## Calculus: Early Transcendentals 8th Edition

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