Originally Posted by

**Abu-Khalil** You may define the partition $\displaystyle x_0<x_1<\ldots<x_{n-1}<x_n$, where $\displaystyle x_k=3+\left(\frac{7-3}{n}\right)k\Rightarrow \Delta x_k=x_k-x_{k-1}=\left(\frac{7-3}{n}\right)$ so when $\displaystyle n\to\infty, \Delta x_k \to 0$.

Finally, $\displaystyle \int_3^7x^2-3x+5dx=\lim_{n\to\infty}\frac{7-3}{n}\sum_{k=1}^nx_k^2-3x_k+5$ and calculate that sum isn't a problem knowing $\displaystyle \sum_{k=1}^nk=\frac{n(n+1)}{2}$ and $\displaystyle \sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}$.