The area of the region bounded by the curves | vedclass

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(B) The curves are $y = x^2$ and $y = |x|$.Since both curves are symmetric about the $y$-axis,the total area is twice the area in the first quadrant.I

Vedclass · Accepted Answer

$1/3$

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(B) The curves are $y = x^2$ and $y = |x|$.Since both curves are symmetric about the $y$-axis,the total area is twice the area in the first quadrant.In the first quadrant,$y = |x|$ becomes $y = x$.The intersection points are found by setting $x^2 = x$,which gives $x(x - 1) = 0$,so $x = 0$ and $x = 1$.The area in the first quadrant is $\int_0^1 (x - x^2) \, dx$.Evaluating the integral: $\int_0^1 (x - x^2) \, dx = [\frac{x^2}{2} - \frac{x^3}{3}]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$.Therefore,the total area is $2 	imes \frac{1}{6} = \frac{1}{3}$.

The area of the region bounded by the curves $y = x^2$ and $y = |x|$ is

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