# M408D SPRING 2015 HW #1 - cheatham (sc36975) HW01 um...

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cheatham (sc36975) – HW01 – um – (53890)1Thisprint-outshouldhave17questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.00110.0 pointsEvaluate the integralI=913x(x+ 2)2dx .1.I=8152.I=353.I=234.I=45correct5.I=1115Explanation:Setu2=x. Then 2u du=dx, whilex= 1=u= 1x= 9=u= 3.In this case,I= 6311(u+ 2)2du= 61u+ 231.ThusI=61513=45.00210.0 pointsThe graph offhas slopedfdx=x2x2+ 1and passes through the point (2,3). Find they-intercept of this graph.1.y-intercept =22.y-intercept =533.y-intercept =834.y-intercept =735.y-intercept =43correctExplanation:The functionfsatisfies the equationsf(x) =x2x2+ 1dx,f(2) = 3.To evaluate the integral setu= 2x2+ 1. Forthendu= 4x dx, in which casef(x) =14u1/2du=16u3/2+C=16(2x2+ 1)3/2+C ,whereChas to be chosen so thatf(2) = 3,i.e.,C+92= 3.Thusf(x) =16(2x2+ 1)3/227+ 3.Consequently, the graph hasy-intercept =43.00310.0 pointsEvaluate the integralI=ln 20ex4ex+ 1dx .1.I=1235correct2.I=343 +5
cheatham (sc36975) – HW01 – um – (53890)23.I=143 +54.I=123 +55.I=34356.I=1435Explanation:Setu= 4ex+ 1. Thendu= 4exdx, whilex= 0=u= 5x= ln 2=u= 9.In this case,I=14951u1/2du=12u1/295.Consequently,I=1235.00410.0 pointsEvaluate the integralI=124x6x3dx.1.I= 62.I=163correct3.I=2034.I= 75.I=463Explanation:Setu=x3. Thendu=dx,so124x6x3dx=91u3udu=91u1/23u1/2duConsequently,I=91u1/23u1/2du=23u3/26u1/291=163.00510.0 pointsDetermine the integralI=11 + 9(x5)2dx .1.I= 3 tan1x53+C2.I= sin13(x5) +C3.I= 3 sin1x53+C4.I=13tan13(x5) +Ccorrect5.I= tan13(x5) +C6.I=13sin13(x5) +CExplanation:Sinceddxtan1x=11 +x2,the substitutionu= 3(x5) is suggested.For thendu= 3dx, in which caseI=1311 +u2du=13tan1u+C ,withCan arbitrary constant. Consequently,I=13tan13(x5) +C.
cheatham (sc36975) – HW01 – um – (53890)3keywords:00610.0 points

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