2024 MIT Integration Bee - Regular Season

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Күн бұрын

Пікірлер: 22

@maxz8807 9 ай бұрын

Well done to all math wizards here - I wouldn't even know where to begin on most of these haha. I'm still stumped by the integral with the floor of the floor of x - would love to see an explanation to the solution of that one

@BPEREZRobertJamesL 2 ай бұрын

Let I be our integral. We can write every real number x as k + t for some integer k and some real 0 < t < 1. Then, we may change our integration into a summation w.r.t. k of integrals w.r.t. t. This gives us the following expression: Sum_(k=1)^9 (Int_0^1 (Floor(k(k+t) dt) We start from k = 1 because for any x such that k = 0, the corresponding integral is zero. Therefore: I = Sum_(k=1)^9 (Int_0^1 (Floor(k^2 + kt) dt) Since k^2 is an integer, we can separate Floor(k^2 + kt) into k^2 + Floor(kt). Thus: I = Sum_(k=1)^9 (Int_0^1 (k^2 + Floor(kt)) dt) By linearity of integration, we have: = Sum_(k=1)^9 (k^2 Int_0^1 (1 dt) + Int_0^1 (Floor(kt) dt)) = Sum_(k=1)^9 (k^2 + Int_0^1 (Floor(kt) dt)) Now, Floor(kt) depends on t, which ranges from 0 to 1. Note that the interval [0, 1] may be decomposed into a union of k disjoint intervals: (0,1) = [0, 1/k) ∪ [1/k, 2/k) ∪ [2/k, 3/k) ∪ ... ∪ [(k-1)/k, 1) We can ignore [0, 1/k) because in that interval, the floor is zero, and therefore, so does the integrand. Thus, we may substitute Floor(kt) = m for some integer 0 < m < k and write the remaining integral as a summation from m = 1 to k-1 of Floor(kt) times the width of each subinterval in the decomposition above, which is 1/k: = Sum_(k=1)^9 (k^2 + Sum_(m=1)^(k-1) (m/k)) = Sum_(k=1)^9 (k^2 + 1/k Sum_(m=1)^(k-1) (m)) By the linearity of summation, we have: = Sum_(k=1)^9 (k^2) + Sum_(k=1)^9 (1/k Sum_(m=1)^(k-1) (m)) Note that the sum of the first n integers is n(n+1)/2, and the sum of the first n integer squares is n(n+1)(2n+1)/6. You may prove this via induction, via telescoping series, or whatever method you come up with. Using these two theorems, we should get: = 9(9 + 1)(18 + 1)/6 + Sum_(k=1)^9 (1/k k(k - 1)/2) = 15*19 + Sum_(k=1)^9 ((k - 1)/2) = 15*19 + 1/2 Sum_(k=1)^9 (k - 1) (distributive property) Now, note that summing the first 9 (k-1)'s (starting from k=1) is the same as summing the first 8 k's. Thus: = 15*19 + 1/2 Sum_(k=1)^8 (k) = 15*19 + 1/2 8(8+1)/2 = 15*19 + 2*9 = 15*19 + 18 = 15*19 + 15 + 3 = 15*(19 + 1) + 3 = 15*20 + 3 = 300 + 3 = 303 Hope this helps.

@BryanDing3000 5 ай бұрын

I'm surprised Problem 16 stumped so many people. A simple integration by parts would've led to the correct answer.

@elmaminsk5411 9 ай бұрын

22:23 - You can see the critical mistake on the board: substitution to 1-x exchanges integration limits, so we shouldn't add minus sign at the beginning of the integral

@sca4723 3 ай бұрын

i got (1/2025) - (1/1013) + (1/2027) it is right?

@CallMeMak. 2 ай бұрын

Feels good to be Able to solve some of them, especially with the time limit! Hope that one day I might be able to do it alongside of them :)

@ayushtripathi2605 7 ай бұрын

Second problem can also be solve by taking x to the power. 4047 from denominator

@Defaulter_4 Ай бұрын

33:38 this guy's smile made my day. So happy to see him :D

@edwardhuang5885 9 ай бұрын

dang that's luke from mathcounts 2018 he was famous in that vid

@mrlove3010 6 ай бұрын

Chirag falor jee advanced air1 2020❤

@maxz8807 9 ай бұрын

Appreciate the subtitles :D

@maxz8807 9 ай бұрын

Also appreciate the man with the purple bucket hat writing his method in the first round! I'm not a math genius, but I do like watching these, and I could actually understand what was happening :)

@maxz8807 9 ай бұрын

Sorry perple bucket hat man, I don't know your name 🥲

@Zach27182 9 ай бұрын

Next time would it be possible to keep the integral on the screen while showing the contestants solving them?

@tvvt005 3 ай бұрын

Wait how did they graph this function? Can someone explain? 13:05

@nakulcodes 2 ай бұрын

ahh so as its in log to the base 43 so from 1 till 43 the function gonna give a value of 0, from 43 to 43 squared its gonna be 2 and so on.

@onichan8557 8 ай бұрын

1:11:03 board 1 is incorrect , the denominator of last 2 terms , they should be whole square and whole cube respectively , but the he wrote 2 cube and 2 square which is obviously different

@antontrygubO_o 7 ай бұрын

I don't understand. Answer is same as correct...