MATH 1830

4.4 Integration by Substitution

4.4 Video

Chain Rule Review.

Find the derivative.

1. $f(x)=\frac{1}{5}\left(x^9+1\right)^5+13$

   $f'(x)=\frac15\times5(x^9+1)^4\times(9x^8)$

   $f'(x)=9x^8(x^9+1)^4$

Find each indefinite integral and check the result by differentiating.

2. $\begin{align}&\int {{{\left( {{x^9}\; + 1} \right)}^4}\;\left( {9{x^8}} \right)dx}\end{align}$

   $u = {x^9} + 1$

   $du = 9{x^8}\;dx$

   $\begin{align}&\int {{u^4}\;du}\end{align}$

   $=\frac{u^5}{5} + C$

   $=\frac{( {{x^9} + 1})^5}{5} + C$

3. $\begin{align}&\int {{{\left( {6{x^3} - 7} \right)}^{ - 6}}\;\left( {18{x^2}} \right)\;dx}\end{align}$

   $u = 6{x^3} - 7$

   $du = 18{x^2}\;dx$

   $\begin{align}&\int {{u^{ - 6}}} du\end{align}$

   $=\frac{u^{ - 5}}{ - 5} + C$

   $=\frac{{{\left( 6{{x}^{3}}-7 \right)}^{-5}}}{-5}+C$

   $=-\frac1{5(6x^3-7)^5}$

4.4 Lecture

4. $\begin{align}&\int {\frac{2}{{9x - 4}}\;\left( 9 \right)\;dx}\end{align}$

   $u = 9x - 4$

   $du = 9\;dx$

   $=\begin{align}&\int {\frac{2}{u}\;du}\end{align}$

   $=2\ln|u| + C$

   $=2\ln| {9x - 4}| + C$

5. $\begin{align}&\int {{{\left( {3t + 5} \right)}^4}\;dt}\end{align}$

   $u = 3t + 5$

   $du = 3\;dt$

   $=\frac{1}{3}\begin{align}&\int {{{\left( {3t + 5} \right)}^4}3dt}\end{align}$

   $=\frac{1}{3}\begin{align}&\int {{u^4}} du\end{align}$

   $=\frac{1}{3}\frac{{{u^5}}}{5} + C$

   $=\frac{1}{15}\left(3t+5\right)^5+C$

6. $\begin{align}&\int {{e^{ - 2x}}\;dx}\end{align}$

   $u = - 2x$

   $du = - 2\;dx$

   $= - \frac{1}{2}\begin{align}&\int {{e^{ - 2x}}( - 2)dx}\end{align}$

   $=- \frac{1}{2}\begin{align}&\int {{e^u}du}\end{align}$

   $=- \frac{1}{2}{e^u} + C$

   $= - \frac{1}{2}{e^{ - 2x}} + C$

4.4 Group Work

7. $\begin{align}&\int {\frac{x}{{5 + {x^2}}}\;dx}\end{align}$

   $u = 5 + {x^2}$

   $du = 2x\;dx$

   $=\frac{1}{2}\begin{align}&\int {\frac{1}{{5 + {x^2}}}} 2xdx\end{align}$

   $=\frac{1}{2}\int {\frac{1}{u}} du$

   $=\frac{1}{2}\ln \left| u \right| + C$

   $=\frac{1}{2}\ln| {5 + {x^2}}| + C$

8. $\begin{align}&\int {\frac{{{t^2}}}{{{{\left( {{t^3}\; - 1} \right)}^4}}}\;dt}\end{align} $

   $u = {t^3} - 1$

   $du = 3{t^2}\;dt$

   $=\begin{align}&\int {\frac{1}{{{{\left( {{t^3} - 1} \right)}^4}}}{t^2}} dt\end{align}$

   $=\frac{1}{3}\begin{align}&\int {\frac{1}{{{{\left( {{t^3} - 1} \right)}^4}}}} 3{t^2}dt\end{align}$

