Integrate functions from point 'a' to point 'a' proof I understand that the integral of any function from and to the same point must equal zero, such;
$$\int_a^a f(x) \,dx= 0$$
It makes sense, area from one point to the same point should be zero. But, how is this shown in a mathematical sense, proving it fully, without just saying 'it makes sense'?
 A: $$\int_a^c f(x) dx = \int_a^b f(x) dx + \int_b^c f(x) dx$$
And if you set $b=c$, you get
$$\int_a^b f(x) dx = \int_a^b f(x) dx + \int_b^b f(x) dx$$
And therefore
$$\int_b^b f(x) dx = 0$$
A: 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*}
A: Paramand Singh's comment comes closest to answering this question so far: The relation is not a statement that can be proved, but a convention that exists in order to fit in with some other true statements.  It reminds me a little bit of this endless “Why is $0! = 1$?” question on Quora. 
For instance, If $a < c < b$ and $f$ is integrable on $[a,b]$, then
$$
    \int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx
\tag{1}
$$
This can be proven from the Riemann sum definition of the definite integral.  The definition
$$
    \int_b^a f(x)\,dx = - \int_a^b f(x)\,dx
\tag{2}
$$
is consistent with the definition, since when you try to partition an interval $[a,b]$ in the opposite order, the differences $\Delta x_i = x_{i} - x_{i-1}$ become negative, and the Riemann sums flip signs.
I looked at Spivak, Courant, and Apostol, and all state (1) as a theorem, make a moral justification for (2) as a definition, and finally define $$\int_a^a f(x)\,dx = 0 \tag{3}$$  These two definitions allow the theorem (1) to be extended to all triples $(a,b,c)$, including $c<a$, $c=a$, $c=b$, and $c>b$.  
Another important fact about the integral is that it makes integrable functions into continuous functions.  If $f$ is integrable on $[a,b]$, and $F(x) = \int_a^c f(t)\,dt$, then $F$ is continuous on $(a,b)$.  If we use definitions (2) and (3), we can extend the domain of $F$ to the domain of $f$, including $x=a$, $x<a$, $x=b$, and $x>b$.  The definitions ensure that the extended function $F$ remains continuous.
It's been said that the best definitions are the ones which make the theorems easy to state and prove.  This is a good example.
A: Suppose $\int^a_a{f(x) dx}$ exists (if it doesn't, then this question doesn't make much sense). From integration rules ($\int^b_a{f(x) dx} = -\int^a_b{f(x) dx}$), we know that the following holds:
$$\int^a_a{f(x) dx} = -\int^a_a{f(x) dx}$$
Now, there is exactly one value that is its own additive inverse: $0$. Therefore:
$$\int^a_a{f(x) dx} = -\int^a_a{f(x) dx} = 0$$
A: 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.
A: May be, you could consider $$I=\int_a^{a+\epsilon} f(x) \,dx$$ and let $F(x)$ to be the antiderivative of $f(x)$.
So $$I=\int_a^{a+\epsilon} f(x) \,dx=F({a+\epsilon})-F(a)$$ Now, use Taylor series around $\epsilon=0$ to get $$F({a+\epsilon})=F(a)+\epsilon  F'(a)+\frac{1}{2} \epsilon ^2 F''(a)+O\left(\epsilon ^3\right)$$ which makes $$I=\epsilon  F'(a)+\frac{1}{2} \epsilon ^2 F''(a)+O\left(\epsilon ^3\right)=\epsilon  f(a)+\frac{1}{2} \epsilon ^2 f'(a)+O\left(\epsilon ^3\right)$$ and look at the limit of $I$ when $\epsilon \to 0$.
A: 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.
A: If you use the de facto property that $$\int_a^bf(x)\ dx=-\int_b^af(x)\ dx$$ then it follows that $$\int_a^a f(x)\ dx=-\int_a^af(x)\ dx$$ hence $$\int_a^a f(x)\ dx=0.$$
A: The Fundamental Theorem of Calculus provides a proof.
Also, you are asking for the area of a line, which is zero.
A: I'm no mathematician but if a and b are complex there may be more than one path from a to b and back to a. If the complex plane includes singularities then there will be a residue when the resulting area is integrated. This is known as a contour integral, if I remember rightly, and the value of the residue is fixed and does not depend on the path, only on the number of singularities enclosed.
