MIT 2006 Integration Bee

What people said

Cites fridge · 2016-03 · Mathematics Stack Exchange · on “How to compute this integral without the use of the error function?”

I was watching this: [this video] And this integral came up. $$\int_{0}^{\infty}{(2x^2+1)e^{-x^2}}dx$$ To which the answer was $\sqrt{\pi}$. They made it clear that you didn't need knowledge about erf$(x)$ to solve this integral, although it looks like one of the competitors tried to solve it using that. How can you solve this without "extra knowledge"?(Assuming you only have mastered the techniques of integral calculus)

Cites Vishu · 2020-04 · Mathematics Stack Exchange · on “Evaluate $\int_0^\pi \frac{\sin\frac{21x}{2}}{\sin \frac x2} dx$ (from MIT Integration Bee)”

I recently watched the MIT Integration Bee ( $2006$ ) video [this video] and stumbled upon this unusual integral: $$\int_0^\pi \frac{\sin\frac{21x}{2}}{\sin \frac x2} dx$$ I thought multiplying up and down by $\cos \frac x2$ would help, after which I got $$ \int_0^\pi \frac{\sin11x + \sin10x}{\sin x}dx = I$$ Now using $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ , $$I=\int_0^\pi \frac{\sin 11x -\sin 10x}{\sin x}$$ and on adding the two we get $$I= \int_0^\pi \frac{\sin 11x}{\sin x}$$ Now there are two paths I could take, either write $\sin 11x$ entirely in terms of $\sin x$ (which is a daunting task) or apply the sine addition rule as $\sin 11x = \sin(10x + x)$ . Doing the latter gives $$I= \int_0^\pi

Cites Rodrigo · 2022-01 · Mathematics Stack Exchange · on “How to prove that $\int_0^\infty\frac{e^{3x}-e^x}{x(e^{3x}+1)(e^x+1)}\text{d}x=\frac{\log3}2$?”

Closed form of definite integral from 2006 MIT Integration Bee The judges claim that $$\int_0^\infty\frac{e^{3x}-e^x}{x(e^{3x}+1)(e^x+1)}\text{d}x=\frac{\text{log}(3)}{2},$$ although no internet calculator except Wolfram|Alpha finds a closed form. WA returns that result. What steps must to be taken to obtain this result? Work from both contestants can be seen in this [this video] .