Question

Let $f\left(x\right)=\underset{x}{\overset{2}{\int }}\frac{dy}{\sqrt{1+y^3}}.$ Then the value of $\underset{0}{\overset{2}{\int }}xf\left(x\right)dx$ is equal to

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

Answer 1

The correct option is D $\frac{2}{3}$

$f\left(x\right)=\underset{x}{\overset{2}{\int }}\frac{dy}{\sqrt{1+y^3}}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{-1}{\sqrt{1+x^3}}$
Now, $I=\underset{0}{\overset{2}{\int }}xf\left(x\right)dx$
$={\left[f\right(x)\cdot \frac{x^2}{2}]}_0^2-\frac{1}{2}\underset{0}{\overset{2}{\int }}{f}^{\prime }\left(x\right)\cdot x^{2}dx$
$=2\cdot f\left(2\right)+\frac{1}{2}\underset{0}{\overset{2}{\int }}\frac{x^2}{\sqrt{1+x^3}}dx$
$=\frac{1}{2}\underset{0}{\overset{2}{\int }}\frac{x^2}{\sqrt{1+x^3}}dx[\because f(2)=0]$
Put $1+x^3=t^2\Rightarrow x^{2}dx=\frac{2}{3}tdt$
Then, $I=\frac{1}{3}\underset{1}{\overset{3}{\int }}dt=\frac{2}{3}$