Definite integral $\int_2^{343} ({x-[x]})^2 dx $ $$ I = \int_2^{343} ({x-[x]})^2 dx $$
Actually , I m stuck in the question of definite integral , 
I know that
  $x-[x]$ is $\{x\}$ which is fractional part of function , but unable to proceed further. Please help.
 A: $I=\sum_{k=2}^{342} \int_{k}^{k+1} (x-k)^2dx=\sum_{k=2}^{342}\frac{1}{3}=\frac{341}{3}$
A: When you are struggling with an integral, it can often help to make a graph of the integrand. (WolframAlpha: graph of $x-\lfloor x \rfloor$ and graph of $(x-\lfloor x \rfloor)^2$. In this case, a graph makes it easy to see that the value of the integral is just 341 times the value of $\int_0^1{x^2}\,dx$, or $341\times \frac{1}{3}$.
To prove it, first show that the integrand $f(x) = (x - [x])^2$ satisfies $f(x+n) = f(x)$ when $n$ is an integer: $$f(x+n) = (x + n - [x + n])^2 =  (x + n - ([x] + n))^2 = (x - [x])^2 = f(x)\,.$$ Next express the integral as a sum of integrals
$$\int_2^{343}{f(x)\,dx} = \int_2^{3}{f(x)\,dx} + \int_3^{4}{f(x)\,dx} + \dots + \int_{342}^{343}{f(x)\,dx} = \sum_{n=2}^{342}{\int_n^{n+1}{f(x)\,dx}}\,.$$
In the rightmost integral, the substitution $u = x - n$ gives
$$\sum_{n=2}^{342}{\int_0^{1}{f(u+n)\,du}}\,.$$
Using the property $f(u+n) = f(u)$ derived earlier, this is
$$\sum_{n=2}^{342}{\int_0^{1}{f(u)\,du}}\,.$$
But in the interval $[0, 1]$, we have $f(u) = (u - [u])^2 = (u - 0)^2 = u^2$. So the integral is
$$\sum_{n=2}^{342}{\int_0^{1}{u^2\,du}} = \sum_{n=2}^{342}{\frac{1}{3}} = 341\times\frac{1}{3}\,.$$
