# Integration - fraction part of square root of x

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.

• It is not a periodic function. Divide the range of integration into subintervals where $[\sqrt{x}]$ is constant
– QED
Dec 1, 2017 at 7:11

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}

• Much better than my answer. Dec 1, 2017 at 7:15
• Yours is more detailed. easier to understand perhaps. Dec 1, 2017 at 7:15
• I deal with the floor function rather than fractional function. Notice that if $x \in [(i-1)^2, i^2)$, then $[\sqrt{x}] = i-1$. Dec 1, 2017 at 7:32
• got it! Thanks! Dec 1, 2017 at 7:37
• And you can get a closed form for a general perfect square upper bound by using en.wikipedia.org/wiki/Faulhaber%27s_formula Dec 1, 2017 at 8:45

$$\{\sqrt x\}=\sqrt x - [\sqrt x]$$

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 $$\int_0^{100}\{\sqrt x\}=\int_0^{100}\sqrt xdx-\left(\int_1^41dx+\int_4^92dx+...+\int_{81}^{100}9dx \right)$$