Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 5 - Applications Of The Definite Integral In Geometry, Science, And Engineering - 5.1 Area Between Two Curves - Exercises Set 5.1 - Page 353: 4

Answer

$$A = \frac{{16}}{3}$$

Work Step by Step

$$\eqalign{ & x = - y \Rightarrow y = - x \cr & x = 2 - {y^2} \Rightarrow y = \sqrt {2 - x} \cr & {\text{From the graph we can note that the area is given by}} \cr & A = \int_{ - 2}^2 {\left( {\sqrt {2 - x} - \left( { - x} \right)} \right)} dx \cr & A = \int_{ - 2}^2 {\left( {{{\left( {2 - x} \right)}^{1/2}} + x} \right)} dx \cr & {\text{Integrate}} \cr & A = \left[ { - \frac{2}{3}{{\left( {2 - x} \right)}^{3/2}} + \frac{1}{2}{x^2}} \right]_{ - 2}^2 \cr & {\text{Evaluate }} \cr & A = \left[ { - \frac{2}{3}{{\left( {2 - 2} \right)}^{3/2}} + \frac{1}{2}{{\left( 2 \right)}^2}} \right] - \left[ { - \frac{2}{3}{{\left( {2 + 2} \right)}^{3/2}} + \frac{1}{2}{{\left( { - 2} \right)}^2}} \right] \cr & {\text{Simplify}} \cr & A = 2 - \left[ { - \frac{{10}}{3}} \right] \cr & A = \frac{{16}}{3} \cr} $$
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