You know that stupid math problems with food that 50-old people share on Facebook? I’m kinda use the answer is yes; those problems are obviously ridiculously easy even if they’re usually presented with captions like “only 1% of the population of the entire multiverse can solve this”. Of course the various math-themed communities on Facebook started to meme on this fact coming out with pearls like the “only 0,000000000001% of the population of the world can solve this” associated with an image featuring the Riemann Hypothesis. Here’s a common *food problem*:

Nothing too hard: it’s clear that ; easy. Parodies of such “problems” exist are usually feature an array of *fruit linear equations* and an integral right below with fruit in it that say “now solve me, you 1% of population genius”. However I’ve always laughed over those parodies-images and never tried to actually solve one till yesterday; and holy shit that was painful. I want to report here the whole thing since I found it really challenging and fun (well, if you’re masochist enough) and I also want to see how WordPress’ LaTeX can handle this:

First let’s do the *system of linear equations*: let’s set and rewrite the whole this as:

This is not so hard so I’m not going to explain what I did step by step, but I can tell you that I used the substitution method starting by finding in the third equation. The results are not so good-looking:

I personally expected more plain results like 1, 3, 5 and even numbers like 6, 8, 12; those fractions made me realize that I was going to suffer a lot to break down that integral. Let’s rewrite it without fruit, the lemon will be our :

Let’s now solve the indefinite integral first:

This thing initially scared me, trig always does, but then I realized that we can use the trigonometric identities and to simplify it:

So the trig went away but this new integral is still problematic: we need to decompose this fraction (if you have any, this is the time to play some epic music like this). This implies that we have to factor and that’s not banal because it have no real roots, it have complex roots but I don’t know if dealing with them is useful. However I found that if we rewrite the thing as we can then substitute to get ; we can now see that easy factor to and thus the whole thing is just : a difference of squares! A difference of squares factor to so equals , undoing the substitution: . So the factoring is done:

It’s now time to decompose this fraction:

We need to find values for so:

We can now solve this system of linear equations to get our answers:

Playing with the first two equations and substituting things is easy to come up with these solutions:

Now:

We can now rewrite the integral applying linearity:

The first thing I noticed at this point is that and that ; that’s very similar to what we have as numerators of our fractions: we just have to multiply the whole integral by :

Considering that we can write the integral as:

The first two integrals are easy: they are in the form that easily equal , where is the natural logarithm of , so:

The third one is a bit tricky; we have to complete the square in order to go on (in doing so we can even apply linearity and split it in two):

Now we can use the rule to solve those two integrals and get that:

Putting everything together we finally have a solution for our initial integral , let’s call it :

One last step: we need to calculate . Due to previous results this ask us to calculate that apparently equals:

That we can approximate to 1.76; probably there are better ways to solve it but I couldn’t figured them out since the last time I did integrals seriously was probably some months ago. However that was fun (and probably also one of the hardest I’ve ever done) and challenging; maybe I’m going to try some other parody-images math problems in the future. If you like math I’ll suggest you to try those things out, they can be pretty cool.

## Credits:

- Chuck Aileron: http://chuckaileron.blogspot.com