Solve This Tricky Integral to get Free Wi-Fi!

Have you ever had to solve a mathematics problem to gain access to Wi-Fi? Let's see what it involves!

Apologies for the poor audio quality in this video, my microphone broke and I had to use the default audio from my camera which echos quite badly!

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For those of you that are new here, hi there 🌞 my name is Ellie and I'm a Cambridge Part III Mathematics Graduate and current Astrodynamics Software Engineer! This channel is where I nerd out about maths, physics, space and coding so if that sounds like something you're interested in, click the subscribe button to follow along ☺️

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[EllieSleightholm]

⚠️ For anyone wondering why the integral was only half of the semi-circle - it’s due to the square root in the integral. Because of the square root, y cannot be negative! I should really have removed the bottom half of the semi-circle but I thought it was clear once I explained that we only needed the top half. Hope that clears up any questions!

[watt_the_border_collie]

You have to appreciate that there was no crazy integration technique involved, just pure mathematical reasoning. Understand the problem and use logical reasonings to get to the result. This is what a mathematician does! 10/10

[oethe]

I love that you used the semi circle, it brought back a long lost memory from Calc AB before I learned trig substitution. Great video!

[theaizere]

I appreciate just how knowledgeable and cohesive you are to be able to hand guide me through this. takes a lot to simplify haha

[NNk-di]

Very briefly explained. I left BSc Math in 2004, but today i gone through such integral solution (by chance) and had tears in my eyes, how we used to solve these equations without any guidance and lack of solution materials notes etc, but internet has made it worldwide source of guidance. Your graphical representation very helpful for solution of such equations.
(From Pakistan)

[TreasureOfure]

"Truly impressed by the way you used mathematical proofs to solve the integration question! Your logical approach and clear presentation made it easy to follow along. I particularly liked everything you did. Keep up the excellent work! Would love to learn more about your techniques"

[marc]

I’m surprised of how understandable this was, considering I never did integration. Good job Ellie, In particular I liked the speed of your video. It made it easier to understand your steps.

[hugoperozo]

Thank you so much Ellie. You are amazing! You break down complex problems into small-sized pieces that anyone can understand. Thanks for sharing your gift and for taking the time to do this. It takes a lot of effort and work to make these types of videos. THANK YOU. You are now my top YouTube channel. Keep up the amazing work! Thank you again.

[Brother_]

Thanks sister now I can learn maths from you with free Wi-Fi 😊

[onphysicsandmathematics]

When I was going to the 9th grade, I was looking forward to learning these things, but I just learned them this year and I love pure mathematics.❤
Thank you Ellie

[TheMathCompany]

With the dx inside one can interpret it as fractional derivative. And the answer is (-4/(3Pi))*(8^(1/2)-1) in that case.

[HyperFocusMarshmallow]

Looking at it quickly.
I := [-2, 2] is an interval that’s symmetric around 0.
So integrals of odd functions [f(x)=-f(-x)] over I will evaluate to zero.
Call the result of the integral Q.
Q = integral over [-2, 2] of q(x)dx,
where

q=(u*v+1/2)*w, where
u(x):=x^3 is odd.
v(x):=cos(x/2) is even.
1/2 is even.
w(x):=sqrt(4-x^2) is even.

for functions
[even] * [even] = [even].
[odd] * [even] = [odd].
[odd] * [odd] = [even].

u*v*w is odd. — it cancels under the integral.
(1/2)*w is even. — it remains.

Q = integral over [-2, 2] of (1/2)sqrt(4-x^2)dx.

Quick change of variables {x=2s, dx=2ds} gives

Q = 2*Integral over [-1, 1] of sqrt(1-s^2)ds.
The integral is the area of the half unit circle. Twice that is the area of the unit circle.

Some might know it as π.

[RY]

I was a little confused as to why the area is only the top part, but its simply because the square root function means only the positive part of the function for anyone else wondering.

[rubhern]

Earned another sub and liked. A great researcher but also a great teacher. Explained it so well and with clarity.

[christianforler]

This was a beautiful solution to the Free WiFi problem. I enjoyed watching this video a lot.

[arthurmorris]

loved your way of explaining, its simple jet effective.

[thomasjefferson]

You know i saw X^3 multipled by cos and i screamed 0, Then I started to do a trig sub in my head, and you showed me a new simpflication i never though of. Great stuff.

[cloverisfan]

fun fact: if you have to solve an integral for wifi password, the answer is either pi or e. trust me

[Tutor-i]

For some reason I remember that this was the first integral I was ever exposed to even before taking a calculus class. It doesn’t require very hard integration techniques and just requires logic.

[iftekharalambhuiyan]

Top tier solving. Process explained beautifully. Earned a sub.