How to solve MIT Integration Bee $2023$: $\int_0^{100} \lfloor x\rfloor x\lceil x\rceil dx$

Question

So I was yet again looking through the homepage of Youtube when I came across this video by Michael Penn which was asking us this question from the $2023$ MIT Integration Bee:$$\text{Integrate: }\int_0^{100}\lfloor x\rfloor x\lceil x\rceil dx$$Which I thought that I might be able to do. Here is my attempt at solving the integral:$$\text{Integrate: }\int_0^{100}\lfloor x\rfloor x\lceil x\rceil dx$$$$\sum_{n=0}^{99}\int_n^{n+1}\lfloor x\rfloor x\lceil x\rceil dx$$$$\sum_{n=0}^{99}n^2+n\int_n^{n+1}xdx$$$$\sum_{n=0}^{99}n^2+n(\frac{x^2}{2}\|_n^{n+1})$$

$$\frac{1}{2}\sum_{n=0}^{99}2n^3+n^2+2n+1$$Or$$24671675$$ My question

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Is the result that I have achieved correct, or what could I do to attain the correct solution/attain it more easily?

Mistakes that I might have made

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1. Sums are pretty difficult for me, so I might have miscalculated a sum or something.
2. Representing integrals as sums incorrectly.

Answer 1

Note that $I(n)=\displaystyle\int_n^{n+1}x^3dx=\frac{(n+1)^4-n^4}{4}=(n+\tfrac 12)(2n^2+2n\color{red}{+1})/2$

On the other hand

$\displaystyle\int_n^{n+1} \lfloor x\rfloor x\lceil x\rceil dx=(n+\tfrac 12)(n^2+n)=I(n)-\color{red}{\tfrac 12(n+\tfrac 12)}=I(n)-\tfrac 12\int_n^{n+1}x\,dx$

It results that

$$\int_0^{100} \lfloor x\rfloor x\lceil x\rceil dx=\int_0^{100}\big(x^3-\tfrac 12x\big)dx=\Bigg[\frac{x^4-x^2}4\Bigg]_0^{100}=\dfrac{10^8-10^4}4$$

Answer 2

It's the correct idea. However, $$ \int_n^{n+1}\lfloor x\rfloor x\lceil x\rceil dx =\int_n^{n+1}nx(n+1) = (n^2+n)\int_n^{n+1}x = (n^2+n)\left(\frac{(n+1)^2 - n^2}{2}\right) $$ Simplifying, you should get $$ \int_n^{n+1}\lfloor x\rfloor x\lceil x\rceil dx =(n^2+n)\left(\frac{(n+1)^2 - n^2}{2}\right) =\frac{n(n+1)(2n+1)}{2} = n^3 + \frac{3n^2}{2} + \frac{n}{2} $$ After summing, I get 24997500.

Answer 3

In general, $$ \int_0^n\lfloor x\rfloor x\lceil x\rceil d x=\frac{n(n+1)^2(n+2)}{4} $$ Proof: $$ \begin{aligned} \int_0^n \lfloor x\rfloor x\lceil x\rceil d x & =\sum_{k=0}^n \int_{k=1}^{k+1}\lfloor x\rfloor x\lceil x\rceil d x \\ & =\sum_{k=1}^n \int_{k+1}^{k+1} k x(k+1) d x \\ & =\sum_{k=1}^n k(k+1) \int_k^{k+1} x d x \\ & =\sum_{k=1}^n k(k+1)\left[\frac{x^2}{2}\right]_k^{k+1} \\ & =\frac{1}{2} \sum_{k=1}^n k(k+1)(2 k+1) \end{aligned} $$ Using the formula for the sum of powers $1,2,3 $ of the first n natural numbers, we have

$$\int_0^n \lfloor x\rfloor x\lceil x\rceil d x = \left[\frac{n(n+1)}{2}\right]^2+\frac{3}{2}\cdot \frac{n(n+1)}{6}(2 n+1)+\frac{1}{4} n(n+1)= \frac{n(n+1)^2(n+2)}{4} $$

In particular, $$ \int_0^{99} \lfloor x\rfloor x\lceil x\rceil d x = \frac{\left.99(100)^2(101\right)}{4}=24997500 $$