Area under a $pV$ diagram What does the area under a Pressure volume diagram equal? 
I read in my textbook it equals 'external' work done, but why is this? 
First of all, what exactly is external work? 
Can you get it external work by the simple formula $W= F\cdot s$?
Also, why does the area under a $pV$ diagram equal 'external' work? 
What is the logic behind that?
 A: The area under the curve in the pV-diagram is the integral
$$
\int p \;\mathrm dV = \int \frac FA A\;\mathrm ds=\int F \;\mathrm ds \equiv W
$$
by definition of pressure as force per area and (infinitesimal) volume as area times distance.
This is the mechanical work done by the system on the environment in case of expansion or by the environment on the system in case of compression, which differ by sign. It is called external to emphasize the interaction with the environment.
A: Why is work done pressure times volume?
The magnitude of the work done when a gas expands is therefore equal to the product of the pressure of the gas times the change in the volume of the gas. By definition, one joule is the work done when a force of one newton is used to move an object one meter.
this is just the simplest way i could define why work done is equal to pressure x volume for those that just want a simplistic answer
A: As I noted from a comment that you are new to (or not yet introduced to) integration, I will try to provide an intuitive answer without relying on integrals.
If your graph axes are volume lengthwise and pressure heightwise, then imagine a rectangular graph. It's area is, as per usual geometry, length-times-height.
That is an amount of pressure, $p$, times an amount of volume change, $\Delta V$:
$$p\cdot \Delta V.$$
Volume is already area times  displacement, and pressure is force per area, so this is easily rewritten to force-times-displacement which is what we already know as work done: $$p\cdot \Delta V=\frac FA\cdot A\cdot \Delta s=F\cdot \Delta s=W.$$
So, if you ever supply a pressure which changes the volume of something, which would correspond to a horizontal curve on a graph (when the pressure is constant), then the work you do is the area under this curve. The area of the rectangular part.

If the pressure is not constant, then the curve is not horizontal. Then the area is harder to calculate than as simply length-times-height. If it is not possible with other means then this is where integration becomes useful.
With integration we imagine the area split into many, many columns that are very, very thin. In fact infinitely many infinitely thin columns. When such a column is very, very thin, then the volume change over this part of the graph is very, very small. Instead of $\Delta V$, we might rather indicate such infinitely small change as $\mathrm dV$. Each such column is very close to a small rectangle, so for each of them we can calculate the work as lenght-times-height:
$$W_\text{column}=p\cdot \mathrm dV.$$
After finding the area of each column, we now just have to sum them all up:
$$W=W_\text{column1}+W_\text{column2}+W_\text{column3}+\cdots$$
$$W=p_1\cdot \mathrm dV+p_2\cdot \mathrm dV+p_2\cdot \mathrm dV+\cdots$$
$$W=\sum p\cdot \mathrm dV.$$
Here I used the summation symbol, $\sum$, to indicate that they are all summed up. Actually, this symbol is typically used for not infinitesimal sizes. When the parts that are summed up are basically infinitely small, then the summation symbol, $\sum$, is typically replaced with an integral symbol, $\int$, instead:
$$W=\int p\cdot \mathrm dV.$$
This is the general version of the work formula that explains the area-under-the-graph idea. And that was a quick intro to integration.
