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(C) Given curves are $y=x^2$ $(i)$ and $y=|x|$ $(ii)$.Since both curves are symmetric about the $y$-axis,the total area is twice the area in the first

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(C) Given curves are $y=x^2$ $(i)$ and $y=|x|$ $(ii)$.Since both curves are symmetric about the $y$-axis,the total area is twice the area in the first quadrant where $x \ge 0$.For $x \ge 0$,$y=|x|=x$.Solving $y=x^2$ and $y=x$,we get $x^2=x$,which implies $x(x-1)=0$,so $x=0$ or $x=1$.The intersection points in the first quadrant are $(0,0)$ and $(1,1)$.The area in the first quadrant is $\int_0^1 (x - x^2) dx$.Total 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{1}{6} ight) = \frac{1}{3} 	ext{ sq. units}$.

The area (in sq. units) enclosed between the curves $y=x^2$ and $y=|x|$ is

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