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

Question

Aicharya · Accepted Answer

$\begin{aligned} 	ext { Required area } & =\int_0^9 \sqrt{x} \mathrm{~d} x-\int_3^9\left(\frac{x-3}{2}ight) \mathrm{d} x \ & =\left[\frac{2 x^{3 / 2}}{3}ight]_0^9-\frac{1}{2}\left[\frac{x^2}{2}-3 xight]_3^9 \ & =\frac{2}{3}(27-0)-\frac{1}{2}(36-18) \ & =9 	ext { sq.units }\end{aligned}$

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

Solution