Single Variable Calculus, Chapter 6, 6.1, Section 6.1, Problem 26

Sketch the region enclosed by the curves $\displaystyle y = |x|$ and $y = x^2 - 2$. Then find the area of the region.







By using a vertical strip,

$\displaystyle A = \int^{x_2}_{x_1} (y_{\text{upper}}, y_{\text{lower}}) dx$

Recall that,

$f(x) = |x| \to f(x) = \begin{array}{cc}
x & \text{ for } x > 0 \\
-x & \text{ for } x < 0
\end{array}$

In order to get the value of the upper and lower limits, we
equate the two functions to get its points of intersection.
So..


$
\begin{equation}
\begin{aligned}

& \text{For } x > 0
\\
\\
& x = x^2 - 2
\\
\\
& x^2 - x - 2 = 0
\end{aligned}
\end{equation}
$


By using Quadratic Formula, we get

$x = 0$ and $x = 2$


$
\begin{equation}
\begin{aligned}

& \text{For } x < 0
\\
\\
& -x = x^2 - 2

\end{aligned}
\end{equation}
$


By using Quadratic Formula, we get

$x = 0$ and $x = -2$

Let $A_1$ and $A_2$ be the area on the left and right part respectively. Thus,


$
\begin{equation}
\begin{aligned}

A_1 =& \int^0_{-2} [-x - (x^2 - 2)] dx
\\
\\
A_1 =& \int^0_{-2} [-x - x^2 + 2] dx
\\
\\
A_1 =& \left[ - \frac{x^2}{2} - \frac{x^3}{3} + 2x \right]^0_{-2}
\\
\\
A_1 =& \frac{10}{3} \text{ square roots }

\end{aligned}
\end{equation}
$


On the other part,


$
\begin{equation}
\begin{aligned}

A_2 =& \int^2_0 [x - (x^2 - 2)] dx
\\
\\
A_2 =& \int^2_0 [x - x^2 + 2] dx
\\
\\
A_2 =& \frac{10}{3} \text{ square roots}

\end{aligned}
\end{equation}
$


Therefore, the total area is $\displaystyle A_1 + A_2 = \frac{20}{3}$