If partials exist and are continuous then they are equal 
Let $f$ be a real valued function on an open subset $E^2$. Prove that if $\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)$ and $\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right)$ exist and are continuous then they are equal. 

So I need to show that if the set $\{(x,y)\in E^2 : x\in[a,b], \ y\in[c,d]\}$ is entirely contained in the set in which $f$ is defined, then $$\int^b_a\left(\int^d_c\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)dy\right)dx=\int^b_a\left(\int^d_c\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right)dy\right)dx.$$
But I am not sure how I can continue the proof.
 A: On the one hand, we have
$$\begin{align}
\int_a^b \left(\int_c^d \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right)(x,y)\,dy\right)\,dx
&= \int_a^b \left(\frac{\partial f}{\partial x}(x,d) - \frac{\partial f}{\partial x}(x,c)\right)\,dx\\
&= \bigl( f(b,d) - f(a,d)\bigr) - \bigl( f(b,c) - f(a,c)\bigr)\tag{1}
\end{align}$$
by applying the fundamental theorem of calculus to the continuous functions $\frac{\partial}{\partial y} \left(\frac{\partial f}{\partial x}\right)$ and $\frac{\partial f}{\partial x}$.
On the other hand, since the continuity of $\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)$ allows changing the order of integration, we have
$$\begin{align}
\int_a^b \left(\int_c^d \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)(x,y)\,dy\right)\,dx
&= \int_c^d \left(\int_a^b \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)(x,y)\,dx\right)\,dy\\
&= \int_c^d \left(\frac{\partial f}{\partial y}(b,y) - \frac{\partial f}{\partial y}(a,y)\right)\,dy\\
&= \bigl( f(b,d) - f(b,c)\bigr) - \bigl(f(a,d) - f(a,c)\bigr)\tag{2}
\end{align}$$
again by applying the fundamental theorem of calculus.
The commutativity of addition reveals $(1)$ and $(2)$ as equal.
For a continuous $g$, we have
$$\lim_{(b,d)\to(a,c)} \frac{1}{(b-a)(d-c)}\int_a^b\int_c^d g(x,y)\,dy\,dx = g(a,c),$$
and thus the equality of $(1)$ and $(2)$ shows
$$\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)(a,c) = \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right)(a,c),$$
and since $(a,c)$ was arbitrary, the equality of the mixed partial derivatives on $E^2$.
