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Q18E

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Found in: Page 369

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Sketch the region enclosed by the given curves andfind its area. 18. $$y = \left| x \right|,y = {x^2} - 2$$.

The $$Area = \frac{{20}}{3}$$.

See the step by step solution

## The sketch of the graph of given equations

As always, draw a picture or use calculator.

A tip for using TI-83 or TI-84 graphing calculator, you can graph the absolute value function by pressing the “Y=” button, then the “MATH” button, then scroll right to the “NUM” category. Select “abs(“ and enter the variable x to graph the function appropriately.

## Find the intersection points

To find where the intersection points are, we must set the equations equal to one another and solve for x. However, because we are interacting with the absolute value function, we must be careful while moving ahead.

Whenever dealing with the absolute value function in these problems, always be aware that there are two cases.

$$\left| x \right| = {x^2} - 2$$

## When x is greater than zero

\begin{aligned}{l}x = {x^2} - 2\\0 = {x^2} - x - 2\\x = - 1\;or\;x = 2\end{aligned}

## When x is less than zero

\begin{aligned}{l} - x = {x^2} - 2\\0 = {x^2} + x - 2\\x = 1\;or\;x = - 2\end{aligned}

## Check derived answer against the graph by finding area

\begin{aligned}{l}A = 2\int\limits_0^2 {(x - ({x^2} - 2))dx} \\ = 2\left( {\left( {\frac{{ - {{\left( 2 \right)}^3}}}{3} + \frac{{{{\left( 2 \right)}^2}}}{2} + 2(2)} \right) - (0)} \right)\\ = 2\left( {\frac{{ - 8}}{3} + 2 + 4} \right)\\ = 2\left( {\frac{{10}}{3}} \right)\\ = \frac{{20}}{3}\end{aligned}