MATH 1830

4.2 Fundamental Theorem of Calculus

Introduction

From 4.1 homework problem #2:

Given the function $f(x)=4-0.16{{x}^{2}}$

Approximate the area under the curve on the interval [0, 6] using 6 right rectangles.

Riemann Sum Exploration

Notes

Evaluate the Definite Integral: Compare your answer to the area you calculated in 4.1 notes.

1. $S(x)=\begin{align}&\int_0^4x^2dx\end{align}$

   $ = (\frac{{{x^3}}}{3} + C)│_0^4$

   $=[\frac{{{4^3}}}{3} + C] - [\frac{{{0^3}}}{3} + C]$

   $=21.33 + C - 0 - C$

   $21.33$

2. $S(x) =\begin{align}&\int_1^4 {2xdx}\end{align}$

   $=(\frac{{2{x^2}}}{2} + C)│_1^4$

   $= ({x^2} + C)|_1^4$

   $=({4^2} + C) - ({1^2} + C)]$

   $=16 + C - 1 - C$

   $=15$

3. $S(x) = \begin{align}&\int_1^4 {{x^3}} dx\end{align}$

   $=( \frac{{{x^4}}}{4} + C)│_1^4$

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

   $=64 + C - \frac{1}{4} - C$

   $=63.75$

4. $S(x) = \begin{align}&\int_{0.5}^2 {\frac{1}{x}} \,dx\end{align} $

   $=(ln \left| x \right| + C)│_{0.5}^2$

   $= (\ln (2) + C) - (\ln (0.5) + C)$

   $= \ln (2) + C - \ln (0.5) - C$

   $= \ln (2) - \ln (0.5)$

   $=\ln (\frac{2}{{0.5}}) = \ln (4)$

   $\approx1.39$

5. $\begin{align}&\int_0^7 8x\;dx\end{align} $

   $= \frac{{8{x^2}}}{2}│_0^7$

   $= 4{x^2}│_0^7$

   $(4{\left( 7 \right)^2} + C) - (4{\left( 0 \right)^2} + C)$

   $= 196 + C - 0 - C$

   $= 196$

6. $\begin{align}&\int_2^3 4{x^3}\;dx\end{align}$

   $= (\frac{{4{x^4}}}{4} + C)│_2^3$

   $= {x^4} + C│_2^3$

   $= ({3^4} + C) - ({2^4} + C)$

   $= 81 + C - 16 - C$

   $= 65$

7. $\begin{align}&\int_0^3 5{e^x}\;dx\end{align}$

   $= 5{e^x}│_0^3$

   $= (5{e^3} + C) - (5{e^0} + C)$

   $= 5{e^3} + C - 5(1) - C$

   $= 5{e^3} - 5$

   $= 5({e^3} - 1)$

   $ \approx 95.43$

8. $\begin{align}&\int_1^{4\;} \frac{7}{x}\;dx\end{align}$

   $= (7\ln| x|)│_1^4$

   $= 7(\ln 4 + C) - 7(\ln 1 + C)$

   $= 7\ln 4 + C - 7\ln 1 - C$

   $= 7\ln 4 - 7\ln 1$

   $\approx 9.70$

9. $\begin{align}&\int_3^7 \;\left( {2 - 4{x^2}} \right)\;dx\end{align}$

   $= (2x - \frac{{4{x^3}}}{3} + C)│_3^7$

   $= [2(7) - \frac{{4{{\left( 7 \right)}^3}}}{3} + C] - [2(3) - \frac{{4{{\left( 3 \right)}^3}}}{3} + C]$

   $= (14 - 457.33 + C) - (6 - 36 + C)$

   $= - 443.33 + 30$

   $= - 413.33$

10. $\begin{align}&\int_4^{25} \frac{5}{{\sqrt x }}\;dx\end{align}$

    $=\begin{align}&\int_{4}^{25}{5{{x}^{-1/2}}dx}\end{align}$

    $= \left(\frac{{5{x^{1/2}}}}{({1/2})} + C \right)│_4^{25}$

    $= (10{x^{1/2}})|_4^{25}$

    $= 10\sqrt {25} + C - (10\sqrt 4 + C)$

    $= 10(5) - 10(2)$

    $= 50 - 20$

    $= 30$

11. Cost: A company manufactures mountaineering 75-liter backpacks. The research department produced the marginal cost function $$B'\left( x \right) = 400 - \;\frac{x}{5}\quad\quad 0\; \le x\; \le 1000$$ Where $B’(x)$ is in dollars and x is the number of backpacks produced per week. Compute the increase in cost when production level increases from 0 backpacks per week to 600 backpacks per week. Set up a definite integral and evaluate it.

