Tutorial

Math V46A Chapter 7 Selected Worked-Out Solutions

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Solutions

7.1 #39.

$$\begin{array}{l}\int \frac{\sqrt{x}+1}{\sqrt[3]{x}}dx=\int \frac{x^{1/2}+1}{x^{1/3}}dx=\int \left(\frac{x^{1/2}}{x^{1/3}}+\frac{1}{x^{1/3}}\right)dx=\int \left(x^{3/6-2/6}+\frac{1}{x^{1/3}}\right)dx=\int \left(x^{1/6}+x^{-1/3}\right)dx\\ =\int x^{1/6}dx+\int x^{-1/3}dx=\frac{x^{7/6}}{\left(\frac{7}{6}\right)}+\frac{x^{2/3}}{\left(\frac{2}{3}\right)}+C=\underset{¯}{\underset{¯}{\frac{6}{7}{x}^{7/6}+\frac{3}{2}{x}^{2/3}+C}}\end{array}$$

7.1 #47.

$$\begin{array}{l}{C}^{\prime }(x)=0.03e^{0.01x}\\ \text{We would like to find}C(x)\text{; use the second formula in the first blue box on page 434 to take the antiderivative:}\\ C(x)=\int 0.03e^{0.01x}dx=0.03\int e^{0.01x}dx\\ C(x)=\frac{0.03}{1}\frac{e^{0.01x}}{0.01}+K=3e^{0.01x}+K\text{(}K\text{is used for constant instead of the usual}C\text{to avoid confusion with}C(x)\text{)}\\ \text{To determine the value of}K\text{, use the fixed cost information: when}x\text{is zero,}C(x)=C(0)=8:\\ C(0)=8=3e^{0.01(0)}+K\\ 8=3e^0+K\\ 8=3(1)+K=3+K\\ 8-3=K\\ K=5\\ \text{So the complete expression for}C(x)\text{is:}\underset{¯}{\underset{¯}{C(x)=3e^{0.01x}+5}}\end{array}$$

7.1 #51.

$$\begin{array}{l}{C}^{\prime }(x)=5x-1/x=5x^1-x^{-1}\\ \text{We would like to find}C(x)\text{; use the power rule (page 430) and the special}{x}^{-1}\text{rule (page 434):}\\ C(x)=\int \left(5x^1-x^{-1}\right)=\int 5x^{1}dx-\int x^{-1}dx=5\int x^{1}dx-\int x^{-1}dx\\ C(x)=5\frac{x^2}{2}-\ln \left|x\right|+K=\frac{5x^2}{2}-\ln \left|x\right|+K\text{(As above,}K\text{is used for constant instead of the usual}C\text{)}\\ \text{To determine the value of}K\text{, use the cost information: when}x\text{is 10,}C(x)=C(10)=94.20:\\ C(10)=94.20=\frac{5{(10)}^2}{2}-\ln \left|10\right|+K\\ \text{In a word problem, we use a decimal approximation for}\ln \left|10\right|\text{to obtain a reasonable value of}K:\\ 94.20\approx \frac{5{(10)}^2}{2}-2.30+K\text{(use a calculator; note that}\left|10\right|=10\text{, so just punch in ln(10))}\\ 94.20\approx 250-2.30+K\\ 94.20\approx 247.70+K\\ K\approx -153.50\\ \text{So the complete expression for}C(x)\text{is:}\underset{¯}{\underset{¯}{C(x)=\frac{5x^2}{2}-\ln \left|x\right|-153.50}}\end{array}$$

7.2 #11.

