MIT Integration Bee!

@laurayzhang

14.0K followers · usually 35.4K views a post

Views

260.9K

Likes

14.5K

Comments

208

Shares

1.4K

SoundTikTok sound · original sound - laurayzhang

Promoted product

MIT Integration Bee

Why it worked

The video likely performed well due to its niche appeal to students and math enthusiasts, the inherent challenge and intrigue of solving a difficult calculus problem, and the creator's relatable persona as an MIT math major.

Summary

A math major at MIT solves a notoriously difficult integration problem from the MIT Integration Bee. The solution involves recognizing a Taylor series expansion for e^x, simplifying the integrand, and then evaluating the definite integral.

Structure

1Introduction to the MIT Integration Bee problem
2Recognizing the Taylor series for e^x
3Simplifying the integrand using the Taylor series
4Evaluating the definite integral
5Confirming the answer on the MIT Integration Bee website

Product placement

The video features a problem from the MIT Integration Bee, which is a math competition. The problem itself is displayed on screen and then solved by the creator. The final answer is confirmed on the MIT Integration Bee website.

On-screen text

MIT Massachusetts Institute of Technology ∫¹/₂₀₂₅₀ (∑∞k=1 (2025x)ᵏe⁻²⁰²⁵ˣ / k!) dx MIT Integration Bee!

Original caption

3, 2, 1, integrate! #mit #calculus #math #integral #integration #LearnOnTikTok #college #school #student #mathnerd #ilovemath #apcalc

More from @laurayzhang

More breakout videos from this creator.

Transcript

MIT Integration Bee! Solve an MIT integration Bee problem with me. I'm a math major at MIT, so hopefully I'll be able to do it. These integration problems are notoriously hard, so let's try it. So my initial thoughts are to try to find a closed form expression for the term inside the parentheses. So for fixed x, you can take out the e to the -2025x term. The summation now really reminds me of the Taylor series for e to the x. The main difference is that we need to go from 2025x to x and then from k=1 instead of k=0. If we write this out, we get that the summation is equal to e to the 2025x minus x to the 0 over 0 factorial, which is just e to the 2025x minus 1. So substituting the summation back into the integral, we get this new simplified integral. We can distribute the e to the -2025x term and then we get the integral from 0 to 1 of 1 minus e to the -2025x dx. Solving the integral, we get x plus 1 over 2025 e to the -2025x evaluated from 0 to 1/2025. Evaluating the expression at 1/2025 and 0 gives us this difference. Which gives us the final answer 1 over 2025e. Looking at the integration Bee website, this seems to be the correct answer.