142PP4W2021T2

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MATH 142
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Dec 19, 2024
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MATH 142 – Calculus II for Management andEconomicsPractice Problems #4 with Full SolutionsPaul Tsopm´en´eEdited by Ben RobideauJanuary 20, 2022Topics: Substitution (continued)1.Find each integral.[Solution onpage 2](a)Z-3x4(x5+ 1)7dx(b)Z-4x+ 43p(x2-2x+ 1)2dx(c)Z2xx2-16dx(d)Zt5t6+ 3dt(e)Z18x2-6x2x3-x2dx(f)Ze1-xdx(g)Zx3ex4dx(h)Zx-1e-x2+2xdx(i)Zep1 +epdp(j)Ze2x-e-2xe2x+e-2xdx(k)Zln(3x)xdx(l)Z(lnz)2zdz(m)Z5xxdx(n)Zxx+ 7dx(o)Zx1 + 2xdx2.LetD0(t) = 60t+52t2+ 5t-1dollars per yearbe the rate at which a company incurs debt, wheretis the amount of time (in years)since the company began.Suppose that the company had accumulated $18860 indebt by the fifth year. Find a formula for the total debt aftertyears.[Solution onpage 8]1
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2Solutions to Practice Problems #41.[Exercise onpage 1](a)Letu=x5+ 1. Thendu= 5x4dx, so thatdu5=x4dx. Therefore,Z-3x4(x5+ 1)7dx=Z-3(x5+ 1)7(x4dx)=Z-3u7du5=-3Z1u7du5Zkf(x)dx=kZf(x)dx=-35Zu-7duZkf(x)dx=kZf(x)dx=-35u-7+1-7 + 1+CZxndx=xn+1n+ 1+C=-35u-6-6+C=110u-6+C=110u6+C1xm=x-m=110(x5+ 1)6+CSubstituteubyx5+ 1(b)First, rearrange the integrand as follows:Z-4x+ 43p(x2-2x+ 1)2dx=Z-4x+ 4(x2-2x+ 1)23dxpxq=xqp=Z(-4x+ 4)(x2-2x+ 1)-23dx1xm=x-mNow, letu=x2-2x+1. Then,du= (2x-2)dx. In the integral, we don’t have(2x-2)dx, but (-4x+4)dxinstead. Multiplying the equationdu= (2x-2)dxby-2, we get-2du= (-4x+ 4)dx. Hence, if we substitutex2-2x+ 1 withuand (-4x+ 4)dxwithduinto the integral, we getZ-4x+ 43p(x2-2x+ 1)2dx2022 Paul Tsopm´en´e All Rights Reserved
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3=Z(-4x+ 4)(x2-2x+ 1)-23dxFrom before=Zu-23(-2du)=-2Zu-23duZkf(x)dx=kZf(x)dx=-2u1313+CZxndx=xn+1n+ 1+C=-231u13+Cabc=acb=-6u13+C=-6(x2-2x+ 1)13+CSubstituteubyx2-2x+ 1(c)We make the substitutionu=x2-16. Then,du= 2x dx. Using this, we haveZ2xx2-16dx=Z1x2-16(2x dx)=Z1udu= ln|u|+CZ1xdx= ln|x|+C=lnx2-16 +CSubstituteubyx2-16(d)Letu=t6+ 3. Then,du= 6t5dtso thatdu6=t5dt. We thus haveZt5t6+ 3dt=Z1t6+ 3(t5du)=Z1udu6=16Z1uduZkf(x)dx=kZf(x)dx=16ln|u|+CZ1xdx= ln|x|+C=16lnt6+ 3 +CSubstituteubyt6+ 32022 Paul Tsopm´en´e All Rights Reserved
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4(e)Letu= 2x3-x2, so thatdu= (6x2-2x)dx. Multiplying both sides of thislatter equation by 3, we get 3du= (18x2-6x)dx. Using this, we haveZ18x2-6x2x3-x2dx=Z12x3-x2(18x2-6x)dx=Z1u(3du)= 3Z1uduZkf(x)dx=kZf(x)dx= 3 ln|u|+CZ1xdx= ln|x|+C=3 ln 2x3-x2+CSubstituteuby 2x3-x2(f)We make the substitutionu= 1-x, the exponent one. Then,du=-dx, so-du=dx. Therefore,Ze1-xdx=Zeu(-du)=-ZeuduZkf(x)dx=kZf(x)dx=-eu+C=-e1-x+CSubstituteuby 1-xAlternate SolutionThe integrandf(x) =e1-x=e-x+1is of the formekx+lwithk=-1 andl= 1. So, by using the formulaRekx+ldx=1kekx+l+C, we getZe1-xdx=1-1e1-x+C=-e1-x+C(g)Letu=x4. Then,du= 4x3dxso thatdu4=x3dx. Therefore,Zx3ex4dx=Zex4(x3dx)=Zeudu4=14ZeuduZkf(x)dx=kZf(x)dx2022 Paul Tsopm´en´e All Rights Reserved
