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Prove that $ f(x) = |x| $ is Riemann integrable on the [-1, 2] using lower and upper integrals

I am confused how to partition ($P_N$) in this interval. Should we consider two different intervals [-1, 0) and [0, 2] or just one interval [-1, 2]? Then how do use the lower and upper integrals? I know that as long as limits of lower and upper integrals are equal, then the function is Riemann integrable.

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For each $n\in\Bbb N$ let$$P_n=\left\{-1,-1+\frac1n,-1+\frac2n,\ldots,-1+\frac{3n}n(=2)\right\}.$$In other words, consider the intervals$$\left[-1,-1+\frac1n\right],\left[-1+\frac1n,-1+\frac2n\right],\ldots,\left[-1+\frac{3n-1}n,2\right].$$There are $3n$ intervals here. The function $f$ is decreasing on the first $n$ of those intervals, that is, on the intervals of the form $\left[-1+\frac{k-1}n,-1+\frac kn\right]$, with $k\in\{1,2,\ldots,n\}$. Therefore, the minimum of $f$ on each of them is attained at $-1+\frac kn$ and that minimum is $1-\frac kn$. And $f$ is increasing on the intervals of the form $\left[-1+\frac{k-1}n,-1+\frac kn\right]$, with $k\in\{n+1,n+2,\ldots,3n\}$. Therefore, the minimum of $f$ on each of them is attained at $-1+\frac{k-1}n$ and that minimum is $-1+\frac{k-1}n$. So,\begin{align}L(f,P_n)&=\sum_{k=1}^n\frac1n\left(1-\frac kn\right)+\sum_{k=n+1}^{3n}\frac1n\left(-1+\frac{k-1}n\right)\\&=\frac1n\left(n-\frac{n+1}2\right)+\frac1n\left(-2n+4n-1\right)\\&=\frac52-\frac3{2n}.\end{align}So, $\lim_{n\to\infty}L(f,P_n)=\frac52$ and a similar computation shows that $\lim_{n\to\infty}U(f,P_n)=\frac52$ too. Therefore, $f$ is Riemann-integrable and $\int_{-1}^2|x|\,\mathrm dx=\frac52$.

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