The area enclosed between the curves y^2=x an | vedclass

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(D) To find the area enclosed between the curves $y^2=x$ and $y=x$, we first find their points of intersection by solving the equations simultaneously

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$\frac{1}{6}$ sq. units

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(D) To find the area enclosed between the curves $y^2=x$ and $y=x$, we first find their points of intersection by solving the equations simultaneously:$y^2 = y \implies y^2 - y = 0 \implies y(y-1) = 0$.Thus, the points of intersection are at $y=0$ and $y=1$. Correspondingly, $x=0$ and $x=1$.The area $A$ is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=1$:$A = \int_0^1 (\sqrt{x} - x) \, dx$$A = \left[ \frac{x^{3/2}}{3/2} - \frac{x^2}{2} ight]_0^1$$A = \left[ \frac{2}{3}x^{3/2} - \frac{1}{2}x^2 ight]_0^1$$A = \left( \frac{2}{3}(1)^{3/2} - \frac{1}{2}(1)^2 ight) - (0 - 0)$$A = \frac{2}{3} - \frac{1}{2} = \frac{4-3}{6} = \frac{1}{6} 	ext{ sq. units}$.

The area enclosed between the curves $y^2=x$ and $y=x$ is

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