tapia (jat4858) – HW02 – clark – (52990) 1This print-out should have 24 questions.Multiple-choice questions may continue o nthe next column or page – find all choicesbefore answering.001 10.0 pointsRewrite the expressionf(x) =4x − 1x2(x − 2)using partial fractions.1. f(x) =1x2−7x − 22. f(x) = −1x2+7x − 23. f(x) =7x−12x2+74(x − 2)4. f(x) =12x−74x2−74(x − 2)5. f(x) = −74x+12x2+74(x − 2)correctExplanation:We have to find A, B, and C so that4x − 1x2(x − 2)=Ax+Bx2+Cx − 2=Ax(x − 2) + B(x −2) + Cx2x2(x − 2).Thus4x − 1 = Ax(x − 2) + B(x − 2) + Cx2.Nowx = 0 =⇒ B =12,whilex = 2 =⇒ C =74.But then on comparing coefficients of x2wesee thatA + C = 0 =⇒ A = −74.Consequently,f(x) = −74x+12x2+74(x − 2).002 10.0 pointsIn the partial fractions decomposi tion ofthe expressionf(x) =x3+ 2x + 3x2+ x − 2,find the term having denominator x + 2.1.1x + 22.3x + 2correct3. −3x + 24. −1x + 25.2x + 26. −2x + 2Explanation:As f(x) = P (x)/Q(x) with degP ≥ degQ,we begin with long divi sion:x −1x2+ x − 2 x3+0x2+2x +3x3+x2−2x−x2+4x +3−x2−x +25x +1Thusf(x) = x − 1 +5x + 1x2+ x − 2.On the other hand,x2+ x − 2 = (x − 1)(x + 2) ,tapia (jat4858) – HW02 – clark – (52990) 2so we look for B and C satisfying5x + 1x2+ x − 2=Bx − 1+Cx + 2.Multiplying through by (x −1) (x + 2) gives5x + 1 = B(x + 2) + C(x − 1) .so after substituting x = 1 and x = −2, wesee t hatB = 2 , C = 3 .Consequently, f (x) has partial fraction de-compositionf(x) = x −1 +2x − 1+3x + 2.keywords: partial fractions003 10.0 pointsDetermine the indefinite integralI =Zx + 8(x + 2)(x − 4)dx .1. I = ln(x − 4)2|x + 2|+ C correct2. I = lnx − 4x + 2+ C3. I = ln(x − 4)2x + 2+ C4. I = ln|x + 2|(x − 4)2+ C5. I = lnx + 2(x − 4)2+ CExplanation:First we have to determine the partial frac-tion decompo sitionx + 8(x + 2) ( x − 4)=Ax + 2+Bx − 4.Multiply through by (x + 2)(x − 4). Thenx + 8 = A( x − 4) + B(x + 2) .Setting x = 4 gives 12 = 6B, i.e, B = 2, whilesetting x = −2 gives 6 = −6A, i.e., A = −1.Thus,I =Z−1x + 2+2x − 4dx= −ln (|x + 2|) + 2 ln (|x − 4|) + C .Consequently,I = ln(x − 4)2|x + 2|+ Cwith C an arbitrary constant.004 10.0 pointsFind the unique function y satisfying theequationsdydx=3(x − 1)(5 − x), y(2) = 0.1. y = 3ln5 − xx − 1− ln(3)2. y =34ln5 − xx − 1− ln(3)3. y = 3lnx − 15 − x+ ln(3)4. y =34lnx − 15 − x+ ln(3)correct5. y =14lnx − 15 − x+ ln(3)Explanation:We first find A, B so that3(x − 1)(5 − x)=Ax − 1+B5 − xby bringing the right hand side to a commondenominator. In this case,3(x − 1)(5 − x)=A(5 − x) + B(x − 1)(x − 1)(5 − x),tapia (jat4858) – HW02 – clark – (52990) 3and soA(5 − x) + B(x −1) = 3.To find the values of A, B particular choices ofx are made When x = 1 , for instance, A =34,while when x = 5, B =34. Thusdydx=341x − 1+15 − x.Hence after integration,y =34(ln (|x − 1|) − ln (|5 − x|)) + C=34lnx − 15 − x+ Cwith C an arbitrary constant. Buty(2) = 