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I'm sorry guys, I know this is probably relatively trivial, but I'm having trouble proving that this function
$$f:[a,b]\to \mathbb{R}$$ given by $f(x)=x$, is integrable on the interval $[a,b]$.

I've been asked to prove this by using the definition of Riemann integral, which I understand just fine I believe. other examples with discrete endpoints like $[0,1]$ or $[0,b]$ make perfect sense to me, but I think since the interval for this question is general it's throwing me off.

I've tried to approach the proof with a partition $$P = \{a, a + (b-a)/n, a + 2(b-a)/n,\ldots, a + n(b-a)/n\}$$ but can't seem to get anything worthwhile. I'm thinking(hoping) that I'm just having some very simple logical oversight. If anybody could give me some assistance or point me in the right direction, it would be greatly appreciated.

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    $\begingroup$ what is the definition of Riemann Integrable you are working from? $\endgroup$ Dec 15 '16 at 1:14
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    $\begingroup$ All the continuous functions are Riemann-Integrable. Your function $f(x)=x$ is continuons, then, Riemann-Integrable. $\endgroup$ Dec 15 '16 at 1:42
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You are trying to show the upper and lower Riemann sums converge to the same value, correct?

Let $\epsilon = (b-a)/n$ be the width of a rectangle when there are $n$ rectangles.

The lower sum is $L = a\epsilon + (a+\epsilon)\epsilon + \ldots + (a+(n-1)\epsilon)\epsilon$.

The upper sum is $U = (a+\epsilon)\epsilon + \ldots + (a+(n-1)\epsilon)\epsilon + (a+n\epsilon)\epsilon$.

What happens to the difference $U-L$ as $n\rightarrow \infty$?

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