   $=\frac{1}{3}\begin{align}&\int {\frac{1}{{{u^4}}}} du\end{align}$

   $=\frac{1}{3}\begin{align}&\int {{u^{ - 4}}} du\end{align}$

   $=\frac{1}{3}\frac{{{u^{ - 3}}}}{{ - 3}} + C$

   $= - \frac{1}{{9{{\left( {{t^3} - 1} \right)}^3}}} + C$

9. $\begin{align}&\int {\frac{{x\; - 1}}{{{x^2}\; - 2x + 5}}\;dx}\end{align}$

   $u = {x^2} - 2x + 5$

   $du = (2x - 2)dx$

   $du = 2(x - 1)\;dx$

   $=\begin{align}&\int {\frac{1}{{{x^2} - 2x + 5}}} (x - 1)dx\end{align}$

   $=\frac{1}{2}\begin{align}&\int {\frac{1}{{{x^2} - 2x + 5}}} 2(x - 1)dx\end{align}$

   $=\frac{1}{2}\begin{align}&\int {\frac{1}{u}} du\end{align}$

   $=\frac{1}{2}\ln \left| u \right| + C$

   $=\frac{1}{2}\ln \left| {{x^2} - 2x + 5} \right| + C$

10. $\begin{align}&\int {\frac{1}{{x\ln x}}\;\;dx}\end{align}$

    $u = \ln x$

    $du = \frac{1}{x}\;dx$

    $=\begin{align}&\int {\frac{1}{{\ln x}}} \cdot \frac{1}{x}dx\end{align}$

    $=\begin{align}&\int {\frac{1}{{u}}} du\end{align}$

    $=\ln| u| + C$

    $=\ln \left| {\ln x} \right| + C$

11. Continuous Money Flow: Suppose money is flowing continuously into a savings account at a rate of \$1000 per year at interest rate of 2%, compounded continuously. The amount that is paid over time, dt, is $$A'(t)=1000e^{0.02t}.$$ What is the accumulation of the amount over the first 5 years?

    $A=\begin{align}&\int_0^5 {1000{e^{0.02t}}\;dt}\end{align}$

    $u = 0.02t$

    $du = 0.02\;dt$

    $A=1000\frac{1}{{0.02}}\begin{align}&\int_0^5 {{e^{0.02t}}0.02\;dt}\end{align}$

    $A=50000\begin{align}&\int_0^5 {{e^u}\;du}\end{align}$

    $A=(50000{e^u}+C)│{_{t=0}^{t=5}}$

    $A=(50000{e^{0.02t}}+C)\left| {_0^5} \right.$

    $A=50000({e^{0.02(5)}} - {e^{0.02(0)}})$

    $A=5258.55$

    The accumulation in the savings account over the first 5 years is approximately $\$5258.55.$

4.4 Additional Practice

Integrate the following.

12. $\begin{align}&\int {{{\left( {5x + 4} \right)}^2}\;dx}\end{align}$

    $u = 5x + 4$

    $du = 5 \;dx$

    $=\frac{1}{5}\begin{align}&\int {{{\left( {5x + 4} \right)}^2}\cdot5\;dx}\end{align}$

    $=\frac{1}{5}\begin{align}&\int {{u^2}\;du}\end{align}$

    $=\frac{1}{5}\cdot\frac{{{u^3}}}{3} + C$

    $=\frac{1}{{15}}{\left( {5x + 4} \right)^3} + C$

13. $\begin{align}&\int{{{\left( 4x-5 \right)}^{1/2}}}\;dx\end{align}$

    $u=4x-5$

    $du=4\;dx$

    $=\frac{1}{4}\begin{align}&\int{{4{\left ( 4x-5 \right)}^{1/2}}}\;dx\end{align}$

    $=\frac14\begin{align}&\int {u^{1/2}}\;du\end{align}$

    $=\frac{1}{4}\cdot\frac{{{u}^{3/2}}}{3/2}+C$

    $=\frac{1}{4}\cdot\frac{2}{3}{{u}^{3/2}}+C$

    $=\frac{1}{6}{{\left( 4x-5 \right)}^{3/2}}+C$

14. $\begin{align}&\int {{{\left( {{x^3} + {x^2}} \right)}^{1/2}}} (3{x^2} + 2x)\;dx\end{align}$

    $u = {x^3} + {x^2}$

    $du = \left(3{x^2} + 2x\right)\;dx$

    $=\begin{align}&\int u^{1/2}\;du\end{align}$

    $=\frac{u^{3/2}}{3/2}+C$

    $=\frac23u^{3/2}+C$

    $=\frac23\left(x^3+x^2\right)^{3/2}+C$

15. $\begin{align}&{{\int{\left( 1+\frac{1}{t} \right)}}^{3}}\frac{1}{{{t}^{2}}}\;dt\end{align}$