0 =⇒ C = −34ln13,i.e.,C =34ln(3).Consequently,y =34lnx − 15 − x+ ln(3).005 10.0 pointsEvaluate the integralI =Z102x2(x + 1)(x2+ 1)dx .1. I = ln(8) −π22. I =12ln(8) −π2correct3. I =π2− ln(2)4. I =12π2− ln(2)5. I =12ln(2) +π26. I = ln(2) +π2Explanation:By partial fractions,2x2(x + 1)(x2+ 1)=Ax + 1+Bx + Cx2+ 1.To determine A, B, and C multiply throughby (x + 1)(x2+ 1): for then2x2= A(x2+ 1) + (x + 1)(Bx + C)= ( A + B)x2+ (B + C)x + (A + C) ,which after comparing coefficients g ivesB = −C , A = −C , B = 1 .ThusI =Z101x + 1+x − 1x2+ 1dx=Z101x + 1dx +Z10xx2+ 1dx−Z101x2+ 1dx,and soI =hln(x + 1) +12ln(x2+ 1)− tan−1(x)i10=12hln((x + 1)2(x2+ 1)) − 2 tan−1(x)i10.Consequently,I =12ln(8) −π2.006 10.0 pointsDetermine the integralI =Zx2− 3x + 5x2− x − 2dx .tapia (jat4858) – HW02 – clark – (52990) 41. I = x − ln(x + 1)3x − 2+ C correct2. I = x − ln(x + 1)3(x − 2)+ C3. I = x − ln(x − 1)3x + 2+ C4. I = x + ln(x + 1)3x − 2+ C5. I = x + ln(x − 1)3(x + 2)+ C6. I = x + ln(x − 1)3x + 2+ CExplanation:By division,x2− 3x + 5x2− x − 2=(x2− x − 2) − 2x + 7x2− x − 2= 1 −2x − 7x2− x − 2.But by partial fractions,2x − 7x2− x − 2=3x + 1−1x − 2.ThusI =Z1 −3x + 1+1x − 2dx .NowZ3x + 1dx = 3 ln |x + 1| + C1,whileZ1x − 2dx = ln |x − 2| + C2.Consequently,I = x − ln(x + 1)3x − 2+ C.keywords: division, partial007 10.0 pointsAt a branch of UT there ar e approximately37000 students. A flu epidemic is spreadingamong these students at a rate proportional tothe product of the number of infected studentsand the number of students who have not beeninfected.Initially 1000 student s were infected, and10 days later 6000 were infected. What per-cent a ge of the student body was infected a fter20 days?(Hint: to simpl ify the algebra, let N(t) bethe number of infected students (in mult iplesof 1000).)1. 3.61 %2. 3.88 %3. 57.42 % correct4. 55.86 %5. 78.07 %Explanation:The product of the number infected andthe number not infected is N(37 − N ), so themodel says thatdNdt= kN(37 − N)where k is a constant. After separating vari-ables and integrating this becomesZ1N(37 − N )dN = kZdt = kt + Cwith C an arbitrary constant. To integratethe left hand side set1N(37 − N )=A1N+A237 − N.tapia (jat4858) – HW02 – clark – (52990) 5But then1N(37 − N )=A1(37 − N) + A2NN(37 − N ),so1 = A1(37−N)+A2N = 37A1+(A2−A1)N.ThusA1=137, A2=137.Consequently,Z1N(37 − N )dN=137Z1NdN +137Z137 − NdN=137lnN37 − N.ThusN37 − N= Ae37kt,so the general solution of (†) is given byN(t) =37AA + e−37ktwhere A is an arbitrary constant.The data(‡) N(0) = 1, N(10) = 6determine the values of A and k. Indeed, fromthe first equation in (‡) it follows that37A = (A + 1), i .e., A =136,soN(t) =371 + 36e−37kt.On the other hand, to determine k we use thesecond equation in (‡), for then it follows that6 =371 + 36e−370k,i.e.,−37k =110lnn1(31)(36)(6)o= −0.1 94129.HenceN(t) =371