Let $f(x) = x - [x]$ for every real number $x$, where $[x]$ is integral part of $x$. then $\int\limits_{-1}^{1} f(x)\, dx$ is

Given $f(x) = x - [x]$

The integral is $\int_{-1}^{1} (x - [x]) \, dx$

This can be separated as $\int_{-1}^{1} x \, dx - \int_{-1}^{1} [x] \, dx$

Since $x$ is an odd function, $\int_{-1}^{1} x \, dx = 0$

For the second integral, we split the limits at the integer value $0$:

$\int_{-1}^{1} [x] \, dx = \int_{-1}^{0} [x] \, dx + \int_{0}^{1} [x] \, dx$

In the interval $(-1, 0)$, $[x] = -1$ and in the interval $(0, 1)$, $[x] = 0$

$\int_{-1}^{1} [x] \, dx = \int_{-1}^{0} (-1) \, dx + \int_{0}^{1} 0 \, dx$

$= -1[x]_{-1}^{0} + 0$

$= -1(0 - (-1)) = -1$

Substituting these back, we get:

$\int_{-1}^{1} (x - [x]) \, dx = 0 - (-1) = 1$

Answer: $1$