    $u = 1 + \frac{1}{t}=1+t^{-1}$

    $du = -\frac{1}{{{t^2}}}\;dt$

    $ = \begin{align}&-\int\left(1+\frac1t\right)^3\cdot\left(-\frac1{t^2}\right)\;dt\end{align}$

    $= - \begin{align}&\int {{u^3}\;du}\end{align}$

    $=\frac{{ - {u^4}}}{4} + C$

    $=- \frac{1}{4}{\left( {1 + \frac{1}{t}} \right)^4} + C$

16. $\begin{align}&\int {\frac{{2x + 2}}{{{x^2} + 2x + 1}}} \;dx\end{align}$

    $u = {x^2} + 2x + 1$

    $du = \left(2x + 2\right)\;dx$

    $=\begin{align}&\int {\frac{1}{u}} \;du\end{align}$

    $=\ln \left| u \right| + C$

    $=\ln \left| {{x^2} + 2x + 1} \right| + C$

17. $\begin{align}&\int {x{e^{{x^2}}}\;dx}\end{align}$

    $u = {x^2}$

    $du = 2x\;dx$

    $=\begin{align}&\int {{e^{{x^2}}}x\;dx}\end{align}$

    $=\frac{1}{2}\begin{align}&\int {{e^{{x^2}}}2x\;dx}\end{align}$

    $=\frac{1}{2}\begin{align}&\int {{e^u}\;du}\end{align}$

    $=\frac{1}{2}{e^u} + C$

    $=\frac{1}{2}{e^{{x^2}}} + C$

18. $\begin{align}&\int {\frac{{{e^{\sqrt x }}}}{{\sqrt x }}} \;dx\end{align}$

    $u=\sqrt x=x^{1/2}$

    $du=\frac12x^{-1/2}\;dx$

    $du = \frac{1}{{2\sqrt x }}\;dx$

    $=2\begin{align}&\int\left(e^\sqrt x\right)\cdot\left(\frac1{2\sqrt x}\right)\;dx\end{align}$

    $=2\begin{align}&\int {{e^u}\;du}\end{align}$

    $=2{e^u} + C$

    $=2{e^{\sqrt x }} + C$

19. $\begin{align}&\int {\frac{{{e^x}}}{{4 - {e^x}}}\;} dx\end{align}$

    $u = 4 - {e^x}$

    $du = - {e^x}\;dx$

    $= - \begin{align}&\int {\frac{{ - {e^x}}}{{4 - {e^x}}}\;dx}\end{align}$

    $=- \begin{align}&\int {\frac{1}{u}\;du}\end{align}$

    $= - \ln \left| u \right| + C$

    $=- \ln \left| {4 - {e^x}} \right| + C$

20. $\begin{align}&\int{\frac{1}{2x\ln x}}\;dx\end{align}$

    $u=\ln x$

    $du=\frac{1}{x}\;dx$

    $=\begin{align}&\int{\frac{1}{2x\ln x}}\;dx\end{align}$

    $=\frac{1}{2}\begin{align}&\int{\frac{1}{u}}\; du\end{align}$

    $=\frac12\ln\left|u\right|+C$

    $= \frac{1}{2}\ln \left| \ln x \right|+C$

21. $\begin{align}&\int {\frac{{{{\left( {\ln x} \right)}^5}}}{x}}\; dx\end{align}$

    $u = \ln x$

    $du = \frac{1}{x}\;dx$

    $=\begin{align}&\int {{{\left( {\ln x} \right)}^5}} \frac{1}{x}\;dx\end{align}$

    $=\begin{align}&\int {{u^5}\;du}\end{align}$

    $=\frac{{{u^6}}}{6} + C$

    $={\frac{{\left( {\ln x} \right)}}{6}^6} + C$

22. The population growth rate of a certain bacteria can be modeled by $$\frac{dP}{dt}=1000{{e}^{{}^{t}/{}_{3}}}$$ where $t$ is measured in hours. What is the population after 3 hours?

