TITLE: Is Riemann integration only concerned with rates of change?
QUESTION [6 upvotes]: Consider the Riemann integrable function $f(x)$. Whenever I see the expression
$$
\int_a^b f(x) \ dx
$$
I read this as
$$
\int_a^b \frac{dF(x)}{dx} \ dx
$$
where $F(x)$ is the antiderivative of $f(x)$. In other words, I consider every Riemann integral to be concerned with integrating a "rate of change", which in this case is $\frac{dF(x)}{dx}$. Is this thinking correct? That is, do all Riemann integrals take in a "rate of change" as an input?

REPLY [9 votes]: No, this is not true in general. For instance, take the step function
$$f(x)=\begin{cases}0&x\leq0\\1&x>0\end{cases}$$
It is Riemann integrable on all intervals, but there is no function $F$ whose derivative is $f$. You can prove this using Darboux's theorem (all derivative functions have the intermediate value property, which $f$ does not have), or using the fundamental theorem of calculus: if an antiderivative of a Riemann integrable function on an interval exists, then it is equal to an integral function up to a constant summand. But you can show that the integral functions of $f$ are not differentiable.

REPLY [7 votes]: No. Say $$f(x)=\begin{cases}1,&(x=0),
\\0,&(x\ne0).\end{cases}$$Then $f$ is Riemann integrable on $[-1,1]$, but there does not exist a function $F$ with $f=F'$.