    $B=\begin{align}&\int_0^{600} {400 - \frac{x}{5}dx}\end{align}$

    $B= (400x - \frac{{{x^2}}}{{5 \cdot 2}} + C)│_0^{600}$

    $B=[400(600) - \frac{{{{600}^2}}}{{10}} + C] - [400(0) - \frac{{{0^2}}}{{10}} + C]$

    $B= 240,000 - 36,000 + C - 0 - C$

    $B= \$ 204,000$

    The cost increased by \$204,000 when production increased from 0 to 600 backpacks per week.

12. Costs of Upkeep of a Marina: Maintenance costs for a marina generally increase as the structures at the marina age. The rate of increase in maintenance costs (in dollars per year) for a particular marina is given approximately by $$M'\left( x \right) = 30{x^2} + 2000$$ where x is the age of marina, in years, and M(x) is the total accumulated costs of maintenance for x years. Write a definite integral that gives the total maintenance costs from the third through the seventh year, and evaluate the integral.

    $M=\begin{align}&\int_3^7 {30{x^2} + 2000 dx}\end{align}$

    $M=(\frac{{30{x^3}}}{3} + 2000x + C)│_3^7$

    $M= 10{x^3} + 2000x + C│_3^7$

    $M= [10{\left( 7 \right)^3} + 2000(7) + C] - [10{\left( 3 \right)^3} + 2000(3) + C]$

    $M= 3430 + 14000 + C - 270 - 6000 - C$

    $M = \$ 11,160$

    The total maintenance costs for the marina for years three through seven will be $\$11,160.$

4.3 Indefinite Integrals

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.

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

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

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

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

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

   $0 - 0 + C = 0$

   $C = 0$

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

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

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

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

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

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

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

   $ - 100 + C = 400$

   $C = 500$

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

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

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

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

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

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

    $5 - 0 + C = - 1$

    $5 + C = - 1$

    $C = - 6$

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

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, $t=52$ and consumption was 8.45 quadrillion Btu.

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

    $5.408 + 3.224 + C = 8.45$

    $8.632 + C = 8.45$

    $C = - 0.182$

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

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

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

    $f(64) = 11.978$

    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?

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

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

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

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

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

    $C = 0$

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

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

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

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

    $t = 23.33$

    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%.

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

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

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

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

    $- 13.4 + C = 82$

    $C = 95.4$

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

    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 is 69.8%.

    The operator efficiency after 8 hours is 49%.

4.4 Integration by Substitution

Introduction

Find the derivative.

1. $f(x)=\left(3x^5+5x\right)^4$

   $f'(x)=4\left(3x^5+5x\right)^3\left(15x^4+5\right)$

   $f'(x)=\left(60x^4+20\right)\left(3x^5+5x\right)^3$

2. $f(x)=\sqrt[5]{\left(3x^2+7\right)^4}$

   $f(x)=\left(3x^2+7\right)^{4/5}$

   $f'(x)=\frac45\left(3x^2+7\right)^{-1/5}\cdot\left(6x\right)$

   $f'(x)=\frac{24x}{5\sqrt[5]{\left(3x^2+7\right)}}$

3. $y=\ln\left(3x^2+5x\right)$

   $y'=\frac1{3x^2+5x}\cdot\left(6x+5\right)$

   $y'=\frac{6x+5}{3x^2+5x}$

4. $f(x)=e^{5x^3-6x+1}$

   $y'=e^{5x^3-6x+1}\left(15x^2-6\right)$

---

Notes

General Indefinite Integral Formulas

Find each indefinite integral and check the result by differentiating.

