2020 Berkeley Integration Bee Finals

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Recommends grand_chat · 2020-04 · Mathematics Stack Exchange · on “How can we evaluate this integral from the 2020 UC Berkeley Integration Bee?”

The correct form of the integral (see the video [this video] ) is: $$ I = \int_1^2 \left( e^{1-\frac1{(x-1)^2}}+ 1\right) + \left( 1+ \frac1 {\sqrt{1-\log(x-1)}} \right)\,dx. $$ This equals $$ \int_1^2 (f(x) + f^{-1}(x))\,dx $$ where $f(x):=\exp\left({1-\frac1{(x-1)^2}}\right)+1$ . Check that $f$ is increasing and $f(1)= 1$ and $f(2)= 2$ , so the value of the integral is $$ 2f(2)-1f(1)=3 $$ using the formula [LINK] $$\int_a^b f(x)\,dx + \int_{f(a)}^{f(b)}f^{-1}(x)\,dx = bf(b)-af(a).$$ This is much easier to see geometrically: Borrowing a picture from this answer [LINK] , you can visualize the integral as the area of an elbow-shaped region: [LINK] which in this case can be decomposed into three sq

Cites zalm · 2020-06 · Mathematics Stack Exchange · on “UC Berkeley Integral Problem: Show that $\int_0^{2\pi} \frac{\min(\sin x, \cos x)}{\max(e^{\sin x},e^{\cos x})}\ {\rm d}x = -4\sinh(1/{\sqrt2})$.”

Show that $$\int_0^{2\pi} \frac{\mathrm{min}(\sin{x},\, \cos{x})}{\mathrm{max}\left(e^{\sin{x}},\, e^{\cos{x}}\right)}\ \mathrm{d}x = -4\sinh\left(\frac{1}{\sqrt{2}}\right).$$ this problem comes from the 2020 UC Berkeley Integration Bee [this video] and was not solved by either of the contestants. Any hints? My initial approach was to compute the maximum and minimum of the specified function by observing the graph for $x\in (0, 2\pi)$ but could not get very far. Thank you!

Cites Rodrigo · 2022-02 · Mathematics Stack Exchange · on “Discrepancy in Closed Form Results for a Complex Integral”

I encountered a complex integral problem and found conflicting results between Wolfram Mathematica and the judges evaluating the work. Specifically, Wolfram Mathematica gives the closed form of the integral: $$-\exp \left(-\cos \left(2 \tan ^{-1}\left(1-\sqrt{2}\right)\right)-\cos \left(2 \tan ^{-1}\left(\sqrt{2}+1\right)\right)\right) \left(2 \sinh \left(\frac{1}{\sqrt{2}}\right) \exp \left(\cos \left(2 \tan ^{-1}\left(1-\sqrt{2}\right)\right)+\cos \left(2 \tan ^{-1}\left(\sqrt{2}+1\right)\right)\right)+e^{\cos \left(2 \tan ^{-1}\left(1-\sqrt{2}\right)\right)}-e^{\cos \left(2 \tan ^{-1}\left(\sqrt{2}+1\right)\right)}\right)$$ However, the judges provided the closed form as: $$ -4 \sinh \lef

Cites Sarvesh Ravichandran Iyer · 2022-06 · Mathematics Stack Exchange · on “Berkeley integration bee 2020 final #1: $\int_{-\infty}^\infty\frac{2020^{-|x|}}{1+5^{\arcsin(\sin^5(x))}}dx$”

[this video] The first integral that is discussed here is the one present in the original post, the answer is apparently $\frac{1}{\ln(2020)}$? AH, OK, It's $2020^{-\mathbf{|x|}}$ in the numerator, NOT $2020^{-x}$ (note the absolute value). [This is not a duplicate] please edit your question accordingly

Cites Kamal Saleh · 2022-11 · Mathematics Stack Exchange · on “Calculate $\int_0^1 \ln^{2020}(x)dx$”

In this video [this video] on mark 17:00 gives the tiebreaker question of the integration bee finals. It asks to calculate $$\int_0^1 \ln^{2020} (x)dx$$ We could create a reduction formula for $\ln^n(x)$ which would be $$I(n,x)=x\ln^n(x)-n\int\ln^{n-1}(x)dx$$ Repeatedly doing this we get this sum: $$I(n,x)=(-1)^nn!+x\sum_{k=0}^{n-1}(-1)^k\ln^{n-k}(x)\binom{n}{k}k!$$ Substituting $n=2020$ and then substituting $x=0$ , I realized that we could just simply substitute $x=1$ to get the answer, which is $2020!$ . This answer is correct, but is my procedure right?