Integrating Floor Functions $\int_0^{1102} \lfloor{x^{1/2}}\rfloor \mathrm{d}x$

Question

I want to find a general formula to calculate the integral of such function (x>0) Could someone kindly help me? $$\int_0^{1102} \lfloor{x^{1/2}}\rfloor \mathrm{d}x$$

Thank you!

Answer 1

Consider that \begin{align} \lfloor \sqrt{x} \rfloor &= 0, \text{ for } 0\leq x<1\\ \lfloor \sqrt{x} \rfloor &= 1, \text{ for } 1\leq x<4\\ \lfloor \sqrt{x} \rfloor &= 2, \text{ for } 4\leq x<9\\ \vdots\\ \lfloor \sqrt{x} \rfloor &= n, \text{ for } n^2\leq x<(n+1)^2. \end{align}

We have

\begin{align} \int\limits_{0}^{1102} \lfloor \sqrt{x}\rfloor dx &= \int\limits_{0^2}^{1^2} \lfloor \sqrt{x}\rfloor dx + \int\limits_{1^2}^{2^2} \lfloor \sqrt{x}\rfloor dx + \int\limits_{2^2}^{3^2} \lfloor \sqrt{x}\rfloor dx + \ldots+ \int\limits_{32^2}^{33^2} \lfloor \sqrt{x}\rfloor dx + \int\limits_{33^2}^{1102} \lfloor \sqrt{x}\rfloor dx\\ &=0\cdot (1^2-0^2)+1\cdot (2^2-1^2)+2\cdot(3^2-2^2)+\ldots+32\cdot (33^2-32^2)+ 33\cdot (1102-33^2)\\ &=\sum\limits_{n=0}^{32} [n((n+1)^2-n^2)] + 33\cdot 13\\ &= \sum\limits_{n=0}^{32} [(2n^2+n)] + 429\\ &= 2\sum\limits_{n=0}^{32} n^2 + \sum\limits_{n=0}^{32} n + 429\\ &= 2\left(\dfrac{32(32+1)(2(32)+1)}{6}\right)+ \left(\dfrac{32(32+1)}{2}\right) + 429\\ &= 2\left(11440\right)+ \left(528\right) + 429\\ &=23837. \end{align}

For general formula: (for $0<x<p$, $p>0$) \begin{align} \int\limits_{0}^{p} \lfloor \sqrt{x}\rfloor dx &=\int\limits_{0^2}^{1^2} \lfloor \sqrt{x}\rfloor dx + \int\limits_{1^2}^{2^2} \lfloor \sqrt{x}\rfloor dx + \ldots+ \int\limits_{(\lfloor \sqrt p\rfloor-1)^2}^{\lfloor \sqrt p\rfloor^2} \lfloor \sqrt{x}\rfloor dx + \int\limits_{\lfloor \sqrt p\rfloor^2}^{p} \lfloor \sqrt{x}\rfloor dx\\ &=\sum\limits_{n=0}^{\lfloor \sqrt p\rfloor-1} [n((n+1)^2-n^2)] + \lfloor \sqrt p\rfloor \cdot ( p-\lfloor \sqrt p\rfloor^2)\\ &= 2\sum\limits_{n=0}^{\lfloor \sqrt p\rfloor-1} n^2 + \sum\limits_{n=0}^{\lfloor \sqrt p\rfloor-1} n + \lfloor \sqrt p\rfloor \cdot (p-\lfloor \sqrt p\rfloor^2)\\ &= 2\left(\dfrac{(\lfloor \sqrt p\rfloor-1)(\lfloor \sqrt p\rfloor-1+1)(2(\lfloor \sqrt p\rfloor-1)+1)}{6}\right)+ \left(\dfrac{(\lfloor \sqrt p\rfloor-1)(\lfloor \sqrt p\rfloor-1+1)}{2}\right) + \lfloor \sqrt p\rfloor \cdot (p-\lfloor \sqrt p\rfloor^2)\\ &= \left(\dfrac{(\lfloor \sqrt p\rfloor-1)(\lfloor \sqrt p\rfloor)(2\lfloor \sqrt p\rfloor-1)}{3}\right)+ \left(\dfrac{(\lfloor \sqrt p\rfloor-1)(\lfloor \sqrt p\rfloor)}{2}\right) + \lfloor\sqrt p\rfloor \cdot (p-\lfloor \sqrt p\rfloor^2)\\ &= (\lfloor \sqrt p\rfloor-1)(\lfloor \sqrt p\rfloor)\left(\dfrac{(2\lfloor \sqrt p\rfloor-1)}{3}+ \dfrac{1}{2}\right) + \lfloor\sqrt p\rfloor \cdot (p-\lfloor\sqrt p\rfloor^2)\\ &= (\lfloor\sqrt p\rfloor-1)(\lfloor\sqrt p\rfloor)\left(\dfrac{(4\lfloor \sqrt p\rfloor+1)}{6}\right) + \lfloor \sqrt p\rfloor \cdot (p-\lfloor \sqrt p\rfloor^2)\\ &= \dfrac{1}{6}(\lfloor\sqrt p\rfloor-1)(\lfloor\sqrt p\rfloor)\left(4\lfloor \sqrt p\rfloor+1\right) + \lfloor \sqrt p\rfloor \cdot (p-\lfloor \sqrt p\rfloor^2)\\ \end{align}