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

   $u = {x^9} + 1$

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

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

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

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

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

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

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

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

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

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

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

   $u = 9x - 4$

   $du = 9\;dx$

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

   $2\ln|u| + C$

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

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

   $u = 3t + 5$

   $du = 3\;dt$

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

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

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

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

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

   $u = - 2x$

   $du = - 2\;dx$

   $ - \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$

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

   $u = 5 + {x^2}$

   $du = 2x\;dx$

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

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

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

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

7. $\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$

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

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

   $du = (2x - 2)dx$

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

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

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

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

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

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

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

   $u = \ln x$

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

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

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

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

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

10. $\begin{align}&\int {\frac{x}{{\sqrt {x + 3} }}\;dx}\end{align}$

    there is an extra x in the problems 10 and 11

    $u = x + 3$

    $x = u - 3$

    $du = dx$

    $\begin{align}&\int {\frac{{u - 3}}{{\sqrt u }}} du\end{align}$

    $\begin{align}&\int\left(u-3\right)u^{1/2}\;du\end{align}$

    $\begin{align}&\int{\left({{u}^{1/2}}-3{{u}^{1/2}}\right)du}\end{align}$

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

    $\frac{2}{3}{{u}^{3/2}}-6{{u}^{1/2}}+C$

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

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

    $u = x + 2$

    $x = u - 2$

    $du = dx$

    $\begin{align}&\int {(u - 2){u^5}du}\end{align}$

    $\begin{align}&\int {\left({u^6} - 2{u^5}\right)} du\end{align}$

    $\frac{{{u^7}}}{7} - \frac{{2{u^6}}}{6} + C$

    $\frac{{{{\left( {x + 2} \right)}^7}}}{7} - \frac{{{{\left( {x + 2} \right)}^6}}}{3} + C$

12. 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 $$A'(t)=1000e^{.02t}.$$ 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 $\$5258.55.$

4.5 Area between Curves

Introduction

Before 1995, the U. S. Census Bureau used the model below to project the number (in thousands) of households in the United States, where $t$ is the number of years after 1990. $${N_1} = 1.35{t^2} + 1078.4t + 92,323\quad$$ For the years 1995-2005, the actual number of households $N$ in the United States can be modeled by $${N_2} = 18.32{t^2} + 1178.3t + 92,099$$

1. Graph ${N_1}$ on the interval $\;5 \le t \le 15$ and shade the area under the curve. What does this shaded area represent?

   This shaded area represents the projected number of households in the United States from 1995 to 2005.

   

   

2. Now graph ${N_2}$ on the interval $\;5 \le t \le 15$ on the same graph and shade the area under the curve. What does this shaded area represent? 

   This shaded area represents the actual number of households in the United States from 1995 to 2005.

3. Did the projection model over-project or under-project the number of households?

   The projection model under-projected the number of households.

4. How would you determine the difference in the number of households given by each model?

   Subtract the bottom equation from the top.

5. Find the difference in the number of households for the two models.  

   $\begin{align}&\int_{5}^{15}{\left((18.32{{x}^{2}}+1178.3x+92,099)-(1.35{{x}^{2}}+1078.4x+92,323)\right)\;dx}\end{align}$

   $=26,134.17$

   The difference between the two models is 26,134 households.

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Notes

Area Bounded by an Interval

Find the area bounded by the graphs of the indicated equations over the given interval. Compute answers to three decimal places.

1. $y_1 = \; - x + 7$ and $y_2 = 0$ on the interval $ - 3 \le x \le 3$

   

   $Area=\begin{align}&\int_{-3}^{3}{\left(y_1-y_2 \right) dx}\end{align}$

   $=\begin{align}&\int_{-3}^{3}{\left((-x+7)-0 \right) dx}\end{align}$

   $=\left(-\frac{{{x}^{2}}}{2}+7x+C\right)│ _{-3}^{3} $

   $=(-\frac{9}{2}+21+C)-(-\frac{9}{2}-21+C)$

   $=-\frac{9}{2}+21+C+\frac{9}{2}+21-C$

   $=42\ unit{{s}^{2}}$

2. $y_1 = \; - \frac{1}{4}{x^3} + 2 \quad$ and $\quad y_2 = 0 \quad$ on the interval $ - 2 \le x \le 1$