    $\frac{dP}{dt}=1000{{e}^{t/3}}$

    $P(t)=\begin{align}&\int_{0}^{3}{1000{{e}^{t/3}}}\;dt\end{align}$

    $u=\frac{t}{3}$

    $du=\frac{1}{3}dt$

    $P(t)=3\begin{align}&\int_{0}^{3}{1000{{e}^{t/3}}}\cdot\frac{1}{3}\;dt\end{align}$

    $=3\begin{align}&\int_{t=0}^{t=3}{1000{{e}^{u}}}\;du\end{align}$

    $=3000{{e}^{u}}│_{t=0}^{t=3}$

    $=3000{{e}^{t/3}}│ _{0}^{3}$

    $=3000{{e}^{3/3}}-3000{{e}^{0/3}}$

    $=3000e-3000(1)$

    $\approx 5155$

    The population will be $5155$ bacteria at the end of 3 hours.

23. The annual marginal revenue for the sale of $x$ iPhones can be modeled by $$R'(x)=20x+700+\frac{200}{x+5}\quad \quad R(0)=0,$$ where $R(x)$ is revenue in dollars and x is number of iPhones in thousands.

    1. Find the revenue function.

       $R(x)=\begin{align}&\int R'(x)\;dx\end{align}$

       $R(x)=\begin{align}&\int\left(20x+700+\frac{200}{x+5}\right)\;dx\end{align}$

       $R(x)=\begin{align}&\int\left(20x+700\right)\;dx\;+\int\frac{200}{x+5}\;dx\end{align}$

       $u=x+5$

       $du=1\;dx$

       $R(x)=\begin{align}&\int\left(20x+700\right)\;dx\;+\;200\int\frac1u\;du\end{align}$

       $R(x)=\frac{20x^2}2+700x+200\ln\left|x+5\right|+C$

       $R(x)=10x^2+700x+200\ln\left|x+5\right|+C$

       Use $R(0)=0$ to calculate the value of C.

       $10(0)^2+700(0)+200\ln\left|0+5\right|+C=0$

       $200\ln\left|5\right|+C=0$

       $321.89+C=0$

       $C=-321.89$

       $R(x)=10x^2+700x+200ln\left|x+5\right|-321.89$

    2. Find the revenue from the sale of 5 million iPhones.

       5,000,000 iPhones =5000 thousand iPhones

       $x=5000$

       $R(500)=10(5000)^2+700(5000)+200\ln(5005)-321.89$

       $R(5000)=253,501,381.70$

       The revenue from the sale of 5 million iPhones is \$253,501,381.70.

24. A city’s population is expected to grow at the rate $$P'(t)=\frac{126{{e}^{16t}}}{1+{{e}^{16t}}},$$ where $t$ is the number of months from now when the population is 30,000.

    1. Find the function for the city’s population.

       $P(t)=\begin{align}&\int\frac{126e^{16t}}{1+e^{16t}}\;dt\end{align}$

       $u=1+e^{16t}$

       $du=16e^{16t}\;dt$

       $P(t)=\left(\frac{126}{16}\right)\begin{align}&\int\frac{16e^{16t}}{1+e^{16t}}\;dt\end{align}\\$

       $=7.875\begin{align}&\int\frac1u\;du\end{align}$

       $P(t)=7.875\ln\left|1+e^{16t}\right|+C$

       Use (0, 30,000) to solve for C.

       $7.875\ln\left|1+e^0\right|+C=30,000$

       $7.875\ln(2)+C=30,000$

       $5.46+C=30,000$

       $C=29,994.54$

       $P(t)=7.875\ln\left|1+e^{16t}\right|+29,994.54$

    2. Find the population 12 months from now.

       $P(t)=7.875\ln\left|1+e^{16t}\right|+29,994.54$

       $P(12)=7.875\ln\left|1+e^{16(12)}\right|+29,994.54$

       $P(12)=31,506.54$

       The population 12 months from now will be approximately 31,506.54 people.

    3. Approximate the increase in the population from month 6 to month 12.

       $\begin{align}&\int_6^{12}\frac{126}{1+e^{16t}}\;dt\end{align}$

       $=7.875\ln\left|1+e^{16t}\right|_6^{12}$

       $=1512-756=756$

       The population increased by approximately 756 people from month 6 to month 12.

