Why does the area under a Maxwell speed distribution graph equal unity? I am learning the "Kinetic Theory of Gases", in which I have now come across the "Maxwell speed distribution" graph, which I have attached here: 

My copy of "Resnick and Halliday" states that the area under each curve has a numerical value of unity. The area under the curve is given by the following integral, yes? 
$$\int_0^{\infty} P(\nu) d\nu$$
My doubt is this; how does this area equal unity? If the y-axis represents the fraction of molecules possessing a given speed(plotted on the x-axis), doesn't the product of the two give the expected value, viz. most probable speed? 
Please do share your insights. MUCH thanks in advance :) Regards. 
 A: The product $P(v)dv$ is the fraction of the molecules with velocities in the range from velocity $v$ to velocity $v+dv$.
Think of it as a histogram. If we divide up the velocity range in groups of some small velocity $dv$ we'd get a histogram looking something like:

For each column the area of column is $P(v)dv$, and this gives the fraction of the molecules in that velocity range. If we add up all the columns we have to get unity because the total fraction of all the molecules is one.
If you now imagine making $dv$ smaller and smaller the columns in the histogram get thinner and in the limint of $dv\rightarrow 0$ we get a smooth curve like the ones you show.
A: Perhaps knowing something about the properties of the probability distribution function $P(v)$ will help?
What you are looking for is a way to describe the distribution of the speeds of the atoms in a gas.
The first thing to note is that you cannot use the probability of an atom having a given speed $v$.
The reason for this is that you can choose a value of the speed $v$ to as many significant figures as you like and then to I hope you realise that you will never find an atom with exactly that speed.  
So you have resort to finding a function $f(v)$ which describes the probability of finding atoms in a certain range of speeds.
Now jump ahead and consider how to make that function $f(v)$, which describes the distribution of the speeds of the atoms in a gas, useful.
It would be useful to be able to plot the graph of the function $f(v)$ against speed (v) and make the area under the graph equal to $1$ .
The reason for doing this in the area under the $f(v)$ against $v$ graph will represent the probability of finding an atom in the range of speeds  $v=0$ to $v=\infty$ which must be $!$.  
Probability has no units but the product $f(v) \;v$ has the units of speed so we define another function $g(v) = \dfrac {f(v)}{1\; \text{m s}^{-1}}$ which has the units s m$^{-1}$ to get over this problem.
Please note the units of $P(v)$ in your graph.
So now the integration $\displaystyle \int^\infty _0 g(v) \;dv= 1$ results in a numerical value with no units. 
Note that the function $g(v)$ has been chosen to have the property that the area is $1$.
The area under the $g(v)$ against $v$ graph can now be used to find the probability of the speed of an atom being between any two speed, $v_1$ and $v_2$ and that is $\displaystyle \int^{v_2} _{v_1} g(v) \;dv$.  
The function $g(v)$ is given a name, the probability distribution function, and a symbol $P(v)$.
Having gone through all this, how to describe the function $P(v)$ in words?
For example what does the value $P(400) = 0.0021 \;\text{s m}^{-1}$ on your $T= 300$ K graph mean?
It means that the probability of finding an atom with a speed between $400 \; \text {m s}^{-1}$ and $401 \; \text {m s}^{-1}$ is approximately $0.0021 \times 1 = 0.0021$ whereas for the same speed range the probability for the gas at $80$ K is approximately $0.0008$.
As the range of speeds gets smaller the approximation gets better and furthermore the ratio of the two probability distribution functions at the two temperatures gives you the ratio of the probabilities of finding the speeds of the atoms at the two temperatures being (around) $400 \; \text {m s}^{-1}$.