   

   $Area=\begin{align}&\int_{-2}^{1}{\left(y_1-y_2 \right) dx}\end{align}$

   $=\begin{align}&\int_{-2}^{1}{\left(\left(-\frac{1}{4}{{x}^{3}}+2\right)-0 \right) dx}\end{align}$

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

   $=(-\frac{{{x}^{4}}}{16}+2x+C\left)│ _{-2}^{1} \right.$

   $=(-\frac{1}{16}+2+C)-(-\frac{16}{16}-4+C)$

   $=-\frac{1}{16}+2+1+4$

   $=6.9375\ unit{{s}^{2}}$

3. $y_1 = \;x\left( {4 + x} \right)\quad $ and $\quad y_2 = 0 \quad$ on the interval $\; - 3 \le x \le - 1$

   

   $Area=\begin{align}&\int_{-3}^{-1}{\left(y_2-y_1 \right) dx }\end{align}$

   $=\begin{align}&\int_{-3}^{-1}{\left(0-(4x+{{x}^{2}})\right)\;dx}\end{align}$

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

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

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

   $=(-2(1)-(-\frac{1}{3})+C)-(-2(9)-(-9)+C)$

   $=-2+\frac{1}{3}+18-9$

   $=7.33\ unit{{s}^{2}}$

4. $y_1 = \; - \frac{2}{x}\quad$ and $\quad y_2 = 0 \quad$ on the interval $1 \le x \le e$

   

   $Area=\begin{align}&\int_{1}^{e}{\left(y_2-y_1 \right) dx }\end{align}$

   $=\begin{align}&\int_{1}^{e}{\left(0-(-\frac{2}{x})\right)\; dx}\end{align}$

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

   $=(2\ln \left| x \right|+C\left)│ _{1}^{e} \right.$

   $=2\ln e-2\ln 1$

   $=2(1)-2(0)$

   $=2\ unit{{s}^{2}}$

5. $y_1 = \; - x - 3\quad$ and $\quad y_2 = 0\quad$ on the interval $ - 7 \le x \le 2$

   

   $Area=\begin{align}&\int_{-7}^{-3}\left(y_1-y_2\right)\;dx\end{align}$$+\begin{align}&\int_{-3}^{2}\left(y_2-y_1\right)\;dx\end{align}$

   $=\begin{align}&\int_{-7}^{-3}\left(\left(-x-3\right)-0\right)\;dx\end{align}$$+\begin{align}&\int_{-3}^{2}\left(0-\left(-x-3\;\right)\right)\;dx\end{align}$

   $=\left(-\frac{x^2}2-3x+C\right)|_{-7}^{-3}+\left(\frac{x^2}2+3x+C\right)|_{-3}^2$

   $=(4.5--3.5)+(8--4.5)$

   $=8+12.5$

   $=20.5\ unit{{s}^{2}}$

6. $y_1 = \; - x + 7\quad$ and $\quad y_2 = - 2\quad$ on the interval $ - 3 \le x \le 3$

   

   $Area=\begin{align}&\int_{-3}^{3}{(y_1-y_2)\;dx}\end{align}$

   $=\begin{align}&\int_{-3}^{3}{\left((-x+7)-(-2)\right)\;dx}\end{align}$

   $=\begin{align}&\int_{-3}^{3}{\left(-x+7+2\right)\;dx}\end{align}$

   $=\begin{align}&\int_{-3}^{3}{\left(-x+9\right)\;dx}\end{align}$

   $=(-\frac{{{x}^{2}}}{2}+9x+C\left)│ _{-3}^{3} \right.$

   $=(-\frac{9}{2}+27+C)-(-\frac{9}{2}-27+C)$

   $=-\frac{9}{2}+27+C+\frac{9}{2}+27-C$

   $=54\ unit{{s}^{2}}$

7. $y_1 = \; - x - 3 \quad$ and $\quad y_2 = 1.5\quad$ for the interval $ - 7 \le x \le 2$

