Question

Area of the region enclosed by $y^2=8x$ and $y=2x$ is

• A. $\frac{4}{3}$
• B. $\frac{3}{4}$
• C. $\frac{1}{4}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $\frac{4}{3}$

Given curves are $y^2=8x$ and $y=2x$
Let's find out their intersection point

$\Rightarrow 4x^2=8x$

$\Rightarrow x^2-2x=0$

$\Rightarrow x(x-2)=0$

$\Rightarrow x=0,2$

The required area is
$A=\stackrel{2}{\underset{0}{\int }}(\sqrt{8x}-2x)dx$
$=\stackrel{2}{\underset{0}{\int }}\sqrt{8x}dx-\stackrel{2}{\underset{0}{\int }}2xdx$
$=\sqrt{8}\times \frac{2}{3}\times [x^{3/2}{]}_0^2-[x^2{]}_0^2$
$=\frac{2\sqrt{8}}{3}\sqrt{8}-4$
$=\frac{16}{3}-4=\frac{4}{3}squnits$