# How to Find the Area Between Two Graphs by Integration

When finding the area between two graphs, there are some things you need to take into account:

1.
The areas’ position in the coordinate system does not matter—that is, it is not important if the graphs are above or below the $x$-axis in this context.
2.
To get a positive area in your answer, you always have to take the top graph minus the bottom one.
3.
Due to Item 2, you must compute an integral for each area.

Then, assuming the graph of $f$ lies above the graph of $g$ between $x=a$ and $x=b$, the area between the graphs is given by:

Formula

### TheAreaBetweenTwoGraphs

$\begin{array}{llll}\hfill A& ={\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.17em}{0ex}}dx-{\int }_{a}^{b}g\left(x\right)\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={\int }_{a}^{b}\phantom{\rule{-0.17em}{0ex}}\left(f\left(x\right)-g\left(x\right)\right)\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Note! You always want to put the top graphs first in the subtraction. If you don’t, the computation yields a negative sign. If you don’t take this into account, you may end up with a completely wrong answer.

Study the following two examples closely and read through them several times until you understand them.

Example 1

Find the area that is bounded by the functions $\begin{array}{llll}\hfill f\left(x\right)& ={x}^{3}+3{x}^{2}-13x-15\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill g\left(x\right)& =10x+7\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

First, you draw the graphs, such that you can see the bounded areas. In this case, there are two areas.

In Area 1, $g\left(x\right)$ is above, and in Area 2, $f\left(x\right)$ is above. That means you have to determine where the graphs intersect—that is, where $f\left(x\right)=g\left(x\right)$:
 ${x}^{3}+3{x}^{2}-13x-15=10x+7$

When solving this equation, you get that $x\approx -6.16$, $x\approx -0.88$ and $x\approx 4.04$. This means that $A=-6.16$, $B=-0.88$ and $C=4.04$, where $g\left(x\right)$ lies above $f\left(x\right)$ between $-6.16$ and $-0.88$, and $f\left(x\right)$ lies above $g\left(x\right)$ between $-0.88$ and $4.04$. Therefore, the total area is:

 $A={A}_{1}+{A}_{2}$

First, find ${A}_{1}$:

$\begin{array}{llll}\hfill {A}_{1}& ={\int }_{-6.16}^{-0.88}{x}^{3}+3{x}^{2}-13x-15\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}-\left(10x+7\right)\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={\int }_{-6.16}^{-0.88}{x}^{3}+3{x}^{2}-23x-22\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\phantom{\rule{-0.17em}{0ex}}\left(\frac{1}{4}{x}^{4}+{x}^{3}-\frac{23}{2}{x}^{2}-22x\right)|{}_{-6.16}^{-0.88}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(\frac{1}{4}{\left(-0.88\right)}^{4}+{\left(-0.88\right)}^{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}-\frac{23}{2}{\left(-0.88\right)}^{2}-22\left(-0.88\right)\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}-\left(\frac{1}{4}{\left(-6.16\right)}^{4}+{\left(-6.16\right)}^{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}-\frac{23}{2}{\left(-6.16\right)}^{2}-22\left(-6.16\right)\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 9.92-\left(-174.63\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 184.55\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill {A}_{1}& ={\int }_{-6.16}^{-0.88}{x}^{3}+3{x}^{2}-13x-15-\left(10x+7\right)\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={\int }_{-6.16}^{-0.88}{x}^{3}+3{x}^{2}-23x-22\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\phantom{\rule{-0.17em}{0ex}}\left(\frac{1}{4}{x}^{4}+{x}^{3}-\frac{23}{2}{x}^{2}-22x\right)|{}_{-6.16}^{-0.88}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\phantom{\rule{-0.17em}{0ex}}\left(\frac{1}{4}{\left(-0.88\right)}^{4}+{\left(-0.88\right)}^{3}-\frac{23}{2}{\left(-0.88\right)}^{2}-22\left(-0.88\right)\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}-\phantom{\rule{-0.17em}{0ex}}\left(\frac{1}{4}{\left(-6.16\right)}^{4}+{\left(-6.16\right)}^{3}-\frac{23}{2}{\left(-6.16\right)}^{2}-22\left(-6.16\right)\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 9.92-\left(-174.63\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 184.55\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Then you find ${A}_{2}$:
$\begin{array}{llll}\hfill {A}_{2}& ={\int }_{-0.88}^{4.04}10x+7\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}-\phantom{\rule{-0.17em}{0ex}}\left({x}^{3}+3{x}^{2}-13x-15\right)\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={\int }_{-0.88}^{4.04}-{x}^{3}-3{x}^{2}+23x+22\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\phantom{\rule{-0.17em}{0ex}}\left(-\frac{1}{4}{x}^{4}-{x}^{3}+\frac{23}{2}{x}^{2}+22x\right)|{}_{-0.88}^{4.04}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(-\frac{1}{4}{\left(4.04\right)}^{4}-{\left(4.04\right)}^{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}+\frac{23}{2}{\left(4.04\right)}^{2}+22\left(4.04\right)\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}-\left(-\frac{1}{4}{\left(-0.88\right)}^{4}-{\left(-0.88\right)}^{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}+\frac{23}{2}{\left(-0.88\right)}^{2}+22\left(-0.88\right)\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 144.04-\left(-9.92\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =153.96\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill {A}_{2}& ={\int }_{-0.88}^{4.04}10x+7-\phantom{\rule{-0.17em}{0ex}}\left({x}^{3}+3{x}^{2}-13x-15\right)\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={\int }_{-0.88}^{4.04}-{x}^{3}-3{x}^{2}+23x+22\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\phantom{\rule{-0.17em}{0ex}}\left(-\frac{1}{4}{x}^{4}-{x}^{3}+\frac{23}{2}{x}^{2}+22x\right)|{}_{-0.88}^{4.04}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\phantom{\rule{-0.17em}{0ex}}\left(-\frac{1}{4}{\left(4.04\right)}^{4}-{\left(4.04\right)}^{3}+\frac{23}{2}{\left(4.04\right)}^{2}+22\left(4.04\right)\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}-\phantom{\rule{-0.17em}{0ex}}\left(-\frac{1}{4}{\left(-0.88\right)}^{4}-{\left(-0.88\right)}^{3}+\frac{23}{2}{\left(-0.88\right)}^{2}+22\left(-0.88\right)\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 144.04-\left(-9.92\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =153.96\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Finally, the total area is
$\begin{array}{llll}\hfill A& ={A}_{1}+{A}_{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 184.55+153.96\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 338.5\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

