A hard integral computed easily
In my previous article on doing integration with the help of inverse functions https://medium.com/cantors-paradise/integration-using-inverse-functions-12094a9bc508 I showed a method for integrating seemingly complicated functions over a finite domain, by simply computing the area of the inverse function and matching it to the area we actually want to compute. In this article, I will apply that same method for solving another interesting integral, the one shown in Figure 1.
First, we must think about the range of the function y = f(x) in the domain x in [0, 1] and what the function looks like. Plugging in a few values of x, we quickly realize that the function is monotonically decreasing, and its range is y = [0, 1] as well. In that sense, we can apply the following formula for computing the definite integral using the inverse function:
This formula is similar to what I presented in my last article, except that one assumed the function was monotonically increasing in the domain [a, b] rather than decreasing. Nevertheless, the procedure is pretty much the same as before. The inverse function in this case is
and this can be expanded into
which is a simple polynomial which can be integrated term by term. Then, we apply Equation 2 to find the desired value:
