (8) = f (8) = 11 8 = 3 . 2 f (t) dt f (t) dt =z 5 f (t) dt +z 8. = (area under y = 6 for 2 x 5) + (area under y = 11 x for 5 x 8) 399 u133 + c (3x + 8)133 + c u = 3x + 8 du = 3 dx. 3 du = dx (b) [5%] write integrand as substitute: u = 3x2 du = 6x dx du = x dx. 1 (c) [5%] method 1: substitute in the de nite integral. u = 2 x = 1 = u = 1. 6 arcsin u + c arcsin(cid:0)3x2(cid:1) + c ( eu) du eu du. Method 2: substitute in associated inde nite integral, then use ftc. u =
