# How do you find the area of the region bounded by the curves y=|x| and y=x^2-2 ?

Sep 25, 2014

This is a definite integral problem

First we need to find the intersections of these 2 functions by setting them equal to each other.

$| x | = {x}^{2} - 2$

This has to be split up into 2 equations.

$x = {x}^{2} - 2$ AND $- x = {x}^{2} - 2$

FIRST
$x = {x}^{2} - 2$
${x}^{2} - x - 2 = 0$
Factor
$\left(x - 2\right) \cdot \left(x + 1\right) = 0$
$x - 2 = 0$
$x = 2$

$x + 1 = 0$
$x = - 1$

SECOND
$- x = {x}^{2} - 2$
${x}^{2} + x - 2 = 0$
Factor
$\left(x + 2\right) \cdot \left(x - 1\right) = 0$
$\left(x + 2\right) = 0$
$x = - 2$

$\left(x - 1\right) = 0$
$x = 1$

The interval is from $\left[- 2 , 2\right]$

The function $| x |$ is greater than the function ${x}^{2} - 2$ over the interval [-2,2].

${\int}_{- 2}^{2} | x | - \left({x}^{2} - 2\right) \mathrm{dx} = {\int}_{- 2}^{2} | x | - {x}^{2} + 2 \mathrm{dx}$

$= {\int}_{- 2}^{2} | x | \mathrm{dx} - {\int}_{- 2}^{2} {x}^{2} \mathrm{dx} + {\int}_{- 2}^{2} 2 \mathrm{dx}$

Note: |x| is an even function and symmetric about the $y$-axis so make this function easier to manage we can assume x>0 and double the area found by multiplying it by 2 and changing the boundaries from 0 to 2.

$= 2 \cdot {\int}_{0}^{2} x \mathrm{dx} - {\int}_{- 2}^{2} {x}^{2} \mathrm{dx} + {\int}_{- 2}^{2} 2 \mathrm{dx}$

$= 2 \cdot {\left[{x}^{2} / 2\right]}_{0}^{2} - 1 \cdot {\left[{x}^{3} / 3\right]}_{- 2}^{2} + 1 \cdot {\left[2 x\right]}_{- 2}^{2}$

$= 2 \cdot \left[{\left(2\right)}^{2} / 2 - {\left(0\right)}^{2} / 2\right] - 1 \cdot \left[{\left(2\right)}^{3} / 3 - {\left(- 2\right)}^{3} / 3\right] + 1 \cdot \left[2 \left(2\right) - 2 \left(- 2\right)\right]$

$= 2 \cdot \left[2 - 0\right] - 1 \cdot \left[\frac{8}{3} + \frac{8}{3}\right] + 1 \cdot \left[4 + 4\right]$

$= 2 \cdot \left[2\right] - \left[\frac{16}{3}\right] + \left[8\right]$

$= \left[4\right] - \left[\frac{16}{3}\right] + \left[8\right]$

$= \left[\frac{12}{3}\right] - \left[\frac{16}{3}\right] + \left[\frac{24}{3}\right]$

$= \left[- \frac{4}{3}\right] + \left[\frac{24}{3}\right]$

$= \left[\frac{20}{3}\right]$

$= 6 \frac{2}{3} \mathmr{and} 6.6667$