# Integrate a family of curves

This is a follow-up question of: I need to define a family (one parameter) of monotonic curves

$x_0$, $y_0$ are known constants. What's the integral of this function?

$f(x, a) = \frac{y_0}{1+\frac{(1-a)x}{a(x_0-x)}}$

A few words on the function itself. For $x_0 = 1, y_0 = 1$, it looks the following diagram. We get a different curve for a different value of $a$, the straight line is when $a = 0.5$, the curves below are have $a < 0.5$ and the ones above have $a > 0.5$.

Basically I need a function that I give it the total area to be covered, and the result is a value of $a$ that produces a curve covering that area. The valid areas are in $[0.0, x_0 * y_0]$. Some examples of this function $g$:

$g(0) = 0.0$ (when $a = 0.0$, the curve has zero area)

$g(x_0 * y_0) = 1.0$ (when $a = 1.0$, the curve has maximum area, the whole rectangle)

$g(0.5 * x_0 * y_0) = 0.5$ (when $a = 0.5$, the curve is a triangle)

I know, from tinkering with it, that the sought function has a sigmoidal shape like this:

(Horizontal axis is the value $a$, from $0.0$ to $1.0$, vertical axis is the area $R$, R for aRea)

My understanding is that I would need to integrate $f(x, a)$ (dx), get a closed formula involving $a, x_0, y_0, R$, and solve for $a$.

I admit that I was probably sleeping/absent when they taught me integration techniques in school and university. Any help much, much appreciated. I would also give a shot to wolfram alpha which has integration examples, but the problem is I have a couple of extra (fixed) parameters, and the syntax of it gives me trouble (or it only supports one symbolic name per function -- but that's unlikely)

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Just to be sure: you want $\int_0^1 f(x,a)\mathrm dx$? –  Ｊ. Ｍ. Oct 10 '11 at 2:56
@J.M., the integration limits should surely be $0$ to $x_0$. However, substituting $u=x/x_0$ will get us back to the $[0,1]$ case while canceling out the use of $x_0$ in the definition. –  Henning Makholm Oct 10 '11 at 3:04
In any event: you have a rational function as the integrand. If you perform partial-fraction decomposition on it, you can integrate and then obtain something involving logarithms. –  Ｊ. Ｍ. Oct 10 '11 at 3:05
@Henning: after seeing the previous question, I realized precisely that. :) –  Ｊ. Ｍ. Oct 10 '11 at 3:06
Sigh. I wish I understood what you are talking about :-/ –  Dimitris Andreou Oct 10 '11 at 3:23

So I guess we want $$\large \int_0^{x_0}\frac{y_0}{1+\frac{(1-a)x}{a(x_0-x)}}dx.$$ Make the substitution $u=x/x_0$, multiply numerator and denominator by $a(1-u)$, $$y_0x_0\int_0^1\frac{a(1-u)}{a+(1-2a)u}du.$$ Now we want to use partial fraction decomposition, so we solve for $A$ and $B$ in $$\frac{a(1-u)}{a+(1-2a)u}=\frac{A}{a+(1-2a)u}+B.$$ We obtain the system $$\begin{cases}A+aB=a \\ B(1-2a)=-a\end{cases}$$ with solutions $$B=\frac{a}{2a-1};\quad A=\frac{a(a-1)}{2a-1}.$$ The integral becomes $$y_0x_0\int_0^1 \left(\frac{A}{a+(1-2a)u}+B\right)du$$ $$=y_0x_0\left(A\left[\frac{\log|a+(1-2a)u|}{1-2a}\right]_0^1+B\right)$$ $$=y_0x_0\left(\frac{a(1-a)}{(2a-1)^2}\log\left(\frac{1-a}{a}\right)+\frac{a}{2a-1}\right)$$ except when $a=1/2$. When $a=1/2$, we don't even need to use decomposition to evaluate $$y_0x_0\int_0^1(1-u)du=\frac{y_0x_0}{2}.$$