   

   $Area=\begin{align}&\int_{-7}^{-4.5}\left(y_1-y_2\right)\;dx\end{align}$$+\begin{align}&\int_{-4.5}^{2}\left(y_2-y_1\right)\;dx\end{align}$

   $\begin{align}&\int_{-7}^{-4.5}\left(\left(-x-3\right)-1.5\right)\;dx\end{align}$$+\begin{align}&\int_{-4.5}^{2}\left(1.5-\left(-x-3\;\right)\right)\;dx\end{align}$

   $\begin{align}&\int_{-7}^{-4.5}\left(-x-4.5\right)\;dx\end{align}$$+\begin{align}&\int_{-4.5}^{2}\left(4.5+x\right)\;dx\end{align}$

   $(\frac{-{{x}^{2}}}{2}-4.5x+C\left)│ _{-7}^{-4.5} \right.$

   $+(4.5x+\frac{{{x}^{2}}}{2}+C\left)│ _{-4.5}^{2} \right.$

   $(10.125-7)+(11--10.125)$

   $3.125+21.125$

   $24.25\ unit{{s}^{2}}$

8. $y_1 = \;\ln (x + e) \quad$ and $\quad y_2 = {e^x} - 5\quad$ on the interval $0 \le x \le 1$

   

   $Area=\begin{align}&\int_{0}^{1}\left(y_1-y_2\right)\;dx\end{align}$

   $=\begin{align}&\int_{0}^{1}\left(\ln (x+e)-({{e}^{x}}-5)\right)\;dx\end{align}$

   $=\begin{align}&\int_{0}^{1}\left(\ln (x+e)-{{e}^{x}}+5\right)\; dx\end{align}$

   $=\begin{align}&\int_{0}^{1}\ln (x+e)\; dx\end{align}$$+\begin{align}&\int_{0}^{1}\left(-{{e}^{x}}+5\right)\; dx\end{align}$

   $u=x+e$

   $du=dx$

   $=\begin{align}&\int_{x=0}^{x=1}ln u\;du +\int_{0}^{1}(-{{e}^{x}}+5)\;dx\end{align}$

   $=\left(u\ln\left|u\right|-u+C\right)\vert_{x=0}^{x=1}+\left(-e^x+5x\right)|_0^1$

   $=\left[\left(x+e\right)\ln\left|x+e\right|-\left(x+e\right)+-e^x+5x\right]|_0^1$

   $=((1+e)\ln \left| 1+e \right|-1-e-e+5)-((0+e)\ln \left| 0+e \right|-0-e-{{e}^{0}}+0)$

   =$3.45-(e\ln e-e-1)$

   =$3.45-(-1)$

   =$4.45\;units^2$

   Define the total area bounded by the two functions. How would you determine the interval, if you were asked to calculate the total area bounded between the two functions?

   Area Not Bounded by an Interval

   Find the area bounded by the graphs of the indicated equations

9. $y_1 = {x^3} + 1 \quad$ and $\quad y_2 = x + 1$

   

   $Area=\begin{align}&\int_{-1}^{0}\left(y_1-y_2\right)\;dx\end{align}$$+\begin{align}&\int_{0}^{1}\left(y_2-y_1\right)\;dx\end{align}$

   $\begin{align}&\int_{-1}^{0}\left(({{x}^{3}}+1)-(x+1)\right)\;dx\end{align}$$+\begin{align}&\int_{0}^{1}\left((x+1)-({{x}^{3}}+1)\right)\;dx\end{align}$

   $\begin{align}&\int_{-1}^{0}{({{x}^{3}}-x)\;dx}\end{align}$$+\begin{align}&\int_{0}^{1}{(x-{{x}^{3}})\;dx}\end{align}$

   $(\frac{{x}^{4}}{4}-\frac{{x}^{2}}{2})|_{-1}^{0}+(\frac{{x}^{2}}{2}-\frac{{x}^{4}}{4})|_{0}^{1}$

