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) $