I am currently studying Partial Differential Equations, specifically the derivation of the heat equation for a one-dimensional rod with constant thermal properties and no heat sources. I am focusing on the total thermal energy between $x=a$ and $x=b$. During the derivation process, I encountered a step that involves the application of the Fundamental Theorem of Calculus, and I am finding it challenging to understand.
Given the constant thermal properties, we have the equation for heat energy density:
\begin{equation} e(x, t)=c \rho u(x, t) \end{equation}
Considering the segment of the rod between $a$ and $b$, we can express the total heat energy contained in this section as:
\begin{equation} \int_a^b c \rho u(x, t) A(x) d x \end{equation}
Under the assumptions of no heat sources and constant thermal properties, the total heat energy is equal to the heat energy flowing through both boundary surfaces. Since the cross-sectional area $A(x)$ changes, it cannot be canceled out:
\begin{equation} \frac{d}{d t} \int_a^b c \rho u(x, t) A(x) d x=\phi(a, t) A(a)-\phi(b, t) A(b) \end{equation}
At this point, I am unsure how to apply the Fundamental Theorem of Calculus to obtain the following expression:
\begin{equation} \frac{d}{d t} \int_a^b \operatorname{c\rho u}(x, t) A(x) d x=-\int_a^b \frac{\partial}{\partial x}[\phi(x, t) A(x)] d x \end{equation}
I have searched for examples and explanations online, but I have not found any that directly address this particular application. Can someone please help me understand the logic behind this step and how the Fundamental Theorem of Calculus is applied here? Thank you :)
Essentially, your equation is given by $$\int\limits_a^b \frac{\partial f(x,t)}{\partial x} dx = f(b,t)-f(a,t)$$
for $f(x,t) = \phi(x,t)A(x)$. This is precisely the second form of the fundamental theorem of calculus you wrote, since in this case, $F' = f$.