integrate with the FLOOR and CEILING | MIT Integration Bee 2023

"CAVENDISH YOU SCOUNDREL! How was it that you were able to come by an integral from the the current year MIT Integration Bee?" ask Muffleford. "Why I obtained the information from a friend of mine", Cavendish began to sweat heavily and Meffleford could see it and gave Cavendish a look of skepticism. Heading off his inquiry, Cavendish quickly added, "...a friend...his...his name is...Chalk."
"CHALK?!", exclaimed Muffleford. "THE Chalk? The...the. one that....", "yes!", Cavendish interrupted. "Yes. the very same...I was desperate and needed a hit of Calculus and I got this integral". Okay, stop reading this nonsense, watch the video all the way through and comment "CAVENDISH YOU SCOUNDREL!"

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[petriachor]

10:55 The first integral from 0 to 1 is indeed one half as we take the antiderivative x²/2 but the second one is just a constant multiplied by the length of the interval, so it should be 1/2 + n instead of 1/2 + n/2

[xenofurmi]

I've been thinking about his video a lot. I love mixing the discrete with the continuous. This is so very cool and I'm so glad you're presenting this.

[daniellosh]

By expanding the sum to powers of n^3, n^2 and n, the answer is 99990000/4 = 24997500. The general result summing from 0 to N is N^2(N^2-1)/4, slightly smaller than n^4/4

[benkelly]

This simplifies a lot if you take the sum from 1 to 100 instead of from 0 to 99. Everything works out nicely in even powers of 10.

[mykolapokhylets]

Once we got to n(n+1)(2n+1)/2 it’s simpler to multiply the terms into polynomial and apply summation formulas to n^3, n^2, n terms

[kjorndogg]

CAVENDISH YOU SCOUNDREL! I was just thinking about if switching summations like what is done here was possible, and that geometric representation is just what I needed to understand it. Very cool!

[alonamaloh]

0*0.5*1 + 1*1.5*2 + 2*2.5*3 + ... is a polynomial of degree 4 because it's the partial sums of a polynomial of degree 3. You can take the first 4 numbers, (0, 3, 15, 42), compute their differences (3, 12, 27), again (9, 15) and again (6). Now the sum we want is the combinatorial numbers multiplied by the first numbers of those rows, like this: 0*n + 3*n*(n-1)/2 + 9*n*(n-1)*(n-2)/6 + 6*n*(n-1)*(n-2)*(n-3)/24. Substitute n=100 and you get 24, 997, 500.

[jkid]

Beautiful illustration of the change of the order of summation

[jay_]

That result at the end isn't the same though (I get 12, 582, 075 when I evaluate it algorithmically). As others have pointed out, the integral from t=0 to t=1 of n dt evaluates to n, not n/2, hence the result being off by about a factor of 2.

[mohammadalkousa]

Very interesting and great vedios you do!

[aronbucca]

No way he's integrating the whole house

[TheEternalVortex]

What's the point of doing the integral substitution? Evaluating it directly you still get (2n+1)/2 as the integral and you can split it up the same way as n+1/2 and continue as in the second method.

[juanixzx]

10:56 I'm think integral from 0 to 1 of n*dt = n, not n/2

[gregtanner-concordia]

It is interesting that the solution is exactly 2, 500 = 100^2/4 less than the integral of x^3 over the same domain (which is 25, 000, 000=100^4/4).

If you let the upper bound of the integral be K, then the result nicely becomes: (K^4-K^2)/4, so the pattern above holds that the integral with the floor and ceiling is always K^2/4 less than the integral where the floor and ceiling are ignored (just replaced with x).

[spacejunk]

What I did was split the integral into smaller integrals over the intervals [n-1 n) = I_n and summed them up from n=1 to n=N>1 (N being Natural).

Since in those intervals floor(I_n) = n-1 and ceil(I_n)= n, I can evaluate them and treat them as constants, so I pull them out of the integral, integrate the remaining x over the interval and then just simplifly the resulting polynomial of n until I get the sum from 1 to N of 1/2*(2n^3-3n^2+n) which is
(N(N+1)/2)^2 - N(N+1)(2N+1)/4 + N(N+1)/4.

With N=100 this is equal to 24, 997, 500

[mohammadalkousa]

For more integrals and solutions. There's a new book "MIT Integration Bee: Solutions of Qualifying Tests from 2010 to 2023"

[alexeyklimenko]

One can use rising factorials (see Wiki) to do it faster. First, put n and (n+1) out of the integral, then the remaining x yields the area of the trapezoid with bases n and (n+1) and height 1. So we arrive at the sum of f(n)=n*(n+1/2)*(n+1) over n=0, ..., 99. Note that Hence to find F(n) such that F(n+1)-F(n)=f(n) one may set Now the integral is F(100)-F(0)=9999*2500-0=24, 997, 500.

[Johnny-twpr]

Ahh I remember I saw a similar integral of x*floor(x). I felt so proud of myself after solving it entirely by hand

[dhruvchaudhary]

£. Integral xdx r*r+1. Limits of each step wise interval is r and r+1

[MrConverse]

2:25, did anyone else notice the counting error?