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(A) The area bounded by the parabola $y=x^{2}$ and the line $y=|x|$ is symmetric about the $y$-axis.The points of intersection of $y=x^{2}$ and $y=|x|

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$1/3$

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(A) The area bounded by the parabola $y=x^{2}$ and the line $y=|x|$ is symmetric about the $y$-axis.The points of intersection of $y=x^{2}$ and $y=|x|$ are found by setting $x^{2}=|x|$.For $x \ge 0$,$x^{2}=x \implies x(x-1)=0$,so $x=0$ or $x=1$.The intersection points are $(0,0)$ and $(1,1)$ in the first quadrant,and $(-1,1)$ in the second quadrant.The required area is $2 	imes \int_{0}^{1} (|x| - x^{2}) dx$.Since $x \ge 0$ in the first quadrant,$|x|=x$.Area $= 2 \int_{0}^{1} (x - x^{2}) dx$$= 2 \left[ \frac{x^{2}}{2} - \frac{x^{3}}{3} ight]_{0}^{1}$$= 2 \left( \frac{1}{2} - \frac{1}{3} ight)$$= 2 \left( \frac{3-2}{6} ight) = 2 \left( \frac{1}{6} ight) = \frac{1}{3} 	ext{ sq. units.}$

Find the area of the region bounded by the parabola $y=x^{2}$ and $y=|x|$.

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