142PP1W2021T2

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University of British Columbia**We aren't endorsed by this school

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MATH 142

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Mathematics

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Dec 19, 2024

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MATH 142 – Calculus II for Management andEconomicsPractice Problems #1 with Full SolutionsPaul Tsopm´en´eJanuary 11, 2022Topics: Antiderivatives and Indefinite Integrals (Introduction)1.Determine whetherF(x) is an antiderivative off(x).[Solution onpage 2](a)f(x) = 9x8,F(x) =x9(b)f(x) =-5x-4,F(x) =x-5(c)f(x) =2x√1+x2,F(x) =√1 +x2(d)f(x) =xex2,F(x) =ex2(e)f(x) = 1 + lnx,F(x) =xlnx2.Determine whether each equality is correct.[Solution onpage 3](a)Z3x2dx=x3+C(b)Ze-xdx=e-x+C(c)Z(1 + 2x)e2xdx=xe2x+C(d)Z-x4-3x2+ 2x(x2+ 1)2dx=x3+ 1x2+ 1+C1

2Solution to Practice Problems #11.[Exercise onpage 1]AntiderivativesLetfbe a function.A functionFis called anantiderivativeoffon anintervalIif the derivative ofF(x) is equal tof(x); that is,F0(x) =f(x)for allxin IFor example, the functionF(x) =x2is an antiderivative off(x) = 2xbecauseF0(x) =ddx[x2] = 2x. So(a)To determine whetherF(x) =x9is an antiderivative off(x) = 9x8, we need tofind the derivative ofF(x) and see if it is equal tof(x).F0(x) =ddxx9= 9x9-1ddx[xn] =nxn-1= 9x8SinceF0(x) =f(x), it follows thatF(x) is an antiderivative off(x).(b)To determine whetherF(x) =x-5is an antiderivative off(x) =-5x-4, we needto find the derivative ofF(x) and see if it is equal tof(x).F0(x) =ddx[x-5] =-5x-5-1ddx[xn] =nxn-1=-5x-6SinceF0(x) is not equal tof(x), it follows thatF(x) is not an antiderivative off(x).(c)We need to find the derivative ofF(x) =√1 +x2and see if it is equal tof(x) =2x√1+x2.F0(x) =ddxh√1 +x2i=ddxh(1 +x2)12i√x=x12=12(1 +x2)12-1ddx[x2]ddx[(g(x))n] =n(g(x))n-1ddx[g(x)]=12(1 +x2)-12(2x)ab-c=a-bcb,ddx[xn] =nxn-1=121(1 +x2)12(2x)x-m=1xm=2x2√1 +x2=x√1 +x22022 Paul Tsopm´en´e All Rights Reserved

3SinceF0(x) is not equal tof(x), it follows thatF(x) is not an antiderivative off(x).(d)We need to find the derivative ofF(x) =ex2and see if it is equal tof(x) =xex2.F0(x) =ddxhex2i=ex2ddx[x2]ddxeg(x)=eg(x)ddx[g(x)]=ex2(2x) = 2xex2SinceF0(x)6=f(x), it follows thatF(x) is not an antiderivative off(x).(e)F0(x) =ddx[xlnx] =ddx[x] lnx+xddx[lnx]Product Rule= (1) lnx+x1xddx[lnx] =1x= lnx+xx= lnx+ 1SinceF0(x) =f(x), it follows thatF(x) is an antiderivative off(x).2.[Exercise onpage 1]Indefinite IntegralsLetfbe a continuous function on an intervalI. IfFis a function such thatF0(x) =f(x) for allxinI, we say that theindefinite integraloff(x) onIisF(x) +C, whereCis a constant (an arbitrary real number), and we writeZf(x)dx=F(x) +CThe symbolRis the integral sign.The functionf(x) is calledintegrand.The termdxindicates that the variable of integration isx.In otherwords,dxtells us the integral is taken with respect tox.The arbitrary real numberCis called anintegration constant.For example, since the derivative ofx2is 2x, it follows thatR2x dx=x2+C.Warning.Do not forget the integration constant when evaluating an indefiniteintegral.(a)The equalityZ3x2dx=x3+Cis correct because the derivative ofx3is 3x2.2022 Paul Tsopm´en´e All Rights Reserved

4(b)The equalityZe-xdx=e-x+Cis not correct because the derivative ofe-xisnot equal toe-x. Indeed,ddx[e-x] =e-xddx[-x] =e-x(-1) =-e-x6=e-x(c)To see if the equalityZ(1 + 2x)e2xdx=xe2x+Cis correct, we need to find thederivative ofxe2xand see if it is equal to (1 + 2x)e2x.ddxxe2x=ddx[x]e2x+xddx[e2x]Product Rule:ddx[f(x)g(x)]=f0(x)g(x) +f(x)g0(x)= (1)e2x+xe2xddx[2x]ddxeg(x)=eg(x)ddx[g(x)]=e2x+xe2x(2)= (1 + 2x)e2xFactor oute2xSinceddx[xe2x] = (1 + 2x)e2x, the following equality is correct.Z(1 + 2x)e2xdx=xe2x+C(d)Letf(x) =-x4-3x2+ 2x(x2+ 1)2andF(x) =x3+ 1x2+ 1We need to find the derivative ofF(x) and see if it is equal tof(x). Rememberthe quotient rule:ddxf(x)g()x=f0(x)g(x)-f(x)g0(x)[g(x)]2Using this, we haveF0(x) =ddxx3+ 1x2+ 1=ddx[x3+ 1](x2+ 1)-(x3+ 1)ddx[x2+ 1](x2+ 1)2=3x2(x2+ 1)-(x3+ 1)(2x)(x2+ 1)23x4+ 3x2-2x4-2x(x2+ 1)2=x4+ 3x2-2x(x2+ 1)2SinceF0(x)6=f(x), the equalityRf(x)dx=F(x) +Cis not correct.2022 Paul Tsopm´en´e All Rights Reserved