## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 5 - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises - Page 338: 63

#### Answer

The area of the region is $32/3$. #### Work Step by Step

$y=x^2-2$ and $y=2$ 1) Draft the graph The graph is enclosed below. From the graph, we choose integration with respect to $x$ to employ here for calculation of the region area. 2) Find the limits of integration: We can find the limits of integration by finding points of intersection between the curve and the line. $$x^2-2=2$$ $$x^2=4$$ $$x=\pm2$$ So, the upper limit is $2$ and the lower one is $-2$. 3) Find the area: Looking at the draft, we see that the region is bounded above by $y=2$ and below by $y=x^2-2$. So, according to definition, the area of the region is $$A=\int^2_{-2}[2-(x^2-2)]dx=\int^2_{-2}(4-x^2)dx$$ $$A=\Big(4x-\frac{x^3}{3}\Big)\Big]^2_{-2}$$ $$A=\Big(8-\frac{8}{3}\Big)-\Big(-8+\frac{8}{3}\Big)$$ $$A=2\Big(8-\frac{8}{3}\Big)$$ $$A=\frac{32}{3}$$ After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.