Showing $\int_{0}^{2022}x^{2}-\lfloor{x}\rfloor\lceil{x}\rceil dx = 674$

Question

It is from the 2022 MIT Integration Bee Question 3 states as follows:

$$\int_{0}^{2022}x^{2}-\lfloor{x}\rfloor\lceil{x}\rceil dx$$

I know that the answer is $674$, but I do not know the process and the steps to derive this solution. Can someone please help?

Welcome to MSE. You're expected to share your work on the problem. Here's how to ask a good question. Follow this guidelines to get help in this forum. — jjagmath, May 23 '22 at 05:02
Hint: write $x=i+f$ where $i$ is an integer. Then $\lfloor x \rfloor \lceil x \rceil=?$ Apply this to each interval in the range. — Ross Millikan, May 23 '22 at 05:03
@RossMillikan It's my understanding that OP is asking how to do this sort of integration as a general rule. Hence this should be made the basis of an answer. Otherwise MSE has completely lost its value. — Prime Mover, May 23 '22 at 05:28
@Prime, I seriously doubt that the value of m.se as a whole depends crucially on Ross Millikan's response to this particular question. But you do have a flair for the dramatic. — Gerry Myerson, May 23 '22 at 05:34
@PrimeMover: we have made a major move to not answering questions of this sort where OP makes no effort. — Ross Millikan, May 23 '22 at 13:37

score 7 · Answer 1 · edited May 23 '22 at 09:08

To keep things simple, let's limit the integration to the interval $(n, n+1)$. where $n \in \mathbb{Z}$.

$$\int_n^{n+1} x^2 - \lfloor x \rfloor \lceil x \rceil dx $$

$$= \int_n^{n+1} x^2 - n (n + 1) dx $$

$$= \left(\frac{1}{3}x^3 - n (n + 1)x\right)\Bigg|_{n}^{n+1} $$

$$= \left(\frac{1}{3}(n+1)^3 - n (n + 1)^2\right) - \left(\frac{1}{3}n^3 - n^2 (n + 1)\right)$$

$$= \frac{1}{3}n^3 + n^2 + n + \frac{1}{3} - n^3 - 2n^2 - n - \frac{1}{3}n^3 + n^3 + n^2 $$

$$= \frac{1}{3} $$

Gee, sure is convenient that all of those $n$'s cancel out and we're left with a constant.

So all we have to do is add up the integral value $\frac{1}{3}$ over $2022$ of these $1$-unit intervals, to get $2022 / 3 = 674.$

score 3 · Answer 2 · answered May 23 '22 at 05:39

$$\begin{aligned} \int^{2022}_0 x^2 - \lfloor x \rfloor \lceil x \rceil \mathrm{d} x &= \sum_{i=0}^{2021}\int_{i}^{i+1} x^2 - \lfloor x \rfloor \lceil x \rceil \mathrm{d} x \\ &= \sum_{i=0}^{2021}\int_{i}^{i+1} x^2 - i (i+1) \mathrm{d} x \\ &= \int_{0}^{2022} x^2 \mathrm{d} x - \sum_{i=0}^{2021}i (i+1). \end{aligned}$$ From here, one just have to compute the integral on the left, and the sum on the right (see Amos's solution for the numerical computation)! The idea is to note that for $i < x < i + 1$, $\lfloor x \rfloor = i$ and for $i < x < i + 1$, $\lceil x \rceil = i+1$. If you are worried about the contribution at $x = i$ and $x = i+1$, note that the value of an integral does not change if it differs at finitely many points (in which for this case, we are not concerned with how $\lfloor x \rfloor \lceil x \rceil$ behaves at the boundary points $x = i$ and $x = i+1$)

Can we do better?

Note that $x^2$ looks just like $\lfloor x \rfloor \lceil x \rceil$, and looks like the discrete approximation of $x^2$. In fact, this intuition can be made rigorous from the previous computation, in which $$\begin{aligned} \int^{2022}_0 x^2 - \lfloor x \rfloor \lceil x \rceil \mathrm{d} x &= \sum_{i=0}^{2021}\int_{i}^{i+1} x^2 - i (i+1) \mathrm{d} x \\ &= \sum_{i=0}^{2021} \left( \frac{(i+1)^3}{3} - \frac{i^3}{3} - i(i+1)(i+1-i)\right) \\ &= \sum_{i=0}^{2021} \frac{1}{3} \\ &= \frac{2022}{3} = \boxed{674}. \\ \end{aligned}$$

You don't have to memorize the formula for infinite sums if you do it this way!

score 2 · Answer 3 · edited May 23 '22 at 09:30

Separate the integral into two halves: $$\int_0^{2022} x^2 dx - \int_0^{2022}\lfloor x \rfloor \lceil x \rceil dx = \frac{2022^3}{3} - \int_0^{2022}\lfloor x \rfloor \lceil x \rceil dx$$

For the second integral, take partition $P={(0,1,2,\cdots,2022)}$ and form lower and upper Darboux sums: $$L(P,f)=U(P,f)=0\cdot 1+1\cdot2+2\cdot3+\ldots+2021\cdot2022=\sum_{x=1}^{2022}x(x-1)$$

which we can compute using formulae for sums of powers of integers if you'd like, or recognizing the summand as $2{x \choose 2}$ we can use the hockey stick identity:

$$\sum_{x = 1}^{2022} 2{x \choose 2} = 2{2023 \choose 3} = \frac{2023 \cdot 2022 \cdot 2021}{3} = \frac{2022^3 - 2022}{3}$$

Subtracting this from our first integral gives

$$\frac{2022^3}{3} - \frac{2022^3 - 2022}{3} = \frac{2022}3 = \boxed{674}$$

That's exactly how I would do it. I wanted to add some intuition here for why we approach it like this - the reason is that we are trying to exploit the property of floor and ceiling functions that, within a certain interval of type $(a,a+1)$, where $a$ is a natural number, we know exactly what the floor and ceil return - $a$ and $a+1$, respectively. This is the motivation for computing the upper and lower Darboux sums as @Amos C N does. — algebroo, May 23 '22 at 05:34

Showing $\int_{0}^{2022}x^{2}-\lfloor{x}\rfloor\lceil{x}\rceil dx = 674$

3 Answers3