$$\begin{array}{l}\int 3x^2{e}^{2x^3}dx\\ \text{There are several possibilities for a}u\text{-substitution (ewww}\mathrm{...}\text{we have to guess!)}\text{.}\\ \text{Possibility 1:}\\ u=3x^2;\text{remainder of integrand (the part that isn't}u\text{) is}{e}^{2x^3}dx\\ \frac{du}{dx}=6x\to du=6xdx\text{, which does not resemble the remainder of the integrand}\to \text{REJECT}\\ \text{Possibility 2:}\\ u=e^{2x^3};\text{remainder of integrand (the part that isn't}u\text{) is}3x^{2}dx\\ \frac{du}{dx}=6x^2{e}^{2x^3}\to du=6x^2{e}^{2x^3}dx\text{, which does not resemble the remainder of the integrand}\to \text{REJECT}\\ \text{Possibility 3:}\\ u=2x^3;\text{remainder of integrand (the part that isn't}u\text{or}e\text{) is}3x^{2}dx\\ \frac{du}{dx}=6x^2\to du=6x^{2}dx\text{, which resembles the remainder of the integrand except for a constant}\to \text{AHA!}\\ \text{Dividing both sides of the}du\text{expression by 2 gives an expression that exactly matches the remainder of the integrand:}\\ \frac{du}{2}=3x^{2}dx\\ \text{Substituting}u\text{and}\frac{du}{2}\text{into the original problem yields}\\ \int 3x^2{e}^{2x^3}dx=\int \left(3x^{2}dx\right)e^{\left(2x^3\right)}=\int \left(\frac{du}{2}\right)e^u=\frac{1}{2}\int e^{u}du=\frac{1}{2}\left(e^u\right)+C=\frac{e^u}{2}+C\\ \text{Restoring the original definition of}u\text{gives:}\\ \int 3x^2{e}^{2x^3}dx=\underset{¯}{\underset{¯}{\frac{e^{2x^3}}{2}+C}}\end{array}$$

7.2 #15.

$$\begin{array}{l}\int \frac{e^{1/z}}{z^2}dz=\int z^{-2}{e}^{1/z}dz\\ \text{There are several possibilities for a}u\text{-substitution (ewww}\mathrm{...}\text{we have to guess!)}\text{.}\\ \text{Possibility 1:}\\ u=z^{-2};\text{remainder of integrand (the part that isn't}u\text{) is}{e}^{1/z}dz\\ \frac{du}{dz}=-2z^{-3}\to du=-2z^{-3}dz\text{, which does not resemble the remainder of the integrand}\to \text{REJECT}\\ \text{Possibility 2:}\\ u=e^{1/z};\text{remainder of integrand (the part that isn't}u\text{) is}{z}^{-2}dz\\ \frac{du}{dz}=-z^{-2}{e}^{1/z}\to du=-z^{-2}{e}^{1/z}dz\text{, which does not resemble the remainder of the integrand}\to \text{REJECT}\\ \text{Possibility 3:}\\ u=1/z;\text{remainder of integrand (the part that isn't}u\text{or}e\text{) is}{z}^{-2}dz\\ \frac{du}{dx}=-z^{-2}\to du=-z^{-2}dz\text{, which resembles the remainder of the integrand except for a negative sign}\to \text{AHA!}\\ \text{Multiplying both sides of the}du\text{expression by}-1\text{gives an expression that exactly matches the remainder of the integrand:}\\ -du=z^{-2}dz\\ \text{Substituting}u\text{and}-du\text{into the original problem yields}\\ \int \frac{e^{1/z}}{z^2}dz=\int z^{-2}{e}^{1/z}dz=\int \left(z^{-2}dz\right)e^{\left(1/z\right)}=\int \left(-du\right)e^u=-\int e^{u}du=-\left(e^u\right)+C=-e^u+C\\ \text{Restoring the original definition of}u\text{gives:}\\ \int \frac{e^{1/z}}{z^2}dz=\underset{¯}{\underset{¯}{-e^{1/z}+C}}\end{array}$$

7.2 #23.

