Admits partials, definite integral of multivariable function is $C^1$. For open $U \subseteq \mathbb{R}^n$, assume $f: [a, b] \times U \to \mathbb{R}$ admits partials $\partial_{x_i}f(t, x_1, \dots, x_n)$ continuous on $[a, b] \times U$. Do we have that $I_f(x_1, \dots, x_n) = \int_a^b f(t, x_1, \dots, x_n)\,dt$ is $C^1$ on $U$, with $\partial_{x_i} I_f = \int_a^b \partial_{x_i} f(t, x)\,dt$?
 A: By here, $\int_a^b \partial_{x_i} f(t, x)\,dt$ is continuous in $x \in U$ since $\partial_{x_i} f$ is continuous on $[a, b] \times U$. Thus, it remains to check that this integrated partial satisfies the condition to be the $i$th partial of $I$. We may fix all $x_j$'s for $j \neq i$ when checking this, and so we are immediately reduced to the case $n=1$. Thus, for $h \in \mathbb{R}$ with $|h|$ small, we need to estimate$$I_f(x+h) - I_f(x) - \int_a^b \partial_xf(t, x)h\,dt = \int_a^b(f(t, x + h) - f(t, x) - \partial_x f(t, x)h)\,dt$$in comparison to $|h|$ as $h \to 0$, with $x \in U \subseteq \mathbb{R}$ fixed.
Since $| \int_a^b F| \le \int_a^b |F|$ for any integrable $F: [a, b] \to \mathbb{R}$, it suffices to prove that for all $\epsilon > 0$,$$|f(t, x + h) - f(t, x) - \partial_x f(t, x)h | \le \epsilon|h|$$for all $t \in [0, 1]$ provided $|h|$ is sufficiently small. Letting $g_{t, h}(s) = f(t, x + sh)$, clearly $g_{t, h}$ is $C^1$ in $s$ with $g_{t, h}'(s) = \partial_x f(t, x+sh)h$ and the Fundamental Theorem gives$$f(t, x+h) - f(t, x) = g_{t, h}(1) - g_{t, h}(0) = \int_0^1 g_{t, h}'(s)\,ds = \int_0^1 \partial_xf(t, x + sh)h\,ds.$$Thus,$$f(t, x + h) - f(t, x) - \partial_xf(t, x)h = \int_0^1 (\partial_xf(t, x+ sh) - \partial_xf(t, x))h\,ds.$$The uniform continuity of $\partial_xf$ on $[a, b] \times[x-c, x+c]$ with $[x-c, x+c] \subseteq U$ ensures that for $|h|$ sufficiently small (in particular, $|h| < c$) we have$$|\partial_xf(t, x + sh) -\partial_x f(t, x)| \le \epsilon$$for all $s \in [0, 1]$. Putting this all together,$$|f(t, x +h) - f(t, x) - \partial_x f(t, x)h| \le \int_0^1 \epsilon|h|\,ds = \epsilon|h|.$$
