The area bounded by the curves y=x^2 and y-6= | vedclass

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(C) The given curves are $y=x^2$ and $y=6-|x|$.Due to symmetry about the $y$-axis,the required area is $2 	imes$ the area in the first quadrant.In th

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$\frac{44}{3}$

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(C) The given curves are $y=x^2$ and $y=6-|x|$.Due to symmetry about the $y$-axis,the required area is $2 	imes$ the area in the first quadrant.In the first quadrant $(x \ge 0)$,the curves are $y=x^2$ and $y=6-x$.To find the intersection point $A$,we set $x^2 = 6-x$,which gives $x^2+x-6=0$.Solving this,$(x+3)(x-2)=0$. Since $x \ge 0$,we have $x=2$.At $x=2$,$y=2^2=4$. So,the point of intersection is $A(2, 4)$.The area in the first quadrant is bounded by $x=0$ to $x=2$ between the curves $y=6-x$ and $y=x^2$.Area $= \int_0^2 ((6-x) - x^2) dx = [6x - \frac{x^2}{2} - \frac{x^3}{3}]_0^2$$= (12 - 2 - \frac{8}{3}) - 0 = 10 - \frac{8}{3} = \frac{30-8}{3} = \frac{22}{3}$.The total area is $2 	imes \frac{22}{3} = \frac{44}{3}$.

The area bounded by the curves $y=x^2$ and $y-6=-|x|$ is

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