$$\begin{array}{l}\int \frac{u}{\sqrt{u-1}}du=\int u{\left(u-1\right)}^{-1/2}du\\ \text{There are several possibilities for a}v\text{-substitution (ewww}\mathrm{...}\text{we have to guess!)}\text{.}\\ \text{Possibility 1:}\\ v=u;\text{remainder of integrand (the part that isn't}v\text{) is}{\left(u-1\right)}^{-1/2}du\\ \frac{dv}{du}=1\to dv=du\text{, which does not resemble the remainder of the integrand}\to \text{REJECT}\\ \text{Possibility 2:}\\ v={\left(u-1\right)}^{-1/2};\text{remainder of integrand (the part that isn't}v\text{) is}udu\\ \frac{dv}{du}=-\frac{1}{2}{\left(u-1\right)}^{-3/2}\to dv=-\frac{1}{2}{\left(u-1\right)}^{-3/2}du\text{, which does not resemble the remainder of the integrand}\to \text{REJECT}\\ \text{Possibility 3:}\\ v=\left(u-1\right);\text{remainder of integrand (the part that isn't}v\text{or a power of}-1/\text{2}\text{) is}udu\\ \frac{dv}{du}=1\to dv=du\text{, which does not resemble the remainder of the integrand; HOWEVER}\dots \\ \dots \text{by adding 1 to both sides of the definition of}v\text{, we obtain}\left(v+1\right)=u\text{, giving a simple expression for the extra}u\text{in terms of}v.\\ \text{Substituting}v\text{,}\left(v+1\right)\text{,}\text{and}dv\text{into the original problem yields}\\ \int \frac{u}{\sqrt{u-1}}du=\int u{\left(u-1\right)}^{-1/2}du=\int \left(v+1\right){\left(v\right)}^{-1/2}dv=\int \left(v{v}^{-1/2}+1v^{-1/2}\right)dv\text{(the last step obtained by distributing)}\\ \int \frac{u}{\sqrt{u-1}}du=\int v{v}^{-1/2}dv+\int 1v^{-1/2}dv=\int v^{1/2}dv+\int v^{-1/2}dv=\frac{v^{3/2}}{\left(\frac{3}{2}\right)}+\frac{v^{1/2}}{\left(\frac{1}{2}\right)}+C=\frac{2}{3}{v}^{3/2}+2v^{1/2}+C\\ \text{Restoring the original definition of}v\text{gives:}\\ \int \frac{u}{\sqrt{u-1}}du=\underset{¯}{\underset{¯}{\frac{2}{3}{\left(u-1\right)}^{3/2}+2{\left(u-1\right)}^{1/2}+C}}\end{array}$$

7.2 #35.

$$\begin{array}{l}\int x{8}^{3x^2+1}dx\\ \text{There are several possibilities for a}u\text{-substitution (ewww}\mathrm{...}\text{we have to guess!)}\text{.}\\ \text{Possibility 1:}\\ u=x;\text{remainder of integrand (the part that isn't}u\text{) is}{8}^{3x^2+1}dx\\ \frac{du}{dx}=dx\to du=dx\text{, which does not resemble the remainder of the integrand}\to \text{REJECT}\\ \text{Possibility 2:}\\ u=8^{3x^2+1};\text{remainder of integrand (the part that isn't}u\text{) is}xdx\\ \frac{du}{dx}=\left(8^{3x^2+1}\right)\left(\ln 8\right)\left(6x\right)\to du=\left(8^{3x^2+1}\right)\left(\ln 8\right)\left(6x\right)dx\text{, which does not resemble the remainder of the integrand}\to \text{REJECT}\\ \text{Possibility 3:}\\ u=3x^2+1;\text{remainder of integrand (the part that isn't}u\text{or}8\text{) is}xdx\\ \frac{du}{dx}=6x\to du=6xdx\text{, which resembles the remainder of the integrand except for a constant}\to \text{AHA!}\\ \text{Dividing both sides of the}du\text{expression by}6\text{gives an expression that exactly matches the remainder of the integrand:}\\ \frac{du}{6}=xdx\\ \text{Substituting}u\text{and}\frac{du}{6}\text{into the original problem (also see page 434 for the antiderivative formula) yields}\\ \int x{8}^{3x^2+1}dx=\int \left(xdx\right)8^{\left(3x^2+1\right)}=\int \left(\frac{du}{6}\right)8^{\left(u\right)}=\frac{1}{6}\int 8^{u}du=\frac{1}{6}\left(\frac{8^u}{\ln 8}\right)+C=\frac{8^u}{6\ln 8}+C\\ \text{Restoring the original definition of}u\text{gives:}\\ \int x{8}^{3x^2+1}dx=\underset{¯}{\underset{¯}{\frac{8^{3x^2+1}}{6\ln 8}+C}}\end{array}$$

