How Do You Calculate the Integral of 1/Floor(x^2) from 1 to 2?

[anemone]

Evaluate $\displaystyle \int_{1}^{2} \dfrac{1}{\left\lfloor{x^2}\right\rfloor}\,dx$ where $\left\lfloor{u}\right\rfloor$ denotes the greatest integer less than or equal to $u$.

--------------------
Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to the following members for their correct solutions::)

1. Olok
2. laura123
3. MarkFL
4. Euge
5. kaliprasad

Solution from Euge:

Spoiler

Partition the interval $[1,2)$ into the subintervals $[1,\sqrt{2})$, $[\sqrt{2}, \sqrt{3})$, and $[\sqrt{3}, 2)$, in which $\lfloor x^2 \rfloor$ has values $1$, $2$, and $3$, respectively. Then

\(\displaystyle \int_1^2 \frac{dx}{\lfloor x^2 \rfloor} = \int_1^{\sqrt{2}} \frac{dx}{1} + \int_{\sqrt{2}}^{\sqrt{3}} \frac{dx}{2} + \int_{\sqrt{3}}^2 \frac{dx}{3} = \sqrt{2} - 1 + \frac{1}{2}(\sqrt{3} - \sqrt{2}) + \frac{1}{3}(2 - \sqrt{3})\)

\(\displaystyle = \frac{1}{6}(6\sqrt{2} - 6 + 3\sqrt{3} - 3\sqrt{2} + 4 - 2\sqrt{3}) = \frac{1}{6}(3\sqrt{2} + \sqrt{3} - 2).\)