25. $\begin{align}&\int {3{t^2}} {\left( {{t^3} + 4} \right)^5}\;dt\end{align}$

    $u = {t^3} + 4$

    $du = 3{t^2}\;dt$

    $=\begin{align}&\int {{{\left( {{t^3} + 4} \right)}^5}} 3{t^2}\;dt\end{align}$

    $=\begin{align}&\int {{u^5}}\; du\end{align}$

    $=\frac{{{u^6}}}{6} + C$

    $=\frac{\left(t^3+4\right)^6}6+C$

26. $\begin{align}&\int{{{x}^{2}}}{{\left( {{x}^{3}}+4 \right)}^{-1/2}}\;dx\end{align}$

    $u={{x}^{3}}+4$

    $du=3{{x}^{2}}\; dx$

    $=\begin{align}&{{\int{\left( {{x}^{3}}+4 \right)}}^{-1/2}}\cdot{{x}^{2}}\;dx\end{align}$

    $=\frac{1}{3}\begin{align}&\int{{{\left( {{x}^{3}}+4 \right)}^{-1/2}}\cdot}\;3{{x}^{2}}\;dx\end{align}$

    $=\frac{1}{3}\begin{align}&\int{{{u}^{-1/2}}}\;du\end{align}$

    $=\frac{1}{3}\cdot\frac{{{u}^{1/2}}}{1/2}+C$

    $=\frac{1}{3}\cdot \frac{2}{1}{{u}^{1/2}}+C$

    $=\frac{2}{3}{{u}^{1/2}}+C$

    $=\frac{2}{3}{{\left( {{x}^{3}}+4 \right)}^{1/2}}+C$

27. $\begin{align}&\int{\frac{{{\left( \sqrt{x}-1 \right)}^{2}}}{\sqrt{x}}}\;dx\end{align}$

    $u=\sqrt x-1=x^{1/2}-1$

    $du=\frac12x^{-1/2}\;dx$

    $du=\frac12\cdot\frac1{\sqrt x}\;dx$

    $=2\begin{align}&\int{{{\left( \sqrt{x}-1 \right)}^{2}}}\frac{1}{2\sqrt{x}}\;dx\end{align}$

    $=2\begin{align}&\int{{{u}^{2}}}\;du\end{align}$

    $=2\frac{{{u}^{3}}}{3}+C$

    $=\frac{2}{3}{{\left( \sqrt{x}-1 \right)}^{3}}+C$

28. $\begin{align}&\int{\frac{x+1}{{{\left( {{x}^{2}}+2x+2 \right)}^{3}}}}\;dx\end{align}$

    $u={{x}^{2}}+2x+2$

    $du=\left(2x+2\right)\;dx=2(x+1)\;dx$

    $\begin{align}&\int{\frac{1}{{{\left( {{x}^{2}}+2x+2 \right)}^{3}}}}\cdot (x+1)\;dx\end{align}$

    $\frac{1}{2}\begin{align}&\int\frac1{\left(x^2+2x+2\right)^3}\cdot2\left(x+1\right)\;dx\end{align}$

    $\frac12\begin{align}&\int{\frac{1}{{{u}^{3}}}\ }du\end{align}$

    $\frac{1}{2}\begin{align}&\int{{{u}^{-3}}}\;du\end{align}$

    $=\frac12\cdot\frac{u^{-2}}{-2}+C$

    $=\frac12\cdot\frac{\left(x^2+2x+2\right)^{-2}}{-2}+C$

    $=\frac{-1}{4{{\left( {{x}^{2}}+2x+2 \right)}^{2}}}+C$

29. $\begin{align}&\int{{{x}^{2}}}{{e}^{-4{{x}^{3}}}}\;dx\end{align}$

    $u=-4{{x}^{3}}$

    $du=-12{{x}^{2}}\;dx$

    $\begin{align}&\int{{{e}^{-4{{x}^{3}}}}\cdot }{{x}^{2}}\;dx\end{align}$

    $=-\frac1{12}\begin{align}&\int\left(e^{-4x^3}\right)\cdot\left(-12x^2\right)\;dx\end{align}$

