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# Fun with integrations

In the first post of this series centered around my university degree I will be talking about integration. Now this topic wasn’t mentioned in our script but I rather stumbled upon it on youtube.

The premise is the following:

We assume that we have a function f(x) which we can integrate over the interval a to b.

Calculate the following integral: $$\int_{2}^{4}\frac{\sqrt{x}}{\sqrt{6-x} + \sqrt{x}} dx$$ Doesn’t look trivial now, does it? But miraculously, it is! This whole integral equates to: $$\frac{b-a}{2} = 1$$.

Let’s take a look at another integral for fun. $$\int_{0}^{\frac{\pi}{2}}\frac{\sqrt{sin x}}{\sqrt{sin x} + \sqrt{cos x}}dx$$

Can you see an emerging pattern? This is yet again equal to: $$\frac{b-a}{2} = \frac{\frac{\pi}{2}}{2} = \frac{\pi}{4}$$.

We can observe that for any integral of the form $$\int_{a}^{b}\frac{f(x)}{f(a+b-x) + f(x)}dx$$ the following holds: $$\int_{a}^{b}\frac{f(x)}{f(a+b-x) + f(x)}dx = \frac{b-a}{2}$$.

Now if I were to just throw this out there I would probably get lynched by Ueli Maurer himself. Thus the proof goes as following: We know that: $$\int_{a}^{b}\frac{f(x)}{f(a+b-x) + f(x)}dx = I$$ This will be our first equation. Now we set \(u = a + b - x\) and \(\frac{du}{dx} = - 1 \implies du = -dx\) by putting that into our integral we receive: $$\int_{b}^{a}\frac{f(a+b-u)}{f(u) + f(a+b-u)} (-du) = I\newline \implies \int_{a}^{b}\frac{f(a+b-u)}{f(u) + f(a+b-u)} du = I \newline \implies \int_{a}^{b}\frac{f(a+b-x)}{f(x) + f(a+b-x)} dx = I $$ Since u is only used as a placeholder variable we may rename it as x. By adding these two integrals (which both equal I) together, we are left with: $$ \int_{a}^{b}\frac{f(a+b-x) + f(x)}{f(x) + f(a+b-x)} dx = 2I\newline \implies \int_{a}^{b} dx = 2I \newline \implies b-a = 2I \newline \implies I = \frac{b-a}{2} \square $$

Now you can solve some integrals very quickly and without the need of much calculation!

This post was largely inspired and based on this “MindYourDecisions” video. Check it out for another example for an integral which you can solve with this method.