Integration of an integral part of x?

Question

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.

Answer 1

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: $$\begin{align} \int_1^2 f(x) \; dx &= \int_1^\sqrt{2} f(x)\; dx + \int_\sqrt{2}^\sqrt{3} f(x)\; dx + \int_\sqrt{3}^2 f(x)\; dx\\ &= \int_1^\sqrt{2} 1 - 1 \; dx + \int_\sqrt{2}^\sqrt{3} 2-1\; dx + \int_\sqrt{3}^2 3- 1\; dx \end{align} $$