Answer 2

The integrand is a step function, so the integral turns into a sum if you integrate over each "step" separately.

$$\int_0^{1102} \lfloor \sqrt{x} \rfloor \, dx = \sum_{n=0}^{32} \int_{n^2}^{(n+1)^2} \lfloor \sqrt{x} \rfloor \, dx + \int_{32^2}^{1102} \lfloor \sqrt{x} \rfloor \, dx$$

Note that for $n^2 \le x < (n+1)^2$ we have $\lfloor \sqrt{x} \rfloor = n$, which makes each integral simple to compute.

Answer 3

Consider using Riemann-Stieltjes integration:

$$ \begin{aligned} \int_a^b\lfloor\sqrt x\rfloor\mathrm dx &=x\lfloor\sqrt x\rfloor|_a^b-\int_a^b x\mathrm d\lfloor\sqrt x\rfloor \\ &=x\lfloor\sqrt x\rfloor|_a^b-\sum_{\sqrt a<n\le\sqrt b}n \\ &=b\lfloor\sqrt b\rfloor-a\lfloor\sqrt a\rfloor-\sum_{\sqrt a<n\le\sqrt b}n \end{aligned} $$

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\int_{0}^{1102}\left\lfloor\,{x^{1/2}}\,\right\rfloor\dd x} \,\,\,\stackrel{x^{1/2}\,\,\, \mapsto\ x}{=}\,\,\, 2\int_{0}^{\root{\vphantom{A^{A}}1102\,}}\left\lfloor\,{x}\,\right\rfloor x\,\dd x \end{align} However, $\ds{\root{1102} = 33\ +\ \overbrace{% \pars{\root{\vphantom{A^{A}}1102} - 33}} ^{\ds{\in \left[0,1\right)}}}$

Then, \begin{align} &\bbox[5px,#ffd]{\int_{0}^{1102}\left\lfloor\,{x^{1/2}}\,\right\rfloor\dd x} \\[5mm] = &\ 2\sum_{n = 0}^{32}\ \int_{n}^{n+ 1}nx\,\dd x + 2\int_{33}^{\root{\vphantom{A^{A}}1102\,}} 33x\,\dd x \\[5mm] = &\ 23408 + 429 = \bbx{23837} \\ & \end{align}

Answer 5

if we look at the square root itself, we can see that: $$\lfloor\sqrt{x}\rfloor=0,0\le x<1$$ $$\lfloor\sqrt{x}\rfloor=1,1\le x<4$$ and it continues in this pattern of: $$\lfloor\sqrt{x}\rfloor=n,n^2\le x<(n+1)^2$$ now all you need to do is find an integer such that $(n+1)^2=1102$, now since $1102$ is not a square number you will have to go up to the closer square below it, then have a final domain, which will be $\lfloor\sqrt{1102}\rfloor=33$. so we know we have to do the following sum: $$\int_0^{1102}\lfloor\sqrt{x}\rfloor dx=\sum_{n=0}^{32}\int_{n^2}^{(n+1)^2}ndx+\int_{33^2}^{1102}33dx$$ $$=\sum_{n=0}^{32}n\left[(n+1)^2-n^2\right]+33\left[1102-33^2\right]$$ $$=\sum_{n=0}^{32}n(2n+1)+429$$ now the rest should be easy :)