intsubsolong_homework solution - Spring 2003 1. Solutions...

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Spring 2003Math 122Integration by Substitution1. Solutions for the integrals:b. For the integral below, we make the substitutionu= 2x2-3, sodu= 4x dx. It followsthat:Zxp2x2-3dx=14Z(2x2-3)12(4x)dx=14Zu12du=16u32+C=16(2x2-3)32+Cd. For the integral below, we make the substitutionu=x2+ 4x-5, sodu= (2x+ 4)dx.It follows that:Zx+ 2(x2+ 4x-5)3dx=12Z(x2+ 4x-5)-3(2x+ 4)dx=12Zu-3du=-14u-2+C=-14 (x2+ 4x-5)2+Cf. For the integral below, we make the substitutionu=-(x2-2x), sodu=-(2x-2)dx.It follows that:Zx-1ex2-2xdx=-12Ze-(x2-2x)(-2(x-1))dx=-12Zeudu=-12eu+C=-12e-x2+2x+Ch. For the integral below, we make the substitutionu= cos(2x), sodu=-2 sin(2x)dx. Itfollows that:Zsin(2x)cos(2x)dx=-12Z1cos(2x)(-2 sin(2x))dx=-12Z1udu=-12ln|u|+C=-12ln|cos(2x)|+C,
i. For the integral below, we make the substitutionu=x12, sodu=12x-12dx. It followsthat:Zexx, dx=2Zex12x-12dx=2Zeudu=2eu+C=2ex+C2. Solutions for the differential equations.

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Term
Fall
Professor
staff
Tags
Calculus, Integrals, Integration By Substitution, Trigraph, 3 g, 5 km, 0 0098 km

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