MATH 1830

4.3 Indefinite Integrals

4.3 Video

Find the indefinite integral. Check by differentiating.

1. $\begin{align}&\int {15{x^2}\;\;dx}\end{align} $

   $=\frac{{15{x^3}}}{3} + C$

   $=5{x^3} + C$

2. $\begin{align}&\int {10\;dx}\end{align}$

   $=10x + C$

3. $\begin{align}&\int {8\;{x^{1/3}}\;\;\;dx}\end{align}$

   $= \frac{{8{x^{4/3}}}}{({4/3})} + C$

   $= \frac{3}{4}(8{x^{4/3}}) + C$

   $= 6{x^{4/3}} + C$

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

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

   $= \frac{{4{x^4}}}{4} + \frac{{2{x^{ - 2}}}}{{ - 2}} + C$

   $= {x^4} - {x^{ - 2}} + C$

   $= {x^4} - \frac{1}{{{x^2}}} + C$

4.3 Lecture

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

   $=\frac{{{x^{ - 3}}}}{{ - 3}} + C$

   $=-\frac{{ {x^{ - 3}}}}{3} + C$

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

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

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

   $= \frac{{{x^3}}}{3} + \frac{{{x^6}}}{6} + C$

7. $\begin{align}&\int {\frac{{1 - x{e^x}}}{x}dx}\end{align}$

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

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

   Find the particular antiderivative of each derivative that satisfies the given condition.

8. $R'\left( x \right) = 600 - 0.6x \quad$ when $R\left( 100 \right) = 70,000$

   $R(x)=\begin{align}&\int {(600 - 0.6x)} dx\end{align}$

   $R(x) = 600x - \frac{{0.6{x^2}}}{2} + C$

   $R(x) = 600x - 0.3{x^2} + C$

   Use $R(100) = 70,000$ to solve for C.

   $600(100) - 0.3({100^2}) + C = 70,000$

   $60,000 - 3000 + C = 70,000$

   $C = 13,000$

   $R(x) = 600x - 0.3{x^2}+13,000$

4.3 Group Work

9. $\frac{{dQ}}{{dt}}\; = \;\;\frac{{100}}{{{t^2}}} \quad$ when $Q\left( 1 \right) = 400$

   $\frac{{dQ}}{{dt}}=100{t^{ - 2}}$

   $Q(t)=\begin{align}&\int {100{t^{ - 2}}} dt\end{align}$

   $Q(t) = \frac{{100{t^{ - 1}}}}{{ - 1}} + C$

   $Q(t) = \frac{{ - 100}}{t} + C$

   Use $Q(1) = 400$ to solve for C.

   $\frac{{ - 100}}{1} + C = 400$

   $ - 100 + C = 400$

   $C = 500$

   $Q(t) = \frac{{ - 100}}{t} + 500$

10. $\frac{{dy}}{{dt}} = 5{e^t} - 4 \quad$ when $y(0)=- 1$

    $y=\begin{align}&\int {\left(5{e^t} - 4\right)dt}\end{align}$

    $y = 5{e^t} - 4t + C$

    Use $y(0) = -1$ to solve for C.

    $5{e^0} - 4(0) + C = - 1$

    $5(1) - 0 + C = - 1$

    $5 - 0 + C = - 1$

    $5 + C = - 1$

    $C = - 6$

    $y = 5{e^t} - 4t - 6$

11. Renewable Energy: According to the Energy Research Institute, in 2012, US consumption of renewable energy was $8.45$ quadrillion Btu (or $8.45 \times 10^{15}$ Btu). Since the 1960’s, consumption has been growing at a rate (in quadrillion Btu per year) given by $$f'(t) = 0.004t + 0.062$$ where t is in years after 1960. Find $f(t)$ and estimate US consumption of renewable energy in 2024.

    $f(t) = \begin{align}&\int {f'(t)dt}\end{align}$

    $f(t) = \begin{align}&\int {\left(0.004t+0.062\right)dt}\end{align}$

    $f(t) = \frac{{0.004{t^2}}}{2} + 0.062t + C$

    $f(t) = 0.002{t^2} + 0.062t + C$

    In 2012 (t=52), consumption was 8.45 quadrillion Btu. Use this information to solve for C.

    $0.002{\left( {52} \right)^2} + 0.062(52) + C = 8.45$

    $5.408 + 3.224 + C = 8.45$

    $8.632 + C = 8.45$

    $C = - 0.182$

    $f(t) = 0.002{t^2} + 0.062t - 0.182$

    Use the completed function $f(t)$ to estimate US consumption of renewable energy in 2024 (t=64).

    $f(64) = 0.002{\left( {64} \right)^2} + 0.062(64) - 0.182$

    $f(64) = 8.192 + 3.968 - 0.182$

    $f(64) = 11.978$

    The estimated US consumption of renewable energy in 2024 will be 11.978 quadrillion BTU.

