Problem Solving the Total Heat in an Insulated Bar I'm having a problem trying to compute the total heat in a fully insulated rod.
Here is the important parts of the problem statement I've been given:
The one-dimensional heat conduction equation, which governs the spatial and temporal evolution
of the temperature in a bar, is given by the following partial differential equation (PDE): 
$$\frac{∂^2u(x,t)}{∂x^2}=D\frac{∂u(x,t)}{∂t} , D>0.$$
for$$0 < x <L$$
The bar is fully insulated. So that would be that the boundaries are insulated, which means $$\frac{∂u}{∂x}=0$$ at $$ x=0$$ and $$ x =L$$
Consider the initial temperature:
$$u(x,0)=f(x)=−x(x−L)$$
Use Vector Calculus to compute the total heat in the bar for all time t.
I know that the integral to solve for the total heat is: $$∫^{L}_{0}u(x,t)dx, t\ge0$$
I've solved for the Sum of the Series for $u(x,t)$ but I'm not sure how to use that or anything else to solve for the total heat.
Any help would be greatly appreciated!
 A: Not sure just how much vector calculus I'm using here, but consider:
Denote the total heat in the bar at time $t \ge 0$ by $Q(t)$; thus
$Q(t) = \int_0^L u(x, t) dx.  \tag{1}$
Then we have
$\dfrac{dQ(t)}{dt} =\dfrac{d}{dt} \int_0^L u(x, t) dx$
$=\int_0^L \dfrac{\partial u(x, t)}{\partial t} dx; \tag{2}$
now since
$\dfrac{\partial^2 u(x,t)}{\partial x^2}=D\dfrac{\partial u(x,t)}{\partial t} \tag{3}$
with $D > 0$, (2) becomes
$\dfrac{dQ(t)}{dt} = D^{-1} \int_0^L \dfrac{\partial^2 u(x, t)}{\partial x^2} dx.  \tag{4}$
The integral on the right of (4) may be evaluated thusly:
$\int_0^L \dfrac{\partial^2 u(x, t)}{\partial x^2} dx = \dfrac{\partial u(L, t)}{\partial x} - \dfrac{\partial u(0, t)}{\partial x} = 0 \tag{5}$
by virtue of the boundary conditions
$\dfrac{\partial u(L, t)}{\partial x} = \dfrac{\partial u(0, t)}{\partial x} = 0.  \tag{6}$
Thus by (4) we see that
$\dfrac{dQ(t)}{dt} = 0, t \ge 0; \tag{7}$
i.e. $Q(t) = Q(0)$ is constant for all $t \ge 0$.  This result is of course in perfect accord with our physical intuition that a completely thermally insulated body should suffer no net gain or loss in total heat energy over the course of time.  Finally, we may compute the (constant) total heat $Q(t)$ by evaluating
$Q(0) = \int_0^L u(x, 0)dx = \int_0^L (Lx - x^2)dx$
$= (\dfrac{Lx^2}{2} - \dfrac{x^3}{3} \mid_0^L = \dfrac{L^3}{2} -  \dfrac{L^3}{3} = \dfrac{L^3}{6}; \tag{8}$
the total heat in the bar is $L^3 / 6$ for all $t \ge 0$.
