Definite Integral of Partial Derivative/Proof 2

Proof
Define $\ds \map G x = \int_a^b \map f {x, y} \rd y$.

The continuity of $f$ ensures that $G$ exists.

Then by linearity of the integral:
 * $\dfrac {\Delta G} {\Delta x} = \dfrac {\map G {x + \Delta x} - \map G x} {\Delta x} = \ds \int_a^b \frac {\map f {x + \Delta x, y} - \map f {x, y} } {\Delta x} \rd y$

We want to find the limit of this quantity as $\Delta x$ approaches zero.

For each $y \in \closedint a b$, we can consider $\map {f_y} x = \map f {x, y}$ as a separate function of the single variable $x$, with $\dfrac {\d f_y} {\d x} = \dfrac {\partial f} {\partial x}$.

Thus by the Mean Value Theorem, there is a number $c_y \in \openint x {x + \Delta x}$ such that:
 * $\map {f_y} {x + \Delta x} - \map {f_y} x = \map {\dfrac {\d f_y} {\d x} } {c_y} \Delta x$

That is:
 * $\map f {x + \Delta x, y} - \map f {x, y} = \map {\dfrac {\partial f} {\partial x} } {c_y, y} \Delta x$

Therefore:
 * $\dfrac {\Delta G} {\Delta x} = \ds \int_a^b \map {\frac {\partial f} {\partial x} } {c_y, y} \rd y$

Now, pick any $\epsilon > 0$ and set $\epsilon_0 = \dfrac {\epsilon} {b - a}$.

Since $\dfrac {\partial f} {\partial x}$ is continuous on the compact set $D$, it is uniformly continuous on $D$.

Hence for each $x$ and $y$:
 * $\size {\map {\dfrac {\partial f} {\partial x} } {x + h, y} - \map {\dfrac {\partial f} {\partial x} } {x, y} } < \epsilon_0$

whenever $h$ is sufficiently small.

Since $x < c_y < x + \Delta x$, it follows that for sufficiently small $\Delta x$:
 * $\size {\map {\dfrac {\partial f} {\partial x} } {c_y, y} - \map {\dfrac {\partial f} {\partial x} } {x, y} } < \epsilon_0$

regardless of our choice of $y$.

So we can say:

But since $\epsilon$ was arbitrary, it follows that:
 * $\ds \lim_{\Delta x \mathop \to 0} \frac {\Delta G} {\Delta x} = \int_a^b \map {\frac {\partial f} {\partial x} } {x, y} \rd y$

and the theorem is proved.