I suppose they mean a Riemann Sum.
You are adding up the area of an infinite number of rectangles using the right side of them as f(x). The width of each rectangle is \(\displaystyle {\Delta}x\).
This is analogous to the dx in your 'regular' integral.
\(\displaystyle {\Delta}x=\frac{b-a}{n}=\frac{3-(-1)}{n}=\frac{4}{n}\)
The right hand method is \(\displaystyle x_{k}=a+k{\Delta}x\Rightarrow -1+\frac{4k}{n}\)
\(\displaystyle \left[\left(-1+\frac{4k}{n}\right)^{2}+9\left(-1+\frac{4k}{n}\right)-2\right]\left(\frac{4}{n}\right)\)
Expanding and taking the sums, gives:
\(\displaystyle \frac{64}{n^{3}}\sum_{k=1}^{n}k^{2}+\frac{112}{n^{2}}\sum_{k=1}^{n}k-\frac{40}{n}\sum_{k=1}^{n}1\)
Note the k and k^2?. Here we sub in the closed form for the sum of consecutive integers and sum of squares.
They are \(\displaystyle \sum_{k=1}^{n}k=\frac{n(n+1)}{2}, \;\ \sum_{k=1}^{n}k^{2}=\frac{n(n+1)(2n+1)}{6}, \;\ \sum_{k=1}^{n}1=n\)
Sub those in.
\(\displaystyle \frac{64}{n^{3}}\cdot \frac{n(n+1)(2n+1)}{6}+\frac{112}{n^{2}}\cdot \frac{n(n+1)}{2}-\frac{40}{n}\cdot n\)
The algebra is a little messy, but not too bad. It whittles down and will be entirely in terms of n. Then, take the limit as \(\displaystyle n\to {\infty}\).
The result is the solution to your definite integral.
