Question

The area under the curve $y=x$ and $y^2=16x$ is

• A. $\frac{64}{3}$
• B. $\frac{32}{3}$
• C. $\frac{16}{3}$
• D. $\frac{128}{3}$

Answer 1

The correct option is D $\frac{128}{3}$


Point of contact
$x^2=16x\Rightarrow x=0;16$
Area enclosed,
$=\underset{0}{\overset{16}{\int }}4\sqrt{x}-xdx=\frac{8}{3}[x^{3/2}{]}_0^{16}-{\left[\frac{x^2}{2}\right]}_0^{16}=\frac{8}{3}\times 64-\frac{16\times 16}{2}=8\times 16(\frac{4}{3}-1)=\frac{128}{3}$