IntBySubstitution_19 - November 18, 2014 initial value)...

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November 18, 2014Worksheet 19: Integration by Substitution (indefinite, definite,initial value)1. FindZx+1x1-1x2dxin two ways:(a) by a substitution;
(b) by multiplying out and integrating term by term. Show that your two answers are thesame.
2. FindZx(x2+ 1)2dxZx(x2+ 1)2dx=Z1u2du2=12Z1u2du=-12u+C=-12(x2+ 1)+C.u=x2+ 1du= 2x dxdu2=x dx3. FindZ41exxdx.Express your answer in terms of whole numbers ande.Z41exxdx=Z21eu(2du)= 2Z21eudu= 2eu21= 2e2-2e.u=xdu=12xdx2du=1xdxwhenx= 1, u=1 = 1whenx= 4, u=4 = 2
4. Find the area under the graph of
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Term
Fall
Professor
Lorenzo Sadun
Tags
Integration By Substitution, Constant of integration, Boundary value problem, dx

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