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Q18E

Expert-verifiedFound in: Page 369

Book edition
2nd

Author(s)
James Stewart

Pages
830 pages

ISBN
9781133112280

**Sketch the region enclosed by the given curves and**

**find its area. 18. \(y = \left| x \right|,y = {x^2} - 2\)**.

The \(Area = \frac{{20}}{3}\).

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.

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.

\(\left| x \right| = {x^2} - 2\)

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

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

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

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