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5=14eu+CZexdx=ex+C=14ex4+CSubstituteubyx4(h)Letu=-x2+ 2x, the exponent one.Then,du= (-2x+ 2)dxso thatdu-2= (x-1)dxandZx-1e-x2+2xdx=Z1eudu-2=Ze-udu-2=-12Ze-uduZkf(x)dx=kZf(x)dx=-121-1e-u+CZekxdx=1kekx+C=12e-u+C=12e-(-x2+2x)+CSubstituteuby-x2+ 2x=12ex2-2x+CDistribute(i)Letu= 1 +ep, the quantity in the denominator. Then,du=epdpandZep1 +epdp=Z1udu= ln|u|+CZ1xdx= ln|x|+C=ln|1 +ep|+CSubstituteuby 1 +ep(j)Letu=e2x+e-2x. Thendu= (2e2x-2e-2x)dxsinceddxeg(x)=eg(x)ddx[g(x)].So,du2= (e2x-e-2x)dxandZe2x-e-2xe2x+e-2xdx=Z1e2x+e-2x(e2x-e-2x)dx=Z1udu22022 Paul Tsopm´en´e All Rights Reserved
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6=12Z1uduZkf(x)dx=kZf(x)dx=12ln|u|+CZ1xdx= ln|x|+C=12lne2x+e-2x+CSubstituteubye2x+e-2x(k)Letu= ln(3x). Then,du=u0(x)dx=33xdx=1xdxsinceddx[ln(g(x))] =ddx[g(x)]g(x).So,du=1xdxandZln(3x)xdx=Zln(3x)1xdx=Zu du=u22+CZxndx=xn+1n+ 1+C=[ln(3x)]22+CSubstituteuby ln(3x)(l)Letu= lnz. Then,du=1zdzandZ(lnz)2zdz=Zu2du=u33+CZxndx=xn+1n+ 1+C=(lnz)33+CSubstituteuby lnz(m)The numerator of the integrand is an exponential function with base 5.Wemake the substitutionu=x, the exponent on 5. Then,du=12xdxso that2du=1xdx. Therefore,Z5xxdx=Z5x1xdx=Z5u(2du)= 2Z5uduZkf(x)dx=kZf(x)dx2022 Paul Tsopm´en´e All Rights Reserved
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7= 25uln 5+CZaxdx=axlna+C=2ln 55u+C=2·5xln 5+CSubstituteubyx(n)Firstly,Zxx+ 7dx=Zx(x+ 7)12dxx=x12Now, letu=x+ 7. Then,du=dxandZxx+ 7dx=Zx(x+ 7)12dxFrom before=Zxu12du=Z(u-7)u12duSolveu=x+ 7 forxto getx=u-7=Z(u32-7u12)duDistribute=Zu32du-Z7u12duR(f(x)±g(x))dx=Rf(x)dx±Rg(x)dx=Zu32du-7Zu12duZkf(x)dx=kZf(x)dx=u5252-7u3232+CZxndx=xn+1n+ 1+C=25u52-723u32+Cabc=acb=25u52-143u32+C=25(x+ 7)52-143(x+ 7)32+CSubstituteubyx+ 7(o)Firstly,Zx1 + 2xdx=Zx(1 + 2x)12dxx=x12=Zx(1 + 2x)-12dx1xm=x-m2022 Paul Tsopm´en´e All Rights Reserved
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8Now, letu= 1 + 2x. Then,du= 2dx, sodu2=dxandZx1 + 2xdx=Zx(1 + 2x)-12dxFrom before=Zxu-12du2=12Zxu-12duZkf(x)dx=kZf(x)dx=12Zu-12u-12duSolveu= 1 + 2xforxto getx=u-12=14Z(u-1)u-12duZkf(x)dx=kZf(x)dx=14Zu12-u-12duDistribute=14Zu12du-Zu-12duR(f(x)±g(x))dx=Rf(x)dx±Rg(x)dx=14Zu12du-14Zu-12duDistribute=14u3232-14u1212+CZxndx=xn+1n+ 1+C=1423u32-1421u12+Cabc=acb=16u32-12u12+C=16(1 + 2x)32-12(1 + 2x)12+CSubstituteuby 1 + 2x2.[Exercise onpage 1]The total debt function,D(t), is given byD(t) =ZD0(t)dt=Z60t+52t2+ 5t-1dtTo find this integral, we make the substitutionu=t2+5t-1. Then,du= (2t+5)duso thatdu2=(t+52)du. Therefore,Z60t+52t2+ 5t-1dt=Z60udu2=602Zu duZkf(x)dx=kZf(x)dx= 30Zu12dux=x122022 Paul Tsopm´en´e All Rights Reserved
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9= 30u3232+CZxndx=xn+1n+ 1+C= 3023u32+Cabc=acb= 20u32+C=20(t2+ 5t-1)32+CSubstituteubyt2+ 5t-1We now need to find the constantC. By the fifth year the company had accumulated$18860 in debt which implies thatD(5) = 18860. So,D(5) = 20(52+ 5(5)-1)32+C= 1886020(49)32+C= 188606860 +C= 18860C= 1200Thus, the total debt function isD(t) = 20(t2+ 5t-1)32+ 12000.2022 Paul Tsopm´en´e All Rights Reserved
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