# Is there a coordinate transformation which splits the integral of a function equally?

If there is a function, (for example, a sine, as in the example), is there a simple way of creating a coordinate system in which the integrals of the function are split equally?

What I am looking to achieve is for the area to be split into equal parts, as in the below image. (not to scale, I drew the lines manually just to illustrate the concept).

Or, even more simply, is there a function f which gives me where I have to split the x axis, given that I want the function g to be split into n equal-area parts?

I'm considering developing my own (possibly needlessly overcomplicated) solution, but before that, I would like to know whether this is a known problem with an already existing solution.

• This is related, but in a more abstract setting. – Giuseppe Negro Mar 17 '18 at 11:23
• Developing your own solution is a nice exercise. To make things easier on yourself, consider first very small areas (large $n$) and function values away from $0$. That way you can approximate your areas with rectangles and see exactly how you would like the width of the rectangle depend on the function value. Trying out with constant or linear $f$ would probably also give you a good idea of what the answer ought to be. – Arthur Mar 17 '18 at 11:24
• @Arthur : I'm a hobby-engineer, not a mathematician, so I might be ignorant of some tools which would make this much easier to do, this is why I'm asking the question. The practical problem I'm looking to solve involves the sine from 0 to pi as in the example image, but I asked this question in a bit more general term hoping to learn about any existing method or trick I was unaware of. – vsz Mar 17 '18 at 11:30
• @Arthur : well, if n is large enough, and the function is not too close to zero, then at an arbitrary point on x, the segment would be approximately the total area divided by the function value at that point. But this all breaks down if we are close to zero, and I need to start the segmentation beginning from zero. – vsz Mar 17 '18 at 12:42

What we actually want here, is to find a relationship between $x_i$ and $x_{i+1}$ so that the area between them is the area of the whole function divided by the number of how many equal divisions we want.

For the sine in the example, the total area is

$$\int_0^\pi sin(x)dx = 2$$

So, the area between $x_i$ and $x_{i+1}$ has to be $2/n$, where $n$ is the number of divisions.

This area is given by

$$\int_{x_i}^{x_{i+1}} sin(x)dx = -cos(x_{i+1}) + cos(x_i)$$

Therefore

$$cos(x_i) - cos(x_{i+1}) = \frac{2}{n}$$

Now we can express the next point in our new coordinate system based on the previous one:

$$x_{i+1} = arccos(cos(x_i)-\frac{2}{n})$$