Riemann Sum of $\int_{-1}^1|x| \ dx$

Question

I am trying to find $\int_{-1}^1f(x) \ dx=\int_{-1}^1|x| \ dx$ using Riemann Sums. I have split $[-1,1]$ into two partitions, $P_n$ and $Q_n$, as the function is decreasing and increasing between $-1$ and $1$. So $P_n$ is a partition of $[-1,0]$ and $Q_n$ is a partition of $[0,1]$. Hence, $P_n=\left\{ -1+\frac kn, \ k=0,\dots,n \right\}$ and $Q_n=\left\{\frac kn, \ k=0,\dots,n \right\}$. So now $R_n=P_n \ \cup \ Q_n$ is the partition of $[-1,1]$. For $Q_n$ the upper and lower sums came out nicely; however, for $P_n$ it is not coming out as nicely. This is where I have got to.

Since $f$ is decreasing on $P_n$,

\begin{align} U(f,P_n)&=\sum_{k=1}^nf(y_{k-1})\frac1n \ \ \ \ (\text{where $y_{k-1}$ are the points in $P_n$}) \\ &=\sum_{k=1}^n\left| -1+\frac{k-1}{n} \right|\frac1n \\ &=\sum_{k=1}^n\left| \frac{k-1-n}{n} \right|\frac1n \end{align}

Now, I am rather stuck!

Answer 1

$$U(f,P_{n})=\sum_{k=1}^n\left| -1+\frac{k-1}{n} \right|\frac1n=\sum_{k=1}^n\left( 1-\frac{k-1}{n} \right)\frac1n$$ $$=\sum_{k=1}^n \frac{n-k+1}{n^{2}}=\sum_{k=1}^n \frac{n}{n^{2}}-\sum_{k=1}^n \frac{k}{n^{2}}+\sum_{k=1}^n \frac{1}{n^{2}}$$ $$1-\frac{n}{n^{2}}\left( \frac{n(n+1)}{2} \right)+\frac{1}{n}$$ $$=1-\frac{n+1}{2n} \rightarrow \frac{1}{2}$$