Let $f$ continuous, show that $\int_0^a (\int_0^x f(x,y) dy) dx = \int_0^a (\int_y^a f(x,y) dx) dy$ $$\int_0^a \int_0^x f(x,y) dy dx = \int_0^a \int_y^a f(x,y) dx dy$$ 
I have tried to help myself, observing the graph and how the area is covered according to the limits, and managed to see that the area of the two integrals are symmetrical. But I am almost certain that that is not necessary to say that the integrals are equal, and I don't know how I could say it.
 A: First get non-variable integration limits and then apply Fubini's theorem.
We have
$$\int_0^a \int_0^x f(x,y)\,dy\,dx = \underbrace{\int_0^a \int_0^a f(x,y) \chi_{\{y \leqslant x\}}\,dy\,dx}_{(1)}, $$
where the indicator function is defined as
$$ \chi_{\{s \leqslant t\}} = \begin{cases}1, & s \leqslant t \\0, & s > t \end{cases}$$
Now you are allowed to interchange integrals when $f$ is absolutely integrable as per Fubini's theorem (this holds when $f$ is continuous) and
$$\int_0^a \int_0^x f(x,y)\,dy\,dx = \underbrace{\int_0^a\int_0^a f(x,y) \chi_{\{y \leqslant x\}}\,dx\,dy}_{\text{after interchanging order of integration in (1)}} = \int_0^a \int_{y}^a f(x,y)\,dx\,dy $$
A: Sketch the region of integration and everything clicks.
The region of integration is a right triangle  with base on the x-axis  from $0$ to $a$ and the hypotenuse is the segment connecting the origin to the point $(a,a)$
Describe the triangle starting with the range of $y$, you have $$0\le y\le a$$ and for each $y$, we have $$y\le x\le a$$ 
That explains the bounds of $$ \int_0^a (\int_0^x f(x,y) dy) dx = \int_0^a (\int_y^a f(x,y) dx) dy$$
A: Hint:
$$LHS:\begin{cases}0\le x\le a\\ 0\le y\le x\end{cases}\Rightarrow \begin{cases}0\le x\le a\\ 0\le y\le x\color{red}{\le a}\end{cases}\Rightarrow RHS:\begin{cases}0\le y\le a\\ y\le x\le a\end{cases}$$
