Integrate functions from point 'a' to point 'a' proof
Consider the definition of Riemann Sum on $[a,b]$
$$ \sum_{k=1}^n f(x_j^\ast)(x_j-x_{j-1}) $$
where $a=x_0 \leqslant x_1 \leqslant \dots \leqslant x_{n-1} \leqslant x_n=b$ is a subdivision of $[a,b]$ and $x^\ast_j\in [x_{j-1},x_j]$, $j=1,\dots,n$, are sample points. Then, if $a=b$, we clearly have $a=x_0=x_1=\dots=x_n=b$, so any Riemann Sum is trivially $0$. It converges to $0$, so $\int_a^b f(x)\; dx=0$.
Note: As noted in @Eric Schmidt's answer, this depends a lot on your choice of definition. In the explanation above, I assumed that points in a subdivision could be equal, which is certainly not universal. If you assume that they have to be different (strict inequality $<$ instead of large ones $\leqslant$), then you cannot define $\int_a^a f(x)\; dx$ at all, since the Riemann Sum itself, hence its convergence, would not be defined.
This will depend on the exact definitions used. It is almost certainly the case that one of the following holds:
- The statement is true by explicit definition.
- The statement follows trivially from the definition. (Technically, option 1 is an instance of this.)
- Integrals where both bounds are the same are not defined.
Consulting texts conveniently at hand, I find the following:
Calculus with Analytic Geometry, 6th ed., by R. E. Larson et al., follows option 1.
Calculus and Analytic Geometry, 9th ed., by G. B. Thomas and R. L. Finney (the cover says Thomas' Calculus Alternate Edition) follows option 1.
Elementary Analysis: The Theory of Calculus by K. A. Ross, appears to follow option 3. (The notation $[a, b]$ is explicitly restricted to the case $a < b$ in this book.)
Principles of Mathematical Analysis, 3rd ed., by W. Rudin, appears to follow option 3 (for the Riemann(-Stieltjes) integral), though if you allow $[a, a]$ as a closed interval you could interpret this as option 2. That the integral is zero holds because, in this case, all the points in a partition must be equal, so the differences between two successive points are zero, so the upper and lower sums over the partition are zero, so the upper and lower integrals are zero, so the integral exists and is zero. But I don't think Rudin means to include this possibility, because Theorem 6.12(c) is written to avoid it.
It's really very simple. In pure mathematical sence:
Suppose, \begin{align*} \int f(x)\ dx&=g(x)+c\\ \implies\int_a^af(x)\ dx&=g(x)+c\big|_{x=a}^{x=a}\\ \implies\int_a^af(x)\ dx&=[g(a)+c]-[g(a)+c]\\ \implies\int_a^af(x)\ dx&=0\end{align*}