Question about exercise 5.11 in Shaum's outline for vector analysis, 2nd edition:
regarding this section of equation:
$\phi(x,y,z)=\int \limits_{x_0}^x F_1(x, y_0, z_0) dx + \int \limits_{y_0}^y F_2(x,y,z_0)~ dy + \int \limits_{z_0}^z F_3(x,y,z) dz$
taking partial derivative with respect to x:
$\frac{\partial}{\partial x}\phi(x,y,z)=\frac{\partial}{\partial x}\bigg(\int \limits_{x_0}^x F_1(x, y_0, z_0) dx + \int \limits_{y_0}^y F_2(x,y,z_0)~ dy + \int \limits_{z_0}^z F_3(x,y,z) dz\bigg)$
$\frac{\partial}{\partial x}\phi(x,y,z)=F_1(x,y_0,z_0) + \frac{\partial}{\partial x}\bigg(\int \limits_{y_0}^y F_2(x,y,z_0)~ dy\bigg) + \frac{\partial}{\partial x}\bigg(\int \limits_{z_0}^z F_3(x,y,z)~ dz\bigg)$
now subtituting: $\frac{\partial F_2}{\partial x} = \frac{\partial F_1}{\partial y}$ and $\frac{\partial F_3}{\partial x} = \frac{\partial F_1}{\partial z}$
$\frac{\partial}{\partial x}\phi(x,y,z)= F_1(x,y_0,z_0) + F_1(x,y,z_0)\bigg|^y_{y_0} + F_1(x,y,z) \bigg|^z_{z_0}$
why does the first term on the right hand side, equal $F_1(x,y_0,z_0)$ instead of $F_1(x,y_0,z_0)\bigg|^x_{x_0}$ with a vertical bar?
They're leaving out the step of moving the derivative under the integral sign.
First, $\dfrac{\partial}{\partial x}\displaystyle\int_{x_0}^x F_1(t,y_0,z_0)\,dt = F_1(x,y_0,z_0)$ by the usual Fundamental Theorem of Calculus.
Next, \begin{align*} \frac{\partial}{\partial x}\int_{y_0}^y F_2(x,t,z_0)\,dt &= \int_{y_0}^y \frac{\partial F_2}{\partial x}(x,t,z_0)\,dt = \int_{y_0}^y \frac{\partial F_1}{\partial y}(x,t,z_0)\,dt \\ &= F_1(x,y,z_0) - F_1(x,y_0,z_0), \end{align*} by the (other) Fundamental Theorem of Calculus.