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The function $f$ is (Riemann) integrable on $I$ if there is some $L\in\mathbb{R}$ such that for every $\epsilon>0$ there is a $\delta> 0$ such that if $P$ is any tagged partition of $I$ with $\|P\|<\delta$, then $|S(f,P) − L|<\epsilon$.

By using this definition show that

$$f(x) =\begin{cases}1,& 0\leq x \leq 1\\2,&1 < x \leq 2\end{cases}$$

Any hint?

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Take $L=1\cdot1+2\cdot1=3$ and for a given $\epsilon>0$ choose $\delta=2$ and $P=\{\{[0,1],\text{ with tag }x=0\}, \{[1,2],\text{ with tag }x=2\}\}$

Then $\|P\|=1<2=\delta$ and since $S(f,P)=f(0)\cdot(1-0)+f(2)\cdot(2-1)=3$ we get $$|S(f,P)-L|=|3-3|=0<\epsilon.$$

Riemann sums approximate the area under the graph by a bunch of rectangles. The reason why we can choose a partition that gives a Riemann sum that is exactly the value of the integral is because the graph of the function is the top of a pair of rectangles.

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