MATH 1830

4.3 Indefinite Integrals

Pre-Class:

• Complete 4.2B Homework assignment: check and correct.
• Take notes on the videos and readings (use the space below).
• Work and check problems #1-4 in the 4.3 NOTES section.
• Complete the 4.3 Pre-Class Quiz.

Introduction

The marginal cost of producing x units of a commodity is given by $$C'\left( x \right) = 3{x^2} + 2x.$$

1. Find the cost function $C\left( x \right)$ for this commodity.

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

   $C(x) = \begin{align}&\int {\left(3x^{2}+2x\right)dx}\end{align}$

   $C(x) = \frac{3x^{3}}{3}+\frac{2x^{2}}{2}+K$

   $C(x) = {x^3} + {x^2} + K$

2. If the fixed costs are \$2000, find the cost of producing 20 units.

   If fixed costs are $\$2000$, $ \quad C(0)=2000$

   $C(x)={x^3} + {x^2} + K$

   $2000 = {0^3} + {0^2} + K$

   $K=2000$

   $C(x) = {x^3} + {x^2} + 2000$

   $C(20) = {20^3} + {20^2} + 2000$

   $C(20) = 8000 + 400 + 2000$

   $C(20) = 10,400$

   The cost of producing 20 units is $10,400.

Notes

Find the indefinite integral. Check by differentiating.

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

   $=10x + C$

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

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

   $=5{x^3} + C$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 $R\left( 0 \right) = 0$

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

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

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

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

   $600(0) - 0.3({0^2}) + C = 0$

   $0 - 0 + C = 0$

   $C = 0$

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

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%.