 $A={A}_{1}+{A}_{2}\approx 184.55+153.96=338.5$

Example 2

Find the area that is bounded by the functions $\begin{array}{llll}\hfill f\left(x\right)& =\frac{1}{2}{x}^{2}-\frac{5}{2}x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill g\left(x\right)& =-5\phantom{\rule{0.17em}{0ex}}cos\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}x-\frac{3}{5}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

for $x\in \phantom{\rule{-0.17em}{0ex}}\left[\text{}-1.23\text{},\text{}6.18\text{}\right]$

First, you draw the graphs so that you can see the bounded areas. In this case, it’s three areas in total:

In Area 1, $f\left(x\right)$ lies above, in Area 2, $g\left(x\right)$ lies above and in Area 3, $f\left(x\right)$ is above again. You therefore have to find where the graphs intersect. That is, where $f\left(x\right)=g\left(x\right)$:
 $\frac{1}{2}{x}^{2}-\frac{5}{2}x=-5\mathrm{cos}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}x-\frac{3}{5}\right)$

When you solve this equation, you get $x\approx -1.23$, $x\approx 1.14$, $x\approx 3.85$ and $x\approx 6.18$. (You also get two other intersections, but these lie outside the relevant interval.)

This means that $A=-1.23$, $B=1.14$, $C=3.85$ and $D=6.18$, where $f\left(x\right)$ lies above between $-1.23$ and $1.14$ and from $3.85$ to $6.18$. The function $g\left(x\right)$ lies above between $1.14$ and $3.85$. The total area is therefore:

 $A={A}_{1}+{A}_{2}+{A}_{3}$

First, you find the area ${A}_{1}$:

$\begin{array}{llll}\hfill {A}_{1}& ={\int }_{-1.23}^{1.14}\frac{1}{2}{x}^{2}-\frac{5}{2}x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{1em}{0ex}}-\phantom{\rule{-0.17em}{0ex}}\left(-5\mathrm{cos}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}x-\frac{3}{5}\right)\right)\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={\int }_{-1.23}^{1.14}\frac{1}{2}{x}^{2}-\frac{5}{2}x+5\mathrm{cos}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}x-\frac{3}{5}\right)\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(\frac{1}{6}{x}^{3}-\frac{5}{4}{x}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}+5\mathrm{sin}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}x-\frac{3}{5}\right)\cdot \frac{2}{3}\right)|{}_{-1.23}^{1.14}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\phantom{\rule{-0.17em}{0ex}}\left(\frac{1}{6}{x}^{3}-\frac{5}{4}{x}^{2}+\frac{10}{3}\mathrm{sin}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}x-\frac{3}{5}\right)\right)|{}_{-1.23}^{1.14}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(\frac{1}{6}{\left(1.14\right)}^{3}-\frac{5}{4}{\left(1.14\right)}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}+\frac{10}{3}\mathrm{sin}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}\left(1.14\right)-\frac{3}{5}\right)\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}-\left(\frac{1}{6}{\left(-1.23\right)}^{3}-\frac{5}{4}{\left(-1.23\right)}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}+\frac{10}{3}\mathrm{sin}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}\left(-1.23\right)-\frac{3}{5}\right)\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 1.61-\left(-4.34\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =5.95\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill {A}_{1}& ={\int }_{-1.23}^{1.14}\frac{1}{2}{x}^{2}-\frac{5}{2}x-\phantom{\rule{-0.17em}{0ex}}\left(-5\mathrm{cos}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}x-\frac{3}{5}\right)\right)\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={\int }_{-1.23}^{1.14}\frac{1}{2}{x}^{2}-\frac{5}{2}x+5\mathrm{cos}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}x-\frac{3}{5}\right)\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\phantom{\rule{-0.17em}{0ex}}\left(\frac{1}{6}{x}^{3}-\frac{5}{4}{x}^{2}+5\mathrm{sin}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}x-\frac{3}{5}\right)\cdot \frac{2}{3}\right)|{}_{-1.23}^{1.14}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\phantom{\rule{-0.17em}{0ex}}\left(\frac{1}{6}{x}^{3}-\frac{5}{4}{x}^{2}+\frac{10}{3}\mathrm{sin}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}x-\frac{3}{5}\right)\right)|{}_{-1.23}^{1.14}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\phantom{\rule{-0.17em}{0ex}}\left(\frac{1}{6}{\left(1.14\right)}^{3}-\frac{5}{4}{\left(1.14\right)}^{2}+\frac{10}{3}\mathrm{sin}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}\left(1.14\right)-\frac{3}{5}\right)\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}-\phantom{\rule{-0.17em}{0ex}}\left(\frac{1}{6}{\left(-1.23\right)}^{3}-\frac{5}{4}{\left(-1.23\right)}^{2}+\frac{10}{3}\mathrm{sin}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}\left(-1.23\right)-\frac{3}{5}\right)\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 1.61-\left(-4.34\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =5.95\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Then you find the area ${A}_{2}$:
$\begin{array}{llll}\hfill {A}_{2}& ={\int }_{1.14}^{3.85}-5\mathrm{cos}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}x-\frac{3}{5}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{1em}{0ex}}-\phantom{\rule{-0.17em}{0ex}}\left(\frac{1}{2}{x}^{2}-\frac{5}{2}x\right)\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={\int }_{1.14}^{3.85}-5\mathrm{cos}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}x-\frac{3}{5}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{1em}{0ex}}-\frac{1}{2}{x}^{2}+\frac{5}{2}x\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(-5\mathrm{sin}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}x-\frac{3}{5}\right)\cdot \frac{2}{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}-\frac{1}{6}{x}^{3}+\frac{5}{4}{x}^{2}\right)|{}_{1.14}^{3.85}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(-\frac{10}{3}\mathrm{cos}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}x-\frac{3}{5}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}-\frac{1}{6}{x}^{3}+\frac{5}{4}{x}^{2}\right)|{}_{1.14}^{3.85}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(-\frac{10}{3}\mathrm{cos}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}\left(3.85\right)-\frac{3}{5}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}-\frac{1}{6}{\left(3.85\right)}^{3}+\frac{5}{4}{\left(3.85\right)}^{2}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}-\left(-\frac{10}{3}\mathrm{cos}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}\left(1.14\right)-\frac{3}{5}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}-\frac{1}{6}{\left(1.14\right)}^{3}+\frac{5}{4}{\left(1.14\right)}^{2}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 12-\left(-1.61\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 13.61\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill {A}_{2}& ={\int }_{1.14}^{3.85}-5\mathrm{cos}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}x-\frac{3}{5}\right)-\phantom{\rule{-0.17em}{0ex}}\left(\frac{1}{2}{x}^{2}-\frac{5}{2}x\right)\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={\int }_{1.14}^{3.85}-5\mathrm{cos}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}x-\frac{3}{5}\right)-\frac{1}{2}{x}^{2}+\frac{5}{2}x\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\phantom{\rule{-0.17em}{0ex}}\left(-5\mathrm{sin}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}x-\frac{3}{5}\right)\cdot \frac{2}{3}-\frac{1}{6}{x}^{3}+\frac{5}{4}{x}^{2}\right)|{}_{1.14}^{3.85}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\phantom{\rule{-0.17em}{0ex}}\left(-\frac{10}{3}\mathrm{cos}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}x-\frac{3}{5}\right)-\frac{1}{6}{x}^{3}+\frac{5}{4}{x}^{2}\right)|{}_{1.14}^{3.85}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\phantom{\rule{-0.17em}{0ex}}\left(-\frac{10}{3}\mathrm{cos}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}\left(3.85\right)-\frac{3}{5}\right)-\frac{1}{6}{\left(3.85\right)}^{3}+\frac{5}{4}{\left(3.85\right)}^{2}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}-\phantom{\rule{-0.17em}{0ex}}\left(-\frac{10}{3}\mathrm{cos}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}\left(1.14\right)-\frac{3}{5}\right)-\frac{1}{6}{\left(1.14\right)}^{3}+\frac{5}{4}{\left(1.14\right)}^{2}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 12-\left(-1.61\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 13.61\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Next, you find the area ${A}_{3}$:
$\begin{array}{llll}\hfill {A}_{3}& ={\int }_{3.85}^{6.18}\frac{1}{2}{x}^{2}-\frac{5}{2}x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}-\phantom{\rule{-0.17em}{0ex}}\left(-5\mathrm{cos}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}x-\frac{3}{5}\right)\right)\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={\int }_{3.85}^{6.18}\frac{1}{2}{x}^{2}-\frac{5}{2}x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}+5\mathrm{cos}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}x-\frac{3}{5}\right)\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(\frac{1}{6}{x}^{3}-\frac{5}{4}{x}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}+5\mathrm{sin}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}x-\frac{3}{5}\right)\cdot \frac{2}{3}\right)|{}_{3.85}^{6.18}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\phantom{\rule{-0.17em}{0ex}}\left(\frac{1}{6}{x}^{3}-\frac{5}{4}{x}^{2}+\frac{10}{3}\mathrm{sin}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}x-\frac{3}{5}\right)\right)|{}_{3.85}^{6.18}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(\frac{1}{6}{\left(6.18\right)}^{3}-\frac{5}{4}{\left(6.18\right)}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}+\frac{10}{3}\mathrm{sin}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}\left(6.18\right)-\frac{3}{5}\right)\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}-\left(\frac{1}{6}{\left(3.85\right)}^{3}-\frac{5}{4}{\left(3.85\right)}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}+\frac{10}{3}\mathrm{sin}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}\left(3.85\right)-\frac{3}{5}\right)\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx -6.12-\left(-12\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 5.88\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill {A}_{3}& ={\int }_{3.85}^{6.18}\frac{1}{2}{x}^{2}-\frac{5}{2}x-\phantom{\rule{-0.17em}{0ex}}\left(-5\mathrm{cos}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}x-\frac{3}{5}\right)\right)\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={\int }_{3.85}^{6.18}\frac{1}{2}{x}^{2}-\frac{5}{2}x+5\mathrm{cos}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}x-\frac{3}{5}\right)\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\phantom{\rule{-0.17em}{0ex}}\left(\frac{1}{6}{x}^{3}-\frac{5}{4}{x}^{2}+5\mathrm{sin}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}x-\frac{3}{5}\right)\cdot \frac{2}{3}\right)|{}_{3.85}^{6.18}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\phantom{\rule{-0.17em}{0ex}}\left(\frac{1}{6}{x}^{3}-\frac{5}{4}{x}^{2}+\frac{10}{3}\mathrm{sin}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}x-\frac{3}{5}\right)\right)|{}_{3.85}^{6.18}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\phantom{\rule{-0.17em}{0ex}}\left(\frac{1}{6}{\left(6.18\right)}^{3}-\frac{5}{4}{\left(6.18\right)}^{2}+\frac{10}{3}\mathrm{sin}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}\left(6.18\right)-\frac{3}{5}\right)\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}-\phantom{\rule{-0.17em}{0ex}}\left(\frac{1}{6}{\left(3.85\right)}^{3}-\frac{5}{4}{\left(3.85\right)}^{2}+\frac{10}{3}\mathrm{sin}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3}{2}\left(3.85\right)-\frac{3}{5}\right)\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx -6.12-\left(-12\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 5.88\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Finally, the total area is
$\begin{array}{llll}\hfill A& ={A}_{1}+{A}_{2}+{A}_{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 5.95+13.61+5.88\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 25.44\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{lll}\hfill A={A}_{1}+{A}_{2}+{A}_{3}\approx 5.95+13.61+5.88\approx 25.44& \phantom{\rule{2em}{0ex}}& \hfill \end{array}$

Previous entry
How to Use Partial Fraction Decomposition for Integration
Next entry
How to Use Solids of Revolution to Find Volumes

# How to Find the Area Between Two Graphs by Integration

Numbers and Quantities
Algebra
Geometry
Statistics and Probability
FunctionsProofs