Integrating a non-function - Work Done using P-V graph 

  The above figure was given to us, with the question saying

An ideal gas is taken from state $A$ to state $B$ via the above three processes. Then for heat absorbed by the gas, which of the following is true?
(A) Greater in $ACB$ than in $ADB$
(B) Least in $ADB$
(C) Same in $ACB$ and $AEB$
(D) Lesser in $AEB$ than in $ADB$

The answer is (D), but I'm not sure how.

A little background for those who have lost contact with Physics:
Heat absorbed is represented by $\Delta Q$, Change in Internal Energy by $\Delta U$, and Work Done by $\Delta W$, with the equation $\Delta Q=\Delta U+\Delta W$ always valid.
Here $\Delta U$ is same for all three processes owing to the fact that they all have the same starting and ending point.
And $\Delta W=\int P\ dV$.
Hence, to compare Heat Absorbed, simply comparing Work Done suffices.
My Question:

In process $ADB$ it is easy to see that the Work done is zero, owing to $dV=0$, but I'm unsure how to calculate Work Done for the other two processes, as they are not functions (of $P$ in $V$)?

Edit: I had missed out on writing that the gas is ideal, I have now added that in.

To describe the three processes, we could as follows
1.Process ADB - Constant volume process. For an ideal gas, this would be a zero work process, and hence all heat is used to change the internal energy


*Process AEB, ACB - Follows a circular process


$$P = P^* + K\sin \theta \\V = V^* + K\cos \theta$$
Hence
$$dV = -K\sin \theta d\theta$$
Now, for AEB
$$W = -\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(KP^*\sin \theta +K^2\sin^2\theta )d\theta $$
$$\implies W_{AEB} = -\frac{K^2\pi}{2} < 0$$
Can you approach ACB in a similar fashion?