12. Sales Analysis: The rate of change of the monthly sales of a newly released video game is given by $$S'(t) = 400{t^{1/3}}$$ $$S(0) = 0$$ where t is the number of months since the game was released and $S(t)$ is the number of games sold each month (in thousands). Find $S(t)$. When will monthly sales reach 20,000,000 games?

    $S(t)=\begin{align}&\int400t^{1/3}\;dt\end{align}$

    $S(t) = \frac{{400{t^{4/3}}}}{{4/3}} + C$

    $S(t) = \frac{3}{4}(400t^{4/3}) + C$

    $S(t) = 300{t^{4/3}} + C$

    Use $S(0) = 0$ to solve for C.

    $300{\left( 0 \right)^{4/3}} + C = 0$

    $C = 0$

    $S(t) = 300{t^{4/3}}$

    Use the completed function $S(t)$ to estimate the month when monthly sales will reach 20,000,000 games.

    $300{t^{4/3}} = 20000$

    ${t^{4/3}} = \frac{{200}}{3}$

    ${\left( {{t^{4/3}}} \right)^{3/4}}= {\left( {\frac{{200}}{3}} \right)^{3/4}}$

    $t = 23.33$

    The monthly sales will reach 20,000,000 games after 23.33 months.

13. Efficiency of a Machine Operator: The rate at which a machine operator’s efficiency Q (in percent) changes with respect to time on the floor without a break is modeled by the function $$\frac{{dQ}}{{dt}} = 0.3t - 7\quad \quad 0 \le t \le 16\,\,hrs$$ where t is the number of house the operator has been working. Find $Q(t)$ given that the operator’s efficiency after working 2 hours is 82%.

    $Q(t)=\begin{align}&\int {\left(0.3t - 7 \right)dt}\end{align}$

    $Q(t) = \frac{{0.3{t^2}}}{2} - 7t + C$

    $Q(t) = 0.15{t^2} - 7t + C$

    Use (2, 82) to solve for C.

    $Q(2) = 0.15{\left( 2 \right)^2} - 7(2) + C = 82$

    $- 13.4 + C = 82$

    $C = 95.4$

    $Q(t) = 0.15{t^2} - 7t + 95.4$

    Find the operators efficiency after 4 hours. After 8 hours.

    $Q(4) = 0.15{\left( 4 \right)^2} - 7(4) + 95.4 = 69.8\%$

    $Q(8) = 0.15{\left( 8 \right)^2} - 7(8) + 95.4 = 49\%$

    The operator efficiency after 4 hours on the floor without a break is 69.8%.

    The operator efficiency after 8 hours on the floor without a break is 49%.

4.3 Additional Practice

Integrate the following.

14. $\begin{align}&\int {(2{x^4} - 5{x^3} - {x^2} + 5x - 8)\;dx}\end{align}$

    $ = \frac{{2{x^5}}}{5} - \frac{{5{x^4}}}{4} - \frac{{{x^3}}}{3} + \frac{{5{x^2}}}{2} - 8x + C$

15. $\begin{align}&\int {\,(3{x^{ - 2}} - 4{x^{ - 3}} + 8)\;dx}\end{align}$

    $= \frac{{3{x^{ - 1}}}}{{ - 1}} - \frac{{4{x^{ - 2}}}}{{ - 2}} + 8x + C$

    $= - 3{x^{ - 1}} + 2{x^{ - 2}} + 8x + C$

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

    $ = 4\ln \left| x \right| + \frac{{3{x^2}}}{2} - 2{e^x} - x + C$

17. $\begin{align}&\int {\left( {2\sqrt[3]{{{x^2}}} - 4\sqrt x + \frac{3}{{{x^2}}} - \frac{5}{x} + 2} \right)} \;dx\end{align}$

    $=\begin{align}&\int\left(2x^{2/3}-4x^{1/2}+3x^{-2}-\frac5x+2\right)\;dx \end{align}$

    $ = \frac{{2{x^{5/3}}}}{{5/3}} - \frac{{4{x^{3/2}}}}{{3/2}} + \frac{{3{x^{ - 1}}}}{{ - 1}} - 5\ln \left| x \right| + 2x + C$

    $= \frac{6}{5}{x^{5/3}} - \frac{8}{3}{x^{3/2}} - 3{x^{ - 1}} - 5\ln \left| x \right| + 2x + C$

18. $\begin{align}&\int {\left( {{e^x} + \frac{2}{x} - \frac{3}{{{x^2}}} - \frac{4}{{\sqrt x }} + 5} \right)\;dx}\end{align}$

    $=\begin{align}&\int\left(e^x+\frac2x-3x^{-2}-4x^{-1/2}+5\right)\;dx\end{align}$

    $ = {e^x} + 2\ln \left| x \right| - \frac{{3{x^{ - 1}}}}{{ - 1}} - \frac{{4{x^{1/2}}}}{{1/2}} + 5x + C$

