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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

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

  1. Sums are pretty difficult for me, so I might have miscalculated a sum or something.
  2. Representing integrals as sums incorrectly.
CrSb0001
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    Well for what it's worth, your method is correct and if it were upto me , I'd do it the same way. As for the answer, you can always check it on Wolfram alpha or any CAS . If it matches then you're good to go. – Mr.Gandalf Sauron May 08 '23 at 18:07
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    I could not follow the step from $\sum_{n=0}^{99}\int_n^{n+1}\lfloor x\rfloor x\lceil x\rceil dx$ to $\sum_{n=0}^{99}n^2+n\int_n^{n+1}xdx$. Just looking at the term for $n=1,$ I get $3$ in the first sum and $5/2$ in the second sum. – David K May 08 '23 at 18:19
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    @DavidK I believe there should be a bracket. I mean $\sum_{n=0}^{99}n(n+1)\int_{n}^{n+1}x,dx$ – Mr.Gandalf Sauron May 08 '23 at 18:21
  • Did you mean to write $\sum_{n=0}^{99}(n^2+n)\int_n^{n+1}xdx$? If so, then that step is OK but I think there is an error further down. – David K May 08 '23 at 18:21

3 Answers3

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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.

Quoka
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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

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

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

Lai
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