$\int_{0}^{100}\left\{ \sqrt{x}\right\} \,dx$

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.

  • $\begingroup$ It is not a periodic function. Divide the range of integration into subintervals where $[\sqrt{x}]$ is constant $\endgroup$ – Abishanka Saha Dec 1 '17 at 7:11


\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}

  • $\begingroup$ Much better than my answer. $\endgroup$ – Piyush Divyanakar Dec 1 '17 at 7:15
  • $\begingroup$ Yours is more detailed. easier to understand perhaps. $\endgroup$ – Siong Thye Goh Dec 1 '17 at 7:15
  • $\begingroup$ I deal with the floor function rather than fractional function. Notice that if $x \in [(i-1)^2, i^2)$, then $[\sqrt{x}] = i-1$. $\endgroup$ – Siong Thye Goh Dec 1 '17 at 7:32
  • $\begingroup$ got it! Thanks! $\endgroup$ – Mohan Sharma Dec 1 '17 at 7:37
  • $\begingroup$ And you can get a closed form for a general perfect square upper bound by using en.wikipedia.org/wiki/Faulhaber%27s_formula $\endgroup$ – orion Dec 1 '17 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 $$\int_0^{100}\{\sqrt x\}=\int_0^{100}\sqrt xdx-\left(\int_1^41dx+\int_4^92dx+...+\int_{81}^{100}9dx \right)$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.