Find the area bounded by the curves {(x, y): | vedclass

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(A) The area bounded by the curves $y = x^2$ and $y = |x|$ is the region where $x^2 \leq y \leq |x|$.The curves intersect where $x^2 = |x|$. Since $x^

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$1/3$

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(A) The area bounded by the curves $y = x^2$ and $y = |x|$ is the region where $x^2 \leq y \leq |x|$.The curves intersect where $x^2 = |x|$. Since $x^2 = |x|^2$,we have $|x|^2 - |x| = 0$,which gives $|x|(|x| - 1) = 0$. Thus,$x = 0, 1, -1$.The region is symmetric about the $y$-axis. We calculate the area in the first quadrant $(x \geq 0)$ and multiply by $2$.In the first quadrant,the curves are $y = x$ and $y = x^2$. The area is given by:$	ext{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 bounded by the curves $\{(x, y): y \geq x^{2} \text{ and } y=|x|\}$.

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