TITLE: Integration - fraction part of square root of x
QUESTION [8 upvotes]: Question
$\int_{0}^{100}\left\{ \sqrt{x}\right\} \,dx$
Where $\{.\}$ denotes the fractional part of x.
My Approach
We know that $\left[x\right]$+$\left\{ x\right\} $=x
$\Longrightarrow$$\left\{ x\right\} =x-\left[x\right]$
$\int_{0}^{100}\left\{ \sqrt{x}\right\} \,dx$ =$\int_{0}^{100}$$\sqrt{x}\,dx$
-$\int_{0}^{100}$$\left[\sqrt{x}\right]\,dx$=$\left[\frac{\sqrt{x^{3}}}{\frac{3}{2}}\right]_{0}^{100}-$$\int_{0}^{100}\left[\sqrt{x}\right]\,dx$.
I cannot solve $\int_{0}^{100}\left[\sqrt{x}\right]\,dx$. But I have
an idea if somehow I can prove that $\left[\sqrt{x}\right]$ is a periodic
function with period $p$ such that $p|100$ then $p\int_{0}^{\frac{100}{p}}\left[\sqrt{x}\right]\,dx$. That
would be easy to solve.

REPLY [16 votes]: Guide:
\begin{align}\int_0^{100} \left[\sqrt{x} \right]\, dx &= \sum_{i=1}^{10} \int_{(i-1)^2}^{i^2} \left[\sqrt{x} \right]\,dx \\
&=\sum_{i=1}^{10}(i^2-(i-1)^2)\sqrt{(i-1)^2}\\
&= \sum_{i=1}^{10} ( 2i+1) (i-1)\end{align}

REPLY [15 votes]: $$\{\sqrt x\}=\sqrt x - [\sqrt x]$$
Now see when $1 \le x < 4$. The integer part $[\sqrt x]=1$ Similarly
$$1 \le x < 4 \implies [\sqrt x]=1\\ 
4 \le x < 9 \implies [\sqrt x]=2 \\
9 \le x < 16 \implies [\sqrt x]=3 \\ ... \\
81 \le x < 100 \implies [\sqrt x]=9\\$$
So you have to do the following integral
$$\int_0^{100}\{\sqrt x\}=\int_0^{100}\sqrt xdx-\left(\int_1^41dx+\int_4^92dx+...+\int_{81}^{100}9dx \right)$$