Question

The area (in square units) bounded by the curves $y=\sqrt{x}$,

$2y-x+3=0$, and lying in the first quadrant is

Answer 1

For $p$ $y=\sqrt{x}$

$y^2=x$

$2y-x+3=0$

$=2y-y^2+3=0$

$=y^2-2y-3=0$

$=y^2-3y+y-3=0$

$y(y-3)+1(y-3)=0$

$y=-1,y=3$

But we need the area in the first quadrant only hence

$y=\sqrt{x}=x=y^2$

$x=1,9$

$P=(9,3)$

$A={\int }_0^9\sqrt{x}dx+{\int }_3^9\frac{x-3}{2}dxA={\int }_0^9\sqrt{x}dx-\frac{1}{2}{\int }_3^9(x-3)dxA={\left[\frac{{2\left(x\right)}^{3/2}}{3}\right]}_0^9+\frac{1}{2}{[\frac{{\left(x\right)}^2}{2}-3x]}_3^{9}A=\frac{2}{3}\times 27-\frac{1}{2}[\frac{81}{2}-27-\frac{9}{2}+9]A=18+18A=36squnits$