Find the area lying between the curves y^{2}= | vedclass

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(A) Given curves are $y^{2}=2x$ and $y=x$. To find the points of intersection,substitute $y=x$ into $y^{2}=2x$: $x^{2}=2x \implies x^{2}-2x=0 \implies

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$ \frac{2}{3} $ sq. units

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(A) Given curves are $y^{2}=2x$ and $y=x$. To find the points of intersection,substitute $y=x$ into $y^{2}=2x$: $x^{2}=2x \implies x^{2}-2x=0 \implies x(x-2)=0$. So,$x=0$ and $x=2$. When $x=0, y=0$ and when $x=2, y=2$. The points of intersection are $(0,0)$ and $(2,2)$. The required area is the integral of the upper curve minus the lower curve from $x=0$ to $x=2$: Area $= \int_{0}^{2} (\sqrt{2x} - x) dx$ $= \int_{0}^{2} (\sqrt{2}x^{1/2} - x) dx$ $= \left[ \sqrt{2} \cdot \frac{x^{3/2}}{3/2} - \frac{x^{2}}{2} ight]_{0}^{2}$ $= \left[ \frac{2\sqrt{2}}{3} x^{3/2} - \frac{x^{2}}{2} ight]_{0}^{2}$ $= \left( \frac{2\sqrt{2}}{3} (2)^{3/2} - \frac{2^{2}}{2} ight) - (0 - 0)$ $= \left( \frac{2\sqrt{2}}{3} \cdot 2\sqrt{2} - \frac{4}{2} ight)$ $= \left( \frac{4 \cdot 2}{3} - 2 ight)$ $= \frac{8}{3} - 2 = \frac{8-6}{3} = \frac{2}{3} 	ext{ sq. units.}$

Find the area lying between the curves $y^{2}=2x$ and $y=x$.

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