Answer
$\int_{2}^{5}x\sqrt{1+x^3}dx$
Work Step by Step
We know the following
$\int_{a}^{b}f(x)dx=\lim\limits_{n \to \infty}\sum\limits_{n=1}^{\infty}f(x_i)\Delta x $
$\Delta x= \frac{b-a}{n}$
$x_i= a+i\Delta x$
In our problem we can see that $f(x)=x\sqrt{1+x^3}$
We are given that the interval of integration is from 2 to 5 therefore these will be our limits of integration
Therefore
$\lim\limits_{n \to \infty}\sum\limits_{n=1}^{\infty}x_i\sqrt{1+x_i^3}\Delta x=\int_{2}^{5}x\sqrt{1+x^3}dx$