    $=-\frac{1}{12}\begin{align}&\int{{{e}^{u}}}\;du\end{align}$

    $=-\frac{1}{12}{{e}^{u}}+C$

    $=-\frac{1}{12}{{e}^{-4{{x}^{3}}}}+C$

30. $\begin{align}&\int {\frac{{1 + {e^{3x}}}}{{{e^{3x}} + 3x}}}\;dx\end{align}$

    $u = {e^{3x}} + 3x$

    $du = \left(3{e^{3x}} + 3\right)dx = 3({e^{3x}} + 1)\;dx$

    $\begin{align}&\int {\frac{1}{{{e^{3x}} + 3x}}(1 + {e^{3x}}} )\;dx\end{align}$

    $=\frac{1}{3}\begin{align}&\int {\frac{1}{u}} \;du\end{align}$

    $=\frac{1}{3}\ln \left| u \right| + C$

    $=\frac{1}{3}\ln \left| {{e^{3x}} + 3x} \right| + C$

31. $\begin{align}&\int{{{e}^{-x}}{{\left( 2{{e}^{-x}}+3 \right)}^{9}}}\;dx\end{align}$

    $u=2{{e}^{-x}}+3$

    $du=-2{{e}^{-x}}\;dx$

    $\begin{align}&\int{{{\left( 2{{e}^{-x}}+3 \right)}^{9}}\cdot {{e}^{-x}}}\;dx\end{align}$

    $=\frac{-1}{2}\begin{align}&\int{{{\left( 2{{e}^{-x}}+3 \right)}^{9}}\cdot -2{{e}^{-x}}}\;dx\end{align}$

    $=-\frac{1}{2}\begin{align}&\int{{{u}^{9}}}\;du\end{align}$

    $=-\frac{1}{2}\frac{{{u}^{10}}}{10}+C$

    $=-\frac{1}{20}{{\left( 2{{e}^{-x}}+3 \right)}^{10}}+C$

32. $\begin{align}&\int{\frac{2\ln x}{x}}\;dx\end{align}$

    $u=\ln x$

    $du=\frac{1}{x}\;dx$

    $2\begin{align}&\int{\ln x\cdot\frac{1}{x}}\;dx\end{align}$

    $=2\begin{align}&\int{u}\;du\end{align}$

    $=2\frac{{{u}^{2}}}{2}+C$

    $={{u}^{2}}+C$

    $={{\left( \ln x \right)}^{2}}+C$

33. $\begin{align}&\int {\frac{{{{\left( {5 + \ln x} \right)}^5}}}{x}}\;dx\end{align}$

    $u = 5 + \ln x$

    $du = \frac{1}{x}\;dx$

    $\begin{align}&\int {{{\left( {5 + \ln x} \right)}^5}} \frac{1}{x}\;dx\end{align}$

    $=\begin{align}&\int {{u^5}}\;du\end{align}$

    $=\frac{{{u^6}}}{6} + C$

    $=\frac{{{{\left( {5 + \ln x} \right)}^6}}}{6} + C$

34. The proportion of disposable income which individuals spend on consumption is known as the marginal propensity to consume (MPC). MPC is the proportion of additional income that an individual consumes. For example, if a household earns one extra dollar of disposable income, and the marginal propensity to consume is 0.65, then of that dollar, the household will spend 65 cents and save 35 cents. For a family of four in 2012, the MPC can be modeled by $$\frac{dQ}{dx}=\frac{0.98}{{{(x-23,049)}^{0.02}}}$$ where x is the family income and Q is the income consumed. Given that the poverty level, in 2012, for a family of four was \$23,050 and all \$23,050 were consumed, use the model to estimate the amount consumed by a family of four whose 2012 income was \$35,000.

    $\begin{align}&\int {\frac{{0.98}}{{{{(x - 23049)}^{0.02}}}}}\; dx\end{align}$

    $u = x - 23049$

    $du = 1\;dx$

    $0.98\begin{align}&\int {{u^{ - 0.02}}} \;du\end{align}$

    $\begin{align}&\int {0.98{{\left( {x - 23049} \right)}^{ - 0.02}}}\; dx\end{align}$

    $0.98\frac{{{u^{0.98}}}}{{0.98}} + C$

    ${u^{0.98}} + C$

    ${\left( {x - 23049} \right)^{0.98}} + C$

    $23050 = {\left( {23050 - 23049} \right)^{0.98}} + C$

    $23050 = {1^{0.98}} + C$

    $C = 23049$

    $Q(x) = {\left( {x - 23049} \right)^{0.98}} + 23049$

    $Q(35000) = {\left( {35000 - 23049} \right)^{0.98}} + 23049$

    $Q(35000) = {\left( {11951} \right)^{0.98}} + 23049$

    $Q(35000)= 9905.04 + 23049$

    $Q(35000)= 32954$

    The estimated amount consumed by a family of four making \$35,000 in 2012 was $32,954.

