Solving ALL THE INTEGRALS from the 2022 MIT Integration Bee Finals

And no, we are not doing the 2021 finals. See you in 2024. Here's the link to my video solving the 2023 finals:

[erfanmohagheghian]

(2cos(x)+3) is always positive, so you don't get +/-! You just get the positive root.

[TEPK-]

4:32 For this, I used different function to solve.
First, let I(a) = integral(0 to inf) (e^(-2x) * sin(ax) / x) dx. (The integral we want to compute will be I(3).)
Observe that I(0) = 0.
Compute I'(a) by integration by part to reach I'(a) = 2 / (a^2 + 4).
So I(a) = arctan(a / 2), then the answer we want is integral(0 to inf) (e^(-2x) * sin(3x) / x) dx = I(3) = arctan(3 / 2).

[ShanBojack]

bro solves the MIT integrals just because he's bored? What a beast!

[sarthak.chatterjee]

Here's a slicker solution to Problem 1 using complex numbers. Let z = e^{ix}. Then the integrand is the modulus of z^20 + 3z^21 + z^22. But | z^20 + 3z^21 + z^22 | = |z|^21 * | z + 1/z + 3 | = 1 * | 3 + 2 cos x | = 3 + 2 cos x. Hence the integral is just 3x + 2 sin x.

[MrWael]

Great job. Thank you for your amazing effort. With my best wishes.

[milessodejana]

You can also use the Laplace transform of sin(3x)/x for the second integral.

[mridulbansal]

Great Video! I enjoyed seeing your solutions to these problem.

I have a much nicer solution to the last problem. We notice that the floor of the log10 of a number is literally just the number of digits (or the exponent when represented in scientific notation). So as long as our asnwer is accurate to the nearest power of 10, we are fine.

We also note that 10^(-x^3) at x = 2022 drops fast. Really fast. Like really really fast. Simply going from x=2022 to x=2023 drops the output by 10 million orders of magnitude. Only a very, very small section of the integral is actually contributing to the relvant part of the integral (again we only care about anything that can contribute to anywhere near the leading digit). In fact, we can stop evaluating the integral when we reach x such that x^3=1+2022^3, as after this point, the rest of the integral will be too small to contribute.

Using the linear approximation for x^3 (the one you used in your video), we find that our upper bound of integration is x=2022 + 1/(3*2022^2), or around 2022+8*10^-8. Evaluating the integral using the trapezoid method gets us I = (Left + Right)*length/2 = = 4.4*10^(-2022^3-8). The trapezoid is a good (over)estimate, and this sliver of the integral is literally contributing orders of magnitude more than the rest of the integral, so we can be confident that this answer is accurate to the power of 10. (To get an appropriate underestimate, I believe that we can use the right rectangle estimate with 3 partitions, but I could be wrong).

Now we take the log and floor it, to get an answer of -2022^3 - 8.

[Starduster]

4:34 can’t we use Laplace here ?
To evaluate the integral from the table
Just asking

[hellobanana]

There is another way of doing the first integral. You can sum sin20x with sin22x and cos20x with cos22x, then more easier to simplify

[krisbrandenberger]

@ 39:58 The denominator should be 3*4*10^6×2, not 3*8*10^6×2. The 2000=2*10^3 is being squared, not cubed. However, I did enjoy the video! Keep up the good work! 👍

[TheMartinbowes]

For problem 4 I thought a u =1/x sub would be a good start. You follow that and quickly find you have to calc integral of 3x^4/(x^3 + 1)^3 in bounds of zero to infinity...how hard could that be? Kept making errors in partial fraction decomp, the third order pole in contour integration is a bitch. I had to dig out my notes on Ostrogradsky method. Several pages later I can confirm you're right! I need a beer.

[chandranshu]

Isn't the first problem super easy if you apply the sine and cosine sum formulae to (sin 20x + sin 22x) and (cos 20x + cos 22x) which nicely reduces the expression under the root to (2 cos x + 3 )² (sin²21x + cos²21x) which immediately reduces to ±(2cos x + 3) from which the result follows immediately.

[Ron_DeForest]

Just wondering what’s the app you’re using in this video?

[krisbrandenberger]

Hey, Professor! The product of two cosines results in a sum of two cosines, not a difference.

cos x*cos y=1/2(cos (x+y)+cos (x-y))

[SonnyBubba]

What software do you use to make these videos?

[romanvolotov]

screw the last problem, I missed like half of your solution 😭😭😭

[ガアラ-hh]

Gamma(1/3 +1)*gamma(2/3+1)/gamma((1/3 +1)(2/3+1))

[emanuellandeholm]

Tomorrow is the big game 505. I'm putting down 5 euro on Man City, will you bet against? :)

[iitaspirant]

What qualifications are required to understand the last question ?