Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 6 - Review - Exercises: 3

Answer

$ \frac{7}{12} $ sq units

Work Step by Step

Step 1 of 3: For a function of the form$ f \left( x \right) =x^{n} $, the integral of such function is given as follows: $ \int _{}^{}f \left( x \right) dx=\frac{x^{n+1}}{n+1} $ Step 2 of 3: Consider the data as given: $ y=1-2x^{2} $,$ y= \vert x \vert $ The region enclosed by the curves$ y=1-2x^{2} $,$ y= \vert x \vert $ is shown below: Step 3 of 3: Let $ f \left( x \right) =1-2x^{2} $, And, $ g \left( x \right) = \vert x \vert $ The area between the curves$ y=f \left( x \right) $ ,$ y=g \left( x \right) $ , and between$ x=a $ and $ x=b $ is given as follows: Thus, $ A= \int _{a}^{b} \vert f \left( x \right) -g \left( x \right) \vert dx $ Here, the area of the region bounded by the curves $ y=1-2x^{2} $ ,$ y= \vert x \vert $ and between $ x=-0.618 $ And $ x=0.618 $ is evaluated as follows: Then, $ A= \int _{-0.618}^{0} \left( 1-2x^{2}- \left( -x \right) \right) dx+ \int _{0}^{0.618} \left( 1-2x^{2}-x \right) dx $ $ = \vert x-\frac{2x^{3}}{3}+\frac{x^{2}}{2} \vert _{-0.618}^{0}+ \vert x-\frac{2x^{3}}{3}-\frac{x^{2}}{2} \vert _{0}^{0.618} $ $ = \left( \left( 0-\frac{2 \left( 0 \right) ^{3}}{3}+\frac{ \left( 0 \right) ^{2}}{2} \right) - \left( \left( -0.618 \right) -\frac{2 \left( -0.618 \right) ^{3}}{3}+\frac{ \left( -0.618 \right) ^{2}}{2} \right) \right) + \left( \left( \left( 0.618 \right) -\frac{2 \left( 0.618 \right) ^{3}}{3}-\frac{ \left( 0.618 \right) ^{2}}{2} \right) - \left( \left( 0 \right) -\frac{2 \left( 0 \right) ^{3}}{3}-\frac{ \left( 0 \right) ^{2}}{2} \right) \right) $ $ =\frac{7}{12} $ Therefore, the required area of the given region is $ \frac{7}{12} $ sq units.
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