Work done when P,V, and T are changing simultaneously
As far as I am aware, there is no generic formula for when all three variables are changing at once.

In this case, the easiest way is to simply use the area of a trapezium.

If you insist on integrating, then you need to know what $P$ is as a function of $V$. This derives from basic principles of integration - if you have a function $f(x)$, then $\int_a^b f(x)\,\mathrm{d}x$ gives the area under the curve of $f(x)$ from $x = a$ to $x = b$.

The line $\mathrm{C \rightarrow A}$ can be easily found since the beginning and end points are given. The gradient of the line is

$$m = \frac{2P_0 - P_0}{2V_0 - V_0} = \frac{P_0}{V_0}$$

and the the intercept is clearly 0 from the diagram (you can work it out anyway). In general the equation of a line $y = y(x)$ is given by $y - y_i = m(x - x_i)$ where $(x_i,y_i)$ is any point on the line.

That was perhaps an unnecessarily long explanation but the point is that you now know the relation

$$P = mV$$

holds (where $m$ was determined earlier). Therefore

$$\begin{align}
w &amp;= -\int_{2V_0}^{V_0} P\,\mathrm{d}V \\
&amp;= -\int_{2V_0}^{V_0} -mV\,\mathrm{d}V \\
&amp;= m\left[\frac{1}{2}V^2\right]_{2V_0}^{V_0} \\
&amp;= m\left(\frac{1}{2}V_0^2 - 2V_0^2\right) \\
&amp;= \frac{P_0}{V_0}\left(-\frac{3}{2}V_0^2\right) \\
&amp;= -\frac{3}{2}P_0V_0
\end{align}$$

which is the same as you found via the first method (you also forgot to put the negative sign in the right-hand side of the equation, but I assume that is just a typo).