    $= {e^x} + 2\ln \left| x \right| + \frac{3}{x} - 8\sqrt x + 5x + C$

19. $\begin{align}&\int {\frac{{2{x^2} + 6x - 7}}{x}} \;dx\end{align}$

    $\begin{align}&\int {(2x + 6 - \frac{7}{x}} )dx\end{align}$

    $= \frac{{2{x^2}}}{2} + 6x - 7\ln \left| x \right| + C$

    $= {x^2} + 6x - 7\ln \left| x \right| + C$

20. $\begin{align}&\int {\,\frac{{4{x^4} - 5{x^3} + 6{x^2} - 7x}}{{{x^2}}}} \;dx\end{align}$

    $\begin{align}&\int {(4{x^2} - 5x + 6 - \frac{7}{x}} )dx\end{align}$

    $= \frac{{4{x^3}}}{3} - \frac{{5{x^2}}}{2} + 6x - 7\ln \left| x \right| + C$

21. The MORF dress in one dress with no buttons, zippers, Velcro, front or back that can be worn 24 different ways. After completing a successful Kickstarter campaign, the company found that the marginal cost for making the dress can be modeled by the function $C'(x) = {x^2} - 16x + 70$ dollars per dress where x is the number of dresses produced.

    
    1. Find the cost function for the MORF dress.

       $C(x) = \frac{{{x^3}}}{3} - \frac{{16{x^2}}}{2} + 70x + C$

       $C(x) = \frac{{{x^3}}}{3} - 8{x^2} + 70x + C$

    2. If $C(0) = 500$ find the total cost for producing 20 dresses.

       $\frac{0}{3} - 8(0) + 70(0) + C = 500$

       $0 + C = 500$

       $C = 500$

       $C(x) = \frac{{{x^3}}}{3} - 8{x^2} + 70x + 500$

       $C(20) = \frac{{8000}}{3} - 3200 + 1400 + 500$

       $C(20) = \$ 1366.67$

    3. What is the fixed cost and what is the total variable cost of 20 dresses?

       fixed cost $= \$ 500$ to produce 20 dresses

       variable cost $= \$ 866.67$ to produce 20 dresses

22. Suppose that when it is t years old, a particular industrial machine generates revenue at the rate $R'(t) = 5000 - 20{t^2}$ dollars per year.

    The operating and servicing costs related to the machine accumulate at the rate of $C'(t) = 2000 + 10{t^2}$ dollars per year.

    Find the profit function given that when the machine is first purchased no profit has been generated from it.

    $R'(t) = 5000 - 20{t^2}$

    $C'(t) = 2000 + 10{t^2}$

    $P'(t) = R'(t) - C'(t)$

    $P'(t) = (5000 - 20{t^2}) - (2000 + 10{t^2})$

    $= 5000 - 20{t^2} - 2000 - 10{t^2}$

    $= 3000 - 30{t^2}$

    $P(t) = \begin{align}&\int {(3000 - 30{t^2})dt}\end{align}$

    $P(t) = 3000t - \frac{{30{t^3}}}{3} + C$

    $P(t) = 3000t - 10{t^3} + C$

    $3000(0) - 10{\left( 0 \right)^3} + C = 0$

    $0 + C = 0$

    $C = 0$

    $P(t) = 3000t - 10{t^3}$

23. The rate of change for the lower limit for the top 5% of earners for all households in the United States from 2000 to 2014 can be modeled by $$I'(t) = 38.58{t^2} - 523.56t + 5416.05$$ where $t$ is the number of years after 2000.

    1. Find a general antiderivative for the lower limit for the top 5% of earners.

       $I'(t) = 38.58{t^2} - 523.56t + 5416.05$

       $I(t)=\begin{align}&\int {(38.85{t^2} - 523.56t + 5416.05)dt}\end{align}$

       $I(t)=\frac{{38.58{t^3}}}{3} - \frac{{523.56{t^2}}}{2} + 5416.05t + C$

       $I(t) = 12.86{t^3} - 261.78{t^2} + 5416.05t + C$

    2. Find a specific antiderivative given that in 2000 the lower limit for the top 5% of earners was $145,220.

       $I(0) = 12.86(0)^3 - 261.78(0)^2 + 5416.05(0) + C=145,220$

       $I(0) = 0 - 0 + 0 + C= 145,220$

       $C = 145,220$

       $I(t) = 12.86{t^3} - 261.78{t^2} + 5416.05t + 145,220$

    3. What will the lower limit for the top 5% of earners be in 2015 using this model?

       $I(15) = 12.86{(15)^3} - 261.78{(15)^2} + 5416.05(15) + 145,220$

       $I(15) = \$ 210,963$

       The lower limit on income for the top 5% of earners in 2015 will be \$210,963.