## Calculus: Early Transcendentals 8th Edition

$\int_{2}^{5}x\sqrt{1+x^3}dx$
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 $\lim\limits_{n \to \infty}\sum\limits_{n=1}^{\infty}x_i\sqrt{1+x_i^3}\Delta x=\int_{2}^{5}x\sqrt{1+x^3}dx$