tapia (jat4858) – HW01 – clark – (52990) 1This print-out should have 22 questions.Multiple-choice questions may continue o nthe next column or page – find all choicesbefore answering.001 10.0 pointsEvaluate the integralI =Z6eln(x)x2dx .1. I =6e−12ln(6) + 12. I =6e−12ln(6) − 13. I =2e−16ln(6) − 14. I =2e−16ln(6) + 1correct5. I =6e+12ln(6) + 16. I =2e+16ln(6) + 1Explanation:After integration by parts,I = −h1xln(x)i6e+Z6e1x2dx .ThusI = −h1xln(x) +1xi6e.Consequentl y,I =2e−16ln(6) + 1.002 10.0 pointsDetermine the integralI =Z4x(ln(x))2dx .1. I = 4x2(ln(x))2+ ln(x) +12+ C2. I = 2x2(ln(x))2+ ln(x) +12+ C3. I = 4x2(ln(x))2− ln(x) +12+ C4. I = 4x2(ln(x))2+ ln(x) −12+ C5. I = 2x2(ln(x))2− ln(x) +12+ Ccorrect6. I = 2x2(ln(x))2− ln(x) −12+ CExplanation:After integration by parts,Zx(ln(x))2dx =12x2(ln(x))2−Zx21xln(x) dx=12x2(ln(x))2−Zx ln(x) dx.But after integration by parts once again,Zx ln(x) dx =12x2ln(x) −12Zx21xdx=12x2ln(x) −12Zx dx=12x2ln(x) −14x2+ C.ThusZx(ln(x))2dx=12x2(ln(x))2−12x2ln(x) +14x2+ C.Consequentl y,I = 2x2(ln(x))2− ln(x) +12+ C.keywords: integration by parts, log function003 10.0 pointstapia (jat4858) – HW01 – clark – (52990) 2Evaluate the definite integralI =Z103e4√xdx .1. I =323e4+ 12. I =383e4+ 1correct3. I =383e4− 14. I =324e4− 15. I =324e4+ 16. I =384e4− 1Explanation:First we use the substitut ion u =√x toeliminate the square root in the integrand.For then12√xdx = du,so dx = 2√xdu = 2u du. ThusI = 6Z10u e4udu.This last integra l we evaluate using integra-tion by parts. HenceI =h32ue4ui10−32Z10e4udu=h32ue4u−38e4ui10,and soI =383e4+ 1.004 10.0 pointsEvaluate the integralI =Zπ/40(4 x − 5) cos 2x dx .1. I =12π + 32. I = π − 73. I =12π −72correct4. I = π + 35. I =12π +32Explanation:After integration by parts,I =12h(4 x − 5) sin 2xiπ/40−12Zπ/40sin 2xddx(4 x − 5) dx=12π −52− 2Zπ/40sin 2x dx=12π −52+ [cos 2x]π/40.Consequentl y,I =12π −72.005 10.0 pointsEvaluate the integralI =Zπ0e−xcos x dx .1. I = eπ− 12. I = −12(eπ+ 1)3. I =12(1 − e−π)4. I = 1 − e−πtapia (jat4858) – HW01 – clark – (52990) 35. I =12(e−π+ 1) correct6. I =12(eπ+ 1)7. I = e−π+ 18. I = eπ+ 1Explanation:After integration by parts,I = −he−xcos xiπ0−Zπ0e−xsin x dx= e−π+ 1 −Zπ0e−xsin x dx .To evaluate this last integral we integrate byparts once again. For thenZπ0e−xsin x dx = −he−xsin xiπ0+Zπ0e−xcos x dx = 0 + I ,in which caseI = e−π+ 1 − I .Consequentl y,I =12(e−π+ 1).006 10.0 pointsDetermine the integralI =Zx(2 sec2x + tan2x) dx .1. I = 4(tan x −ln |sec x|) − x + C2. I = 4(tan x + ln |sec x|) − x2+ C3. I = 3(x tan x − ln |sec x| ) −12x2+ Ccorrect4. I = 3( ta n x − ln |sec x|) − x + C5. I = 3(x tan x + ln |sec x|) −12x2+ C6. I = 4(x tan x + ln |sec x|) −12x + CExplanation:Astan2x = sec2x − 1 ,the integrand can be rewritten asx(2 sec2x + (sec2x − 1)) = x(3 sec2x − 1) .ThusI =Zx(3 sec2x − 1), dx= 3Zx sec2x dx −12x2.On the other hand,ddx(tan x) = sec2x ,so integration by parts is suggested for evalu-ating this last integral, since then3Zx sec2x dx = 3(x tan