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