   $[(0-0)-(\frac{1}{4}-\frac{1}{2})]+[(\frac{1}{2}-\frac{1}{4})-(0-0)]$

   $\frac{1}{4}+\frac{1}{4}$

   $\frac{1}{2}\;units^2$

10. $y_1 = - {x^3} - 2{x^2} + 1\quad$ and $\quad y_2 = - x - 1$

    

    $Area=\begin{align}&\int_{-2}^{-1}\left(y_2-y_1\right)\;dx\end{align}$$+\begin{align}&\int_{-1}^{1}\left(y_1-y_2\right)\;dx\end{align}$

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

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

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

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

    $(1.0833-0.667)+(1.5833--1.0833)$

    $0.4167+2.6667$

    $3.08\;units^2$

11. $y_1 = - 3 + {x^2}$ and $y_2 = - {e^{2x - 3}} + 2$

    

    The functions intersect at $x=1.79$ and $x=-2.24$, therefore the area is bound in the interval [-2.24, 1.79].

    Area=$\begin{align}&\int_{-2.24}^{1.79}(y_2-y_1)\;dx\end{align}$

    $=\begin{align}&\int_{-2.24}^{1.79}{(-{{e}^{2x-3}}+2)-(-3+{{x}^{2}}})\;dx\end{align}$

    $=\begin{align}&\int_{-2.24}^{1.79}\left(-e^{2x-3}-x^2+5\right)\;dx\end{align}$

    $u=2x-3$

    $du=2dx$

    $=\frac{1}{2}\begin{align}&\int_{x=-2.24}^{x=1.76}{-{{e}^{u}}\;du}+\int_{-2.24}^{1.76}{-{{x}^{2}}+5\; dx}\end{align}$

    $=\frac{1}{2}(-{{e}^{u}})\left| _{x=-2.24}^{x=1.79} \right.+(\frac{-{{x}^{3}}}{3}+5x\left)| _{-2.24}^{1.79} \right.$

    $=(\frac{-{{e}^{2x-3}}}{2}-\frac{{{x}^{3}}}{3}+5x\left)| _{-2.24}^{1.79} \right.$

    $=6.15-(-7.45)$

    $=13.6\;units^2$

12. The useful life of a piece of rental equipment is the duration for which the equipment will be profitable to the rental business, and not how long the equipment will actually last. Many factors affect a piece of equipment’s useful life, including the frequency of use, the age when acquired and the repair policy and certain environmental conditions. The change in revenue for renting the equipment, over time, is modeled by $$R'\left( t \right) = 4t{e^{ - 0.2{t^2}}},$$ in thousands of dollars per year. The change in cost for routine maintenance of the equipment is modeled by $$C'\left( t \right) = .5t,$$ also in thousands of dollars per year. Find the area between the graphs of $C'$ and $R'$ over the interval from the time the equipment is purchased until the equipment reaches the end of its useful life. Interpret the results.

    Note: The end of the useful life is t where $C'\left( t \right) = R'(t)$

    Area =$\begin{align}&\int_{0}^{3.22}R'(t)-C'(t)\;dt\end{align}$

    $=\begin{align}&\int_{0}^{3.22}{(4t{{e}^{-0.2{{t}^{2}}}}-0.5t})\;dt\end{align}$

    $u=-0.2{{t}^{2}}$

    $du=-0.4t\;dt$

    $=\frac{4}{-0.4}\begin{align}&\int_{t=0}^{t=3.22}{e^u}\;du+\int_{0}^{3.22}-0.5t\;dt\end{align}$

    $=-10\begin{align}&\int_{t=0}^{t=3.22}{{{e}^{u}}}\;du +\int_0^{3.22}-0.5t\;dt\end{align}$

    $=-10{{e}^{u}}\left| _{t=0}^{t=3.22} \right.+\frac{-0.5{{t}^{2}}}{2}\left| _{0}^{3.22} \right.$

    $=-10{{e}^{-0.2{{t}^{2}}}}-0.25{{t}^{2}}\left| _{0}^{3.22} \right.$

    $=-3.85-(-10)$

    $=6.15$

    The useful life of this piece of equipment is 6.15 years.