Definite Integral of Partial Derivative/Proof 1

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Theorem

Let $f \left({x, y}\right)$ and $\dfrac {\partial f} {\partial x} \left({x, y}\right)$ be continuous functions of $x$ and $y$ on $D = \left[{x_1 \,.\,.\, x_2}\right] \times \left[{a \,.\,.\, b}\right]$.

Then:

$\displaystyle \frac \d {\d x} \int_a^b f \left({x, y}\right) \rd y = \int_a^b \frac {\partial f}{\partial x} \left({x, y}\right) \rd y$

for $x \in \left[{{x_1} \,.\,.\, {x_2}}\right]$.


Proof

From Leibniz Integral Rule:

$\displaystyle \frac \d {\d y} \int_{a \left({y}\right)}^{b \left({y}\right)} f \left({x, y}\right) \rd x = f \left({y, b \left({y}\right)}\right) \frac {\d b} {\d y} - f \left({y, a \left({y}\right)}\right) \frac {\d a} {\d y} + \int_{a \left({y}\right)}^{b \left({y}\right)} \frac {\partial} {\partial y} f \left({x, y}\right) \rd x$

where $a \left({y}\right)$, $b \left({y}\right)$ are continuously differentiable.

Setting $a \left({y}\right)$ and $b \left({y}\right)$ constant, so that $\exists a, b \in \R: a \left({y}\right) = a, b \left({y}\right) = b$:

$\dfrac {\d a} {\d y} = \dfrac {\d b} {\d y} = 0$

from which the result follows immediately.

$\blacksquare$