MAT126 Fall 2009Practice FinalThe actual Final exam will consist of twelve problems1

Problem 11. EvaluateRπ/2π/3sin3(x) cos2(x)dx2. EvaluateRsin2(x)dxSolution:1.First, recognize that the integrand consists of sin(x) raised to an odd power,multiplied by cos(x) raised to an even power. Such a function is a prime candidate forthe substitutionu=cos(x) (so that one of the sin(x) can be used for the change ofvariable, and the rest may be expressed in terms of cos(x) via the identity sin2(x)=1-cos2(x)).Therefore we substituteu=cos(x), wherebydu=-sin(x)dx, and we can write(noting thatu=cos(π/3)=1/2 whenx=π/3, andu=cos(π/2)=0 whenx=π/2)Zx=π/2x=π/3sin3(x) cos2(x)dx=Zx=π/2x=π/3(-sin2(x) cos2(x))(-sin(x)dx)=Zx=π/2x=π/3(-1+cos2(x))(cos2(x))(-sin(x)dx)=Zu=0u=1/2(u2-1)(u2)du=Zu=0u=1/2(u4-u2)du=u55-u33u=0u=1/2=(0)55-(0)33!-(1/2)55-(1/2)33!=-2-55+2-332.There are two ways to approach this problem. The first is to use the identitycos(2x)=cos2(x)-sin2(x)=1-2 sin2(x)to getsin2(x)=12(1-cos(2x))2

This implies thatZsin2(x)dx=Z12(1-cos(2x))dx=Z12dx-Z12cos(2x)dx=12x-12Zcos(2x)dx=12x-12sin(2x)2+C=12x-14sin(2x)+CThe second approach is to use an integration by parts, settingu=sinxanddv=sinxdx; this means thatdu=cosxdxandv=-cosx. Therefore, integration by partsgivesZsin2(x)dx=(sinx)(-cosx)-Z(-cosx)(cosxdx)=-sinxcosx+Zcos2(x)dxNow we plug in the identity cos2(x)=1-sin2(x) to getZsin2(x)dx=-sinxcosx+Z(1-sin2(x))dx=-sinxcosx+Zdx-Zsin2(x)dxAddingRsin2(x)dxto both sides of the equation gives2Zsin2(x)dx=-sin(x) cos(x)+Zdx=-sin(x) cos(x)+x+Cso thatZsin2(x)dx=12x-12sin(x) cos(x)+CNotice that both approaches give the same answer, because of the identity sin(2x)=2 sin(x) cos(x).3

Problem 21. Estimate the integral8Z7dxln(x)using three rectangles and(a) right endpoints(b) left endpoints(c) Are your answers in 1a and 1b over- or under-estimates of the actual inte-gral?2. Do the same for the under-integral functionf(t)=et3Solution: