Multiple Areas (College Board AP® Calculus AB): Revision Note

The figure below shows a sketch of the curve with equation $y=x(x-1)(x+2)$.

Find the total area of the shaded regions.

Answer:

The zeros are $x=0,1,-2$

Find the definite integral from $x=-2$ to $x=0$

$\begin{array}{rcl}{\int }_{-2}^{0}x\left(x-1\right)\left(x+2\right)dx& =& {\int }_{-2}^0\left(x^3+x^2-2x\right)dx\\ & =& {\left[\frac{1}{4}{x}^4+\frac{1}{3}{x}^3-x^2\right]}_{-2}^0\\ & =& \left(0\right)-\left(\frac{1}{4}·{\left(-2\right)}^4+\frac{1}{3}·{\left(-2\right)}^3-{\left(-2\right)}^2\right)\\ & =& 0-\left(4-\frac{8}{3}-4\right)\\ & =& \frac{8}{3}\end{array}$

Find the definite integral from $x=0$ to $x=1$

$\begin{array}{rcl}{\int }_0^{1}x\left(x-1\right)\left(x+2\right)dx& =& {\left[\frac{1}{4}{x}^4+\frac{1}{3}{x}^3-x^2\right]}_0^1\\ & =& \left(\frac{1}{4}·{\left(1\right)}^4+\frac{1}{3}·{\left(1\right)}^3-{\left(1\right)}^2\right)-\left(0\right)\\ & =& \left(\frac{1}{4}+\frac{1}{3}-1\right)-0\\ & =& -\frac{5}{12}\end{array}$

Subtract the negative integral from the positive one

$\frac{8}{3}-\left(-\frac{5}{12}\right)=\frac{37}{12}$

$\frac{37}{12}$ units squared

The shaded region in the diagram below is the region enclosed by the curves with equations$y=f\left(x\right)$ and $y=g\left(x\right)$ where

$f\left(x\right)=(x-2){(x-3)}^2$

$g\left(x\right)=x^2-5x+6$  

Find the total area of the shaded region.

Answer:

Note first of all that for calculating areas between two curves, it doesn't matter that the areas are partially above and partially below the $x$-axis

Start by finding the $x$-values of the points of intersection of the two curves

$\begin{array}{rcl}f\left(x\right)& =& g\left(x\right)\\ \left(x-2\right){\left(x-3\right)}^2& =& x^2-5x+6\\ \left(x-2\right){\left(x-3\right)}^2& =& \left(x-2\right)\left(x-3\right)\end{array}$

This can be solved by moving everything to one side of the equation and factoring further

$\begin{array}{rcl}\left(x-2\right){\left(x-3\right)}^2-\left(x-2\right)\left(x-3\right)& =& 0\\ \left(x-2\right)\left(x-3\right)\left(\left(x-3\right)-1\right)& =& 0\\ \left(x-2\right)\left(x-3\right)\left(x-4\right)& =& 0\end{array}$

$x=2,x=3,x=4$

Those will be the integration limits for the two integrals we need

Between $x=2$ and $x=3$, $y=f\left(x\right)$ is the 'top' function, so

$\begin{array}{rcl}{\int }_2^3\left(f\left(x\right)-g\left(x\right)\right)dx& =& {\int }_2^3\left(\left(x-2\right){\left(x-3\right)}^2-\left(x^2-5x+6\right)\right)dx\\ & =& {\int }_2^3\left(x^3-9x^2+26x-24\right)dx\\ & =& {\left[\frac{1}{4}{x}^4-3x^3+13x^2-24x\right]}_2^3\\ & =& \left(\frac{81}{4}-81+117-72\right)-\left(4-24+52-48\right)\\ & =& \frac{1}{4}\end{array}$

That is the area for the part of the region on the left

Between $x=3$ and $x=4$, $y=g\left(x\right)$ is the 'top' function, so

$\begin{array}{rcl}{\int }_3^4\left(g\left(x\right)-f\left(x\right)\right)dx& =& {\int }_3^4\left(\left(x^2-5x+6\right)-\left(x-2\right){\left(x-3\right)}^2\right)dx\\ & =& {\int }_3^4\left(-x^3+9x^2-26x+24\right)dx\\ & =& {\left[-\frac{1}{4}{x}^4+3x^3-13x^2+24x\right]}_3^4\\ & =& \left(-64+192-208+96\right)-\left(-\frac{81}{4}+81-117+72\right)\\ & =& \frac{1}{4}\end{array}$

That is the area for the part of the region on the right

Add together to get the total area

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

The total area is $\frac{1}{2}$ units squared

Even if you are using your calculator to work out the value of an integral on the exam

• you should write down the integral you are evaluating, to make sure your answer gets full credit

The shaded region in the diagram below is the region enclosed by the $x$-axis and the graph of the function $y=x^3-12x^2+35x$.

Find the total area of the shaded region.

Answer:

Start by finding the points where the graph intersects the $x$-axis

$\begin{array}{rcl}{x}^3-12x^2+35x& =& 0\\ x\left(x-5\right)\left(x-7\right)& =& 0\end{array}$

$x=0,x=5,x=7$

To work this out by hand we'd need to find the two areas separately, between $x=0$ and $x=5$ (the bit above the $x$-axis), and between $x=5$ and $x=7$ (the bit below the $x$-axis)

Instead we can work out the entire area between $x=0$ and $x=7$ with one integral using absolute value

$\mathrm{Area}={\int }_0^7\left|x^3-12x^2+35x\right|dx$

That integral can be evaluated using your graphing calculator

$\begin{array}{rcl}{\int }_0^7\left|x^3-12x^2+35x\right|dx& =& \frac{407}{4}=101.75\end{array}$

Total area = 101.75 units squared

The shaded region in the diagram below is the region enclosed by the curves with equations$y=f\left(x\right)$ and $y=g\left(x\right)$ where

$f\left(x\right)=(x-2){(x-3)}^2$

$g\left(x\right)=x^2-5x+6$  

Find the total area of the shaded region.

Answer:

Start by finding the $x$-values of the points of intersection of the two curves

$\begin{array}{rcl}\left(x-2\right){\left(x-3\right)}^2& =& x^2-5x+6\\ \left(x-2\right){\left(x-3\right)}^2& =& \left(x-2\right)\left(x-3\right)\\ \left(x-2\right){\left(x-3\right)}^2-\left(x-2\right)\left(x-3\right)& =& 0\\ \left(x-2\right)\left(x-3\right)\left(\left(x-3\right)-1\right)& =& 0\\ \left(x-2\right)\left(x-3\right)\left(x-4\right)& =& 0\end{array}$

$x=2,x=3,x=4$

To work this out by hand we'd need to find the two areas separately, between $x=2$ and $x=3$ (where $y=f\left(x\right)$ is on top), and between $x=3$ and $x=4$ (where $y=g\left(x\right)$ is on top)

Instead we can work out the entire area between $x=2$ and $x=4$ with one integral using absolute value

$\begin{array}{rcl}\mathrm{Area}& =& {\int }_2^4\left|f\left(x\right)-g\left(x\right)\right|dx\\ & =& {\int }_2^4\left|\left(x-2\right){\left(x-3\right)}^2-\left(x^2-5x+6\right)\right|dx\end{array}$

That integral can be evaluated using your graphing calculator

$\begin{array}{rcl}{\int }_2^4\left|\left(x-2\right){\left(x-3\right)}^2-\left(x^2-5x+6\right)\right|dx& =& \frac{1}{2}=0.5\end{array}$

Total area = 0.5 units squared