If it is the shaded area that you want to solve by integration, then here is one way.

An element of the area, dA = [(V of curve BC) -(V of curve DA)]*dP

dA = [1.8/P -0.72/P]dP

dA = [1.08/P]dP

According to the figure, P goes from 0.6 to 1.8,

so, integrating from P = 0.6 to P = 1.8,

A = INT(0.6 to 1.8)[1.08/ P]dp

A = (1.08)[ln(P)]|(0.6 to 1.8)

A = (1.08)[ln(1.8) -ln(0.6)]

A = 1.1865 -----------------**

That is if P and V are of the same power of 10.

But according to the figure,

P is in 10^5

while V is in 10^(-3)

So if you multiply 10^5 by 10^(-3), you will get 10^2

Therfore, A must be multiplied by 10^2

So, A = 1.1865 *10^2 = 118.65 joules