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I wasn't really sure how to title this, but the way the problem is phrased makes me think of improper integrals so I thought that might be a good title.

Let $f\colon[a,b]\to\mathbb{R}, a be Riemann integrable on $[a,b]$. Show that $ \int_a^b f(x) dx = \lim_{T\to b}\int_a^T f(x)dx $ I don't really see that there's a whole lot to do here, which is part of my confusion; it feels like it should be harder than this. Let $F$ denote the antiderivative of $f$ which exists by assumption, then apply the fundamental theorem of calculus: $ \lim_{T\to b}\int_a^T f(x)dx = \lim_{T\to b } F(T)-F(a) = F(b)-F(a) = \int_a^b f(x) dx $ The only thing that feels a little shaky is justifying $\displaystyle\lim_{T\to b } F(T)=F(b)$. We're only given that $f$ is Riemann integrable, which says nothing about the continuity of the function. What prevents something weird happening at $b$?

Edit: The fundamental theorem on calculus, Rudin version: If $f$ is Riemann integrable on $[a,b]$ and if there is a differentiable function $F$ on $[a,b]$ such that $F'=f$, then $ \int_a^b f(x) dx = F(b)-F(a) $

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    @chris: I'm going to post a bigger answer.2012-10-23

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In Rudin there are two relevant theorems. Here is one of them:

6.20 Theorem Let $f$ be Riemann integrable on $[a,b]$. For $a \le x \le b$, put $F(x)=\int_a^x f(t) \, dt.$ Then $F$ is continuous on $[a,b]$; furthermore, if $f$ is continuous at a point $x_0$ of $[a,b]$, then $F$ is differentiable at $x_0$, and $F'(x_0)=f(x_0).$

You have already stated the other theorem (6.21) in your edit, but you do not need that theorem to answer your question. (In fact, it does not apply here since $f$ may not be continuous.) If $f$ is Riemann integrable then $F$ (as defined in Theorem 6.20) is continuous on $[a,b]$. Then $\lim_{x \rightarrow b} F(x) = F(b)$, i.e. $\lim_{x \rightarrow b} \int_a^x f(t) \, dt = \int_a^b f(t) \, dt.$

Your proof states "Let $F$ denote the antiderivative of $f$ which exists by assumption", which is simply not true!

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    Thank you. I think what I really need to do is lay out all of the theorems that I have at my disposal and get everything organized. I completely forgot this existed, and as you noted, the FTC isn't particularly useful here, precisely because of objections that I noted earlier. Thanks for all of your help.2012-10-23