The area of the region bounded by the curve y | vedclass

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(C) To find the area bounded by the curve $y^2=9x$ and the line $y=3x$,we first find the points of intersection.Substituting $y=3x$ into $y^2=9x$,we g

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$\frac{1}{2}$ sq.units

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(C) To find the area bounded by the curve $y^2=9x$ and the line $y=3x$,we first find the points of intersection.Substituting $y=3x$ into $y^2=9x$,we get $(3x)^2=9x$,which implies $9x^2=9x$,so $x^2-x=0$,giving $x(x-1)=0$. Thus,$x=0$ and $x=1$.For $x=0$,$y=0$. For $x=1$,$y=3$.The area is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=1$.$	ext{Area} = \int_0^1 (\sqrt{9x} - 3x) dx$$= \int_0^1 (3\sqrt{x} - 3x) dx$$= 3 \left[ \frac{2}{3}x^{3/2} - \frac{x^2}{2} ight]_0^1$$= 3 \left( \frac{2}{3}(1)^{3/2} - \frac{(1)^2}{2} ight) - 0$$= 3 \left( \frac{2}{3} - \frac{1}{2} ight)$$= 3 \left( \frac{4-3}{6} ight) = 3 	imes \frac{1}{6} = \frac{1}{2} 	ext{ sq.units}$.

The area of the region bounded by the curve $y^2=9x$ and the line $y=3x$ is

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