Question

Find the area of the region bounded by the curve $y^2=x$ and the lines $x=4,x=9$ and the $x-$axis in the first quadrant.

Answer 1

AB line x = 4

CD line x = 9

$y^2=x$

Let we need to calculate

${\int }_4^{9}y.dx$

as $y^2=x\to y=\pm \sqrt{x}$

As first Quadrant $y=\sqrt{x}$

${\int }_4^{9}y.dx={\int }_4^9\sqrt{x}.dx$

$={\int }_4^9{x}^{1/2}.dx$

$={\left[\frac{x^{3/2}}{\frac{3}{2}}\right]}_4^9$

$=\frac{2}{3}{\left[x^{3/2}\right]}_4^9$

$=\frac{2}{3}\left[\right(9^{1/2}{)}^3-{\left(\right(4{)}^{1/2})}^3]$

$=\frac{2}{3}[3^3-2^3]$

$=\frac{2}{3}[27-8]$

$=\frac{2}{3}\times 19=\frac{38}{3}$

Area = $\frac{38}{3}=12.6$ Square unit