# Definite Integral of Partial Derivative/Proof 2

## Theorem

Let $\map f {x, y}$ and $\map {\dfrac {\partial f} {\partial x} } {x, y}$ be continuous functions of $x$ and $y$ on $D = \closedint {x_1} {x_2} \times \closedint a b$.

Then:

- $\displaystyle \frac \d {\d x} \int_a^b \map f {x, y} \rd y = \int_a^b \map {\frac {\partial f} {\partial x} } {x, y} \rd y$

for $x \in \closedint {x_1} {x_2}$.

## Proof

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

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

Then by linearity of the integral:

- $\dfrac {\Delta G} {\Delta x} = \dfrac {\map G {x + \Delta x} - \map G x} {\Delta x} = \displaystyle \int_a^b \frac {\map f {x + \Delta x, y} - \map f {x, y} } {\Delta x} \rd y$

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:

- $\dfrac {\Delta G} {\Delta x} = \displaystyle \int_a^b \map {\frac {\partial f} {\partial x} } {c_y, y} \rd y$

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:

\(\displaystyle \size {\lim_{\Delta x \mathop \to 0} \frac {\Delta G} {\Delta x} - \int_a^b \map {\frac {\partial f} {\partial x} } {x, y} \rd y}\) | \(=\) | \(\displaystyle \lim_{\Delta x \mathop \to 0} \size {\int_a^b \map {\frac {\partial f} {\partial x} } {c_y, y} - \frac {\partial f}{\partial x} \rd y}\) | |||||||||||

\(\displaystyle \) | \(\le\) | \(\displaystyle \lim_{\Delta x \mathop \to 0} \int_a^b \size {\map {\frac {\partial f} {\partial x} } {c_y, y} - \map {\frac {\partial f} {\partial x} } {x, y} } \rd y\) | |||||||||||

\(\displaystyle \) | \(\le\) | \(\displaystyle \int_a^b \epsilon_0 \rd y\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \epsilon\) |

But since $\epsilon$ was arbitrary, it follows that:

- $\displaystyle \lim_{\Delta x \mathop \to 0} \frac {\Delta G} {\Delta x} = \int_a^b \map {\frac {\partial f} {\partial x} } {x, y} \rd y$

and the theorem is proved.

$\blacksquare$