The area (in sq. units) bounded by the curves $y=\sqrt{x}, 2 y-x+3=0$, $\mathrm{X}$-axis and lying in the first quadrant, is

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Aicharya · Accepted Answer

$\begin{aligned} & \int_0^9 \sqrt{x} \mathrm{~d} x-\frac{1}{2} 	imes 6 	imes 3 \ & =\frac{2}{3}\left[x^{3 / 2}ight]_0^9-9\end{aligned}$ $=\frac{2}{3} 	imes 27-9=9$

The area (in sq. units) bounded by the curves $y=\sqrt{x}, 2 y-x+3=0$, $\mathrm{X}$-axis and lying in the…

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