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.
