What is $\int_n^{n+1}(x - [x])\, dx$, where $[\cdot]$ is the greatest integer function and $n$ is a natural number?

The given integral is $I = \int_n^{n+1}(x - [x])\, dx$.

We know that $x - [x] = \{x\}$, which is the fractional part function.

The fractional part function is periodic with a period of $1$.

Therefore, the integral over any interval of length $1$ is equal to the integral over $[0, 1]$.

$I = \int_0^1 (x - [x])\, dx$

In the interval $[0, 1)$, $[x] = 0$.

$I = \int_0^1 x\, dx = \left[ \dfrac{x^2}{2} \right]_0^1 = \dfrac{1}{2}$

Alternatively, in the interval $[n, n+1)$, $[x] = n$.

$I = \int_n^{n+1} (x - n)\, dx = \left[ \dfrac{(x-n)^2}{2} \right]_n^{n+1} = \dfrac{1^2}{2} - 0 = \dfrac{1}{2}$

Answer: $\dfrac{1}{2}$