7.5 #21.

$$\begin{array}{l}y=x^{4/3},y=2x^{1/3}\\ \text{First order of business: find the limits on the integral used to find the area}\text{.}\\ \text{In previous problems, these limits were provided as}x\text{values}\text{. In this case, however, we have to find them ourselves}\text{.}\\ \text{The problem authors have selected curves that cross each other at two locations}\text{.}\\ \text{Each such crossing occurs at a point (ordered pair) on the coordinate system where the curves are graphed}\text{.}\\ \text{The limits are the}x\text{coordinates of the two points (ordered pairs) where these crossings occur}\text{.}\\ \text{To find these}x\text{coordinates, we set the equations equal to each other and solve for}x\text{:}\\ x^{4/3}=2x^{1/3}\\ \text{To solve this for}x\text{, we isolate zero on one side of the equation, then factor}\text{. Since the product of the resulting factors is zero,}\\ \text{we conclude that either the first factor must be zero, or the second must be zero (similar to solving a quadratic by factoring)}\text{.}\\ x^{4/3}-2x^{1/3}=0\to \left(x^{1/3}\right)\left(x-2\right)=0\\ \text{From the first parentheses:}{x}^{1/3}=0\to {\left(x^{1/3}\right)}^3={\left(0\right)}^3\to \underset{¯}{\underset{¯}{x=0}}\\ \text{From the second parentheses:}x-2=0\to \underset{¯}{\underset{¯}{x=2}}\\ \text{The limits of integration are therefore}x=0\text{and}x=2.\text{(Whew! All that work just to get the limits}\text{.)}\\ \text{Second order of business: determine which curve is higher on the graph}\text{.}\\ \text{When we find the area between two curves, we subtract the area beneath the lower curve from the area beneath the upper curve}\text{.}\\ \text{If we accidentally subtract backwards (lower minus upper), we get a negative answer}\text{.}\\ \text{To find the higher curve, we calculate the}y\text{value associated with some}x\text{value located between the limits of integration}\text{.}\\ \text{The higher curve is the one with the larger}y\text{value}\text{.}\\ \text{As the limits of integration are}x=0\text{and}x=2\text{, a simple choice would be}x=1.\\ \text{For the}y=x^{4/3}\text{curve, if}x=1\text{, then}y=1.\\ \text{For the}y=2x^{1/3}\text{curve, if}x=1\text{, then}y=2.\text{So this is the higher curve}\text{.}\\ \text{Third order of business: find the area between the curves by integrating (finally, some calculus!)}\text{.}\\ \text{Area}A={\int }_a^b\left[f(x)-g(x)\right]dx={\int }_0^2\left[2x^{1/3}-x^{4/3}\right]dx={\int }_0^{2}2x^{1/3}dx-{\int }_0^2{x}^{4/3}dx=2{\int }_0^2{x}^{1/3}dx-{\int }_0^2{x}^{4/3}dx\\ A={\left[2\frac{x^{4/3}}{\left(4/3\right)}-\frac{x^{7/3}}{\left(7/3\right)}\right]}_0^2={\left[2\left(\frac{3}{4}\right)x^{4/3}-\left(\frac{3}{7}\right)x^{7/3}\right]}_0^2={\left[\frac{3}{2}{x}^{4/3}-\frac{3}{7}{x}^{7/3}\right]}_0^2\\ A=\left[\frac{3}{2}{\left(2\right)}^{4/3}-\frac{3}{7}{\left(2\right)}^{7/3}\right]-\left[\frac{3}{2}{\left(0\right)}^{4/3}-\frac{3}{7}{\left(0\right)}^{7/3}\right]=\left[\frac{3}{2}{\left(2\right)}^{4/3}-\frac{3}{7}{\left(2\right)}^{7/3}\right]-0=\underset{¯}{\underset{¯}{\frac{3}{2}{\left(2\right)}^{4/3}-\frac{3}{7}{\left(2\right)}^{7/3}}}\end{array}$$