Saturday, March 10, 2012

Single Variable Calculus, Chapter 5, 5.5, Section 5.5, Problem 48

Find the definite integral 40x1+2xdx

Let u=1+2x, then du=2dx, so dx=du2. When x=0,u=1 and when x=4,u=9. Thus,



40x1+2xdx=40u12udu240x1+2xdx=40u12udu240x1+2xdx=1440u1udu40x1+2xdx=1440uu1udu40x1+2xdx=1440u12u12du40x1+2xdx=14[u12+112+1u12+112+1]4040x1+2xdx=14[u3232u1212]4040x1+2xdx=14[2(9)3232(9)12]14[2(1)3232(1)12]40x1+2xdx=14(12)14(43)40x1+2xdx=3+41240x1+2xdx=3+1340x1+2xdx=9+1340x1+2xdx=103

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