35. Target’s market research department has determined that the marginal price for a particular brand of deodorant can be modeled by the function $$p'(x)=-0.016{{e}^{-0.02x}},$$ where $x$ is the number deodorant sticks sold per week in thousands.

    1. Find the price-demand equation if the weekly demand is 40,000 deodorant sticks when the price is \$3.99.

       $p(x)=\begin{align}&\int-0.016e^{-0.02x}\;dx\end{align}$

       $u=-0.02x$

       $du=-0.02\;dx$

       $p(x)=\left(\frac{-0.016}{-0.02}\right)\begin{align}&\int-0.02e^{-0.02x}\;dx\end{align}$

       $= (.8)\begin{align}&\int e^u\;du\end{align}$

       $p(x)=.8e^{-0.02x}+C$

       $p(40)=3.99$

       $.8e^{-0.02(40)}+C=3.99$

       $.8e^{-0.8}+C=3.99$

       $0.36+C=3.99$

       $C=3.63$

       $p(x)=.8e^{-0.02x}+3.63$

    2. Find the demand when the price of the deodorant is \$3.79 per stick.

       $p(x)=.8e^{-0.02x}+3.63$

       $.8e^{-0.02x}+3.63=3.79$

       $.8e^{-0.02x}=0.16$

       $e^{-0.02x}=0.2$

       $ln e^{-0.02x}=ln0.2$

       $-0.02x=ln0.2$

       $x=\frac{\ln0.2}{-0.02}$

       $x=80.47$

       When the price is \$3.79 the demand for the deodorant is 80,470 deodorant sticks per week.

4.4 In-class Practice

Evaluate each indefinite integral.

1. $\begin{align}&\int-12x^2\left(-4x^3+2\right)^{-3}\;dx\end{align}$

   $u=-4x^3+2$

   $du=-12x^2\;dx$

   $\begin{align}&\int u^{-3}\;du\end{align}$

   $=\frac{u^{-2}}{-2}+C$

   $=\frac{-1}{2\left(-4x^3+2\right)^2}+C$

   $=\frac{\left(-4x^3+2\right)^{-2}}{-2}+C$

2. $\begin{align}&\int\left(e^{4x}-4\right)^{1/3}\cdot8e^{4x}\;dx\end{align}$

   $u=e^{4x}-4$

   $du=4e^{4x}\;dx$

   $\frac84\int\left(e^{4x}-4\right)^{1/3}\cdot4e^{4x}\;dx$

   $2\begin{align}&\int u^{1/3}\;du\end{align}$

   $=2\cdot\frac{u^{4/3}}{4/3}+C$

   $=\frac{6}4u^{4/3}+C$

   $=\frac{3}2\left(e^{4x}-4\right)^{4/3}+C$

   $=\frac32\left(e^{4x}-4\right)^{4/3}+C$

3. $\begin{align}&\int5\sqrt{2x+3}\;dx\end{align}$

   $u=2x+3$

   $du=2\; dx$

   $\frac52\begin{align}&\int\sqrt{2x+3}\cdot2\;dx\end{align}$

   $\frac52\begin{align}&\int u^{1/2}\;du\end{align}$

   $\frac52\frac{u^{3/2}}{3/2}+C$

   $\frac52\cdot\frac23\cdot u^{3/2}+C$

   $\frac53\left(2x+3\right)^{3/2}+C$

4. $\begin{align}&\int\frac{12x^2}{x^3+2}\;dx\end{align}$

   $u=x^3+2$

   $du=3x^2\;dx$

   $\frac{12}3\begin{align}&\int\frac{3x^2}{x^3+2}\;dx\end{align}$

   $4\begin{align}&\int\frac1u\;du\end{align}$

   $4\ln\left|u\right|+C$

   $4\ln\left|x^3+2\right|+C$

5. $\begin{align}&\int\frac{5e^{-3+\ln3x}}x\;dx\end{align}$

   $u=-3+\ln\left(3x\right)$

   $du=\frac1{3x}\cdot3\;dx$

   $du=\frac1x\;dx$

   $5\begin{align}&\int e^u\;du\end{align}$

   $=5e^u+C$

   $=5e^{-3+\ln3x}+C$