integration of floor function

Question

How to Integrate the following? $$\int_{2}^{343}\{x-\lfloor{x}\rfloor\}^2dx = \int_{2}^{3}x^2dx + \int_{3}^{4}x^2dx+\cdots+\int_{342}^{343}x^2dx$$

This is what I did, How can I proceed further with this? Any help is appreciated.

Answer 1

The idea of separating the integral in integer intervalls was correct.

You have

$$ \int_n^{n+1} \left(x-\lceil x \rceil \right)^2dx=\int_n^{n+1} (x-n)^2dx=\frac{1}{3} $$ so $$ I=\sum_{n=2}^{342} \frac{1}3=\frac{341}3 $$

Edit:

As noted in the comments of the OP, as the fractional part function is periodic with period 1 would have been enought to calculate $$ \int_0^1 x^2 dx $$