What is the volume of $A'EF-ABD$? 
$ABCD-A'B'C'D'$ is a cube with a edge length of $6$.
$E$ is the midpoint of $A'B'$ and $F$ is the point on $A'D'$ where $|A'F|=2|D'F|$.
The question is: what is the volume of $A'EF-ABD$ ?

I have two methods, one is integral.

Let $x$ be the length of $AA_2$, I get $$V_{A'EF-ABD}=\int_{0}^{6}\frac{1}{2}(6-\frac{x}{2})(6-\frac{x}{3})dx=69$$
The other method is to divide it into two parts.

\begin{align*}
V_{A'EF-ABD} &= V_{D-A'EF}+V_{D-A'EBA} \\ 
 &= \frac{1}{2}\times 3\times 4\times 6\times \frac{1}{3}+(3+6)\times 6\times \frac{1}{2}\times 6\times \frac{1}{3}\\ 
 &= 12+54\\ 
 &= 66\\ 
\end{align*}
It's obvious that $69\neq 66$, what causes the difference?

The quadrilateral $BEFD$ is not planar. This means you are missing some of the volume because the line $ED$ runs through your solid rather than on it.
