An Ode to Some Beautiful Integrals

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MIT Integration Bee 2023 – Finals Problem 5

Are we doing okay so far? Let’s move on to the last problem of the competition, an absolute cracker of a problem: $\displaystyle I = \int_0^1 \left( \sum_{n=1}^{\infty} \frac{ \lfloor 2^n x \rfloor }{3^n} \right)^2 \: \mathrm{d}x .$ This problem, I have no shame in admitting, had me stumped. In this post, I will…

April 14, 2023April 14, 2023

MIT Integration Bee 2023 – Finals Problem 4

Problem 4 of the bee asks us to find the value of $\displaystyle C = \Bigg\lfloor 10^{20} \int_2^{\infty} \frac{x^9}{x^{20} – 48x^{10} + 575} \: \mathrm{d}x \Bigg\rfloor .$ In many ways, this was the simplest problem of the bee, with only the floor function proving to induce a bit of complexity but also making it…

April 13, 2023April 13, 2023

MIT Integration Bee 2023 – Finals Problem 3

On to Problem 3, which looks like this: $\displaystyle I = \int_{-1/2}^{1/2} \sqrt{x^2 + 1 + \sqrt{x^4+x^2+1}} \: \mathrm{d}x .$ We try and express the integrand as a sum of two radicals, i.e., $\sqrt{x^2 + 1 + \sqrt{x^4+x^2+1}} = \sqrt{a} + \sqrt{b}.$ Squaring both sides, $latex x^2 + 1 + \sqrt{x^4+x^2+1} =…