x −Ztan x dx) .ButZtan x dx = ln |sec x| + C .Consequentl y,I = 3(x tan x − ln |sec x|) −12x2+ Cwith C an arbitrar y constant.keywords: integration by parts, trig function007 10.0 pointsThe base of a solid S is the region enclosedby the graphs ofy =pln(x), x = e, y = 0 .tapia (jat4858) – HW01 – clark – (52990) 4If the cross sections of S perpendicular to thex-axis are squares, determine the volume, V ,of S.1. V =23cu. units2. V = 1 cu. unit correct3. V = 2(e3− 1) cu. units4. V =12cu. units5. V =13(e3− 1) cu. unitsExplanation:The base of the solid S is given by1 21eSince the square cross-section has side-lengthy, its area is given by A(y) = y2= ln(x).ThusV =Ze1A(y) dy=Ze1y2dx =Ze1ln(x) dx .After Integration by Parts, therefore,V =hx ln(x) − xie1.Consequentl y,volume = e − (e − 1) = 1 cu. unit .008 10.0 pointsDetermine the integralI =Z(2 sin(θ) + sin3(θ)) dθ .1. I = 3 cos(θ) +13cos3(θ) + C2. I = −3 cos(θ) +13cos3(θ) + C correct3. I = 3 cos(θ) −13sin3(θ) + C4. I = −3 cos(θ) −13cos3(θ) + C5. I = −3 cos(θ) −13sin3(θ) + C6. I = 3 cos(θ) +13sin3(θ) + CExplanation:After simplification, we see that2 sin(θ) + sin3(θ) = sin(θ)(2 + sin2(θ)) .On the other hand,sin2(θ) = 1 − cos2(θ) .Thus the integrand can be rewritten assin(θ)(2 + (1 − cos2(θ)))= sin(θ)(3 − cos2(θ)) .As this is now of the form sin(θ) f(cos(θ)), thesubsitution x = cos(θ) is suggested. For thendx = −sin(θ) dθ ,in which caseI = −Z(3 − x2) dx = −3x −13x3+ C .Consequentl y,I = −3 cos(θ) +13cos3(θ) + Cwith C an arbitrar y constant.keywords: trig identity, trig function, integraltapia (jat4858) – HW01 – clark – (52990) 5009 10.0 pointsThe shaded region inπ2π3π2xyis bounded by the graph off(x) = 5 cos3(x)on [0, 3π/2] and the the x-axis. Find the areaof this region.1. area = 1 5π2. area = 1 0π3. area =1524. area = 1 0 correct5. area =152π6. area = 1 5Explanation:The area of the shaded region is given byI =Z3π/20|5 cos3(x)|dxwhich as the graph shows can in turn be writ-ten asI =Zπ/205 cos3(x) dx−Z3π/2π/25 cos3(x) dx .Sincecos2(x) = 1 − sin2(x) ,we thus see thatI =nZπ/20−Z3π/2π/2o5 cos(x)(1−sin2(x)) dx .To evaluate these integrals, set u = sin(x).For thendu = cos(x) dx ,in which caseZπ/205 cos(x)(1 − sin2(x)) dx= 5Z10(1 − u2) du= 5hu −13u3i10=103,whileZ3π/2π/25 cos(x)(1 − sin2(x)) dx= 5Z−11(1 − u2) du= 5hu −13u3i−11= −203.Consequentl y, the shaded region hasarea = 10 .keywords:010 10.0 pointsEvaluate the integralI =Zπ0p1 + sin(θ) dθ .1. I =√3 − 12. I = 3 −√33. I = 1tapia (jat4858) – HW01 – clark – (52990) 64. I =√3 + 15. I = 4 correct6. I = 2Explanation:By the double angle formula,sin(2x) = 2 sin(x) cos(x) ,while by the Pythagorean ident ity,cos2(x) + sin2(x) = 1 .Thus1 + sin(2x) = cos2(x) + 2 sin(x) cos(x) + sin2(x)= (cos(x) + sin(x))2.Taking x = θ/2, we thus see thatp1 + sin(θ) =q(cos(θ/2) + sin(θ/2))2= cos(θ/2) + sin(θ/2)whenevercos(θ/2) + sin(θ/2) ≥ 0 ,hence for 0 ≤ θ ≤ π. ThusI =Zπ0cosθ2+ sinθ2dθ= 2hsinθ2− cosθ2iπ0.Consequentl y,I = 4 .011 10.0 pointsDetermine the integralI