# Integration of an integral part of x?

If $f(x)={\left\lfloor x^2\right\rfloor -\left\lfloor x\right\rfloor ^2}$,where ${\left\lfloor x\right\rfloor }$ denotes the greatest integer $\le x$ then $\int_1^2 f(x)dx?$ please give some hint. thank you.

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Break the interval into regions where $1<x^2<2$, $2<x^2<3$ and $3<x^2<4$. –  Eckhard Mar 19 '13 at 17:07

On the interval $(1,2)$ you have $[x] = 1$. On the interval $(1,\sqrt2)$ you have $[x^2] = 1$ and on the interval $(\sqrt{2}, \sqrt{3})$ you have $[x^2] = 3$ and on the interval $(\sqrt{3}, 4)$ you have $[x^2] = 3$. So you can break on the integral into: \begin{align} \int_1^2 f(x) \; dx &= \int_1^\sqrt{2} f(x)\; dx + \int_\sqrt{2}^\sqrt{3} f(x)\; dx + \int_\sqrt{3}^2 f(x)\; dx\\ &= \int_1^\sqrt{2} 1 - 1 \; dx + \int_\sqrt{2}^\sqrt{3} 2-1\; dx + \int_\sqrt{3}^2 3- 1\; dx \end{align}