BEWARE: Monster Integral

Рет қаралды 21,491

Жыл бұрын

Parched, struggling to move forward - chalk is on the run again because a Monster Integral is out for revenge. Will chalk overcome the predicament he's in? Only the end of the credits will tell.
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Пікірлер: 83

@rain_deer Жыл бұрын

That's a spooky place to stop.

@yoscot629 Жыл бұрын

disintegration of parts

@bsmith6276 Жыл бұрын

This is my favorite kind of monster integral. Everything you need to solve it would be things taught in a high school AP Calc class, or taught in a college freshman semester of Calc 2. But this integral is far more intricate than anything you would see in a textbook and needs to be processed through multiple layers of manipulation, including non-obvious substitutions you would not likely see in an average elementary Calc class.

@names9769 Жыл бұрын

He really woke up one day and decided to write the funniest descriptions.

@nuranichandra2177

There is no way one would think of such grave manipulation of the given integrand in an exam situation.

@toadjiang7626

Just saw this video, and I was totally blown away by how easy this integral is once I let x^2 = tan(t), and this whole thing turned into a nice beta function (√2/16)*int_(0 to pi/2) 2*(sin(2t))^(1/2) (cos(2t))^(-1/2) d(2t) = (√2/16)*B(3/4, 1/4) = pi/8. It only took like 2 minutes.

@ShoeboxInAShoebox Жыл бұрын

Gotta say, really don’t like the newer video titles and thumbnails. Makes each video feel more like clickbait, so I don’t watch them as much. I prefer to see thumbnails relating much more directly to the problem addressed in the video

@technoviking Жыл бұрын

You can also do this as a contour integral using a quarter circle. Taking care to account for a sign change due to the branch cuts, we get the contour integral, as the radius of the circle -> oo and the paths around the branch cuts approach the branch cuts, has value 2*I + 2i*I, where I is the value of our original integral. But by the Cauchy residue theorem and noting the residue at 1/sqrt(2)+i/sqrt(2) is 1/8-i/8, this is 2*pi*i*(1/8-i/8) = pi/4 + i*pi/4. It follows I = pi/8.

@danielevilone Жыл бұрын

Integration is a simple game, you do huge calculations and in the end π appears.

@user-mp7bt5bw5l

u can solve it also with the substitution: u=sqrt(tanx) , this will lead to the integral of sqrt(tanx) from 0 to pi/2 which we already know how to solve this integral.

@nevoitzhak2092 Жыл бұрын

@Notthatkindofdr Жыл бұрын

It turns out you can do this with a single substitution: u=sqrt(1-x^4)/x. (Hard to think of or motivate, but then so were the substitutions that Michael used!) It is then routine to check that the original integral is equal to the integral of 1/(u^4+4) from u=0 to infinity, and it is a challenging but classic partial fractions problems to show this equals pi/8.

@CTJ2619

I love these gnarly integrals !

@goodplacetostop2973 Жыл бұрын

19:15

@sitobubble5697 Жыл бұрын

Well done, it's one of the most beautiful integrations I've seen in my life.

@RAG981 Жыл бұрын

Well done! Like it.

@mohammadabdulla8601 Жыл бұрын

Imagine having such a question in an exam ! Nice problem 👍

@manucitomx Жыл бұрын

Wow

@ANTONIOMARTINEZ-zz4sp Жыл бұрын

Michael uses the Sophie Germain's identity, but he makes a little mistake when he writes it on the blackboard. Anyway, thank you very much for such an awesome exercise.