Calculate an integral using Riemann sum

Question

Using the Upper and lower Riemann sum, calculate : $\int^3_1$ $x$

Upper sum : $\sum\limits^n _{i=1} =supremum_k * (x_{k-1} , x_k)$

Lower sum: $\sum\limits^n _{i=1}=infimum_k * (x_{k-1} , x_k)$

I have tried using a equipartition $(1+2k)\over n$ and $3k\over n$but I couldn't arrive to the final result which is 4

Answer 1

Suppose you divide $[1,3]$ into $n$ identical intervals: $1,1+\frac2n,1+\frac4n,\ldots,1+\frac{2n}n=3$. Then the upper sum will be$$\frac2n\sum_{k=1}^n\left(1+\frac{2k}n\right)=2+2\frac{n+1}n$$and $\inf_{n\in\mathbb N}2+2\frac{n+1}n=4$. Can you do the same with the lower sum?