Question

lf $f\left(x\right)=\underset{n\to \infty }{lim}{n}^2(x^{1/n}-x^{1/(n+1)}),x>0$, then $\int xf\left(x\right)dx$ is equal to

• A. $\frac{x^2}{2}+c$
• B. $0$
• C. $x^2\log x-\frac{1}{2}{x}^2+c$
• D. $\frac{x^2}{2}\log x-\frac{x^2}{4}+c$

Answer 1

The correct option is D $\frac{x^2}{2}\log x-\frac{x^2}{4}+c$

$f\left(x\right)=\underset{n\to \infty }{lim}{n}^2(x^{1/n}-x^{1/(n+1)}),x>0$
Let $m=\frac{1}{n}\therefore \text{as}n\to \infty \Rightarrow m\to 0$
$\Rightarrow f\left(x\right)=\underset{n\to 0}{lim}\frac{x^m-x^{m/(m+1)}}{m^2}=\underset{n\to 0}{lim}\frac{x^{m/m+1}(x^{m-m/m+1}-1)}{m^2}$
$=\underset{n\to 0}{lim}\frac{x^{m/m+1}(x^{m^2/m+1}-1)}{m^2/m+1}(m+1)=x^0.\log x.(0+1)=\log x$
Hence $\int xf\left(x\right)dx=\int x\log xdx=\log x.\frac{x^2}{2}-\int \frac{1}{x}\frac{x^2}{2}dx$
$=\frac{x^2}{2}\log x-\frac{x^2}{4}+c$