As long as you have the cumulative distribution, and most importantly, an inverse for the cumulative, this is all you need. For example, suppose you want to find the points where you can divide that area into 3 equal parts, given a normal distribution.
Here, I'll use a normal distribution with mean 3, and variance 4, so standard deviation 2. I want three equal parts, so the break points in the area will be:
areabreaks = linspace(0,1,4)
There must always be one extra break point. Now, we can ignore the first and last such point, because they must live at -inf and +inf.
Now, use the inverse cumulative distribution, for a normal with those parameters. If you don't know how to use norminv, then READ THE HELP.
Now, we want to find the corresponding points where the cumulative area is 1/3 and 2/3, in our case, because I chose 3 area segments. In fact, norminv is smart enough to know that 0 maps to -inf, and 1 maps to +inf.
z = norminv(areabreaks,mu,sigma)
intervals = [z(1:end-1);z(2:end)]
Nothing difficult. And only a couple of lines of code. The same will apply to your problem. TRY IT!
What is the corresponding inverse cumulative distribution function for a lognormal?