The area of the region bounded by the curves | vedclass

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(B) Given curves are $y^{2}=8x$ $(i)$ and $y=x$ (ii).On solving equations $(i)$ and (ii),we substitute $y=x$ into $y^{2}=8x$:$x^{2}=8x \Rightarrow x^{

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$\frac{32}{3}$

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(B) Given curves are $y^{2}=8x$ $(i)$ and $y=x$ (ii).On solving equations $(i)$ and (ii),we substitute $y=x$ into $y^{2}=8x$:$x^{2}=8x \Rightarrow x^{2}-8x=0 \Rightarrow x(x-8)=0$.Thus,the points of intersection are $x=0$ and $x=8$.For $x=0$,$y=0$ and for $x=8$,$y=8$.The required area is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=8$:$	ext{Area} = \int_{0}^{8} (\sqrt{8x} - x) dx$$= \int_{0}^{8} (2\sqrt{2}x^{1/2} - x) dx$$= \left[ 2\sqrt{2} \cdot \frac{x^{3/2}}{3/2} - \frac{x^{2}}{2} ight]_{0}^{8}$$= \left[ \frac{4\sqrt{2}}{3} x^{3/2} - \frac{x^{2}}{2} ight]_{0}^{8}$$= \left( \frac{4\sqrt{2}}{3} (8)^{3/2} - \frac{8^{2}}{2} ight) - (0 - 0)$$= \frac{4\sqrt{2}}{3} (16\sqrt{2}) - \frac{64}{2}$$= \frac{4 	imes 16 	imes 2}{3} - 32$$= \frac{128}{3} - 32 = \frac{128 - 96}{3} = \frac{32}{3}$.Therefore,the area is $\frac{32}{3}$ square units.

The area of the region bounded by the curves $y^{2}=8x$ and $y=x$ is

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