$ABCD-A'B'C'D'$ 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 objective of the following argumentation is to show that the rigid body at stake is not a not a truncated pyramid. The calculation based on the integration of the surface area of right triangles is all right. What is the problem then?
Let $A$ be the origin of our spatial coordinate system. Also, let $\overrightarrow{AB}$, $\overrightarrow{AD}$, and $\overrightarrow{AA'}$ be the $x$, the $y$, and the $z$ axes, respectively. Then, looking from above (if we cut the $BDEF$ surface with planes parallel to the $xy$ plane) we see straight lines whose equations are:
$$y=-\frac{1}{18}(z+18)x+6-\frac{z}{3},\ 0\le z\le 6,$$
where the slope and the $y$ crossing depend on $z$.
In terms of $x,y,z$ this is not an equation of a plain.
That is, $BDEF$ is not a plain. The equation of the $BDEF$ surface is $$-y-\frac{1}{18}zx-x-\frac{z}{3}+6=0$$
or in an explicit form: $$z(x,y)=\frac{6-x-y}{\frac{1}{3}+\frac{x}{18}}.$$
Whit the little help from Alpha and after some cosmetics, here is the surface at stake:
EDITED
Clearly, the part we are interested in looks like a plane. The Alpha figure is not convincing though. In the figure the surface looks, from above, like a convex surface. If it was convex from above then the pyramidal calculations ought to have given a greater volume than the (correct) integration. This is why I compared two curves: (1) the straight line joining the points $E$ and $D$ and (2) the line running on the surface right below $DE$:
It is clear that the surface is concave from above (at least along the line examined.)