Answer should be 0.50 .
a
$\left [ x \right ] =$ Largest interger less than or equall to x
$\begin{bmatrix} \left [ 0 \leq x < 1 \Rightarrow [x] = 0 \right ] \\ \left [ 1 \leq x < 2 \Rightarrow [x] = 1 \right ] \\ \left [ 2 \leq x < 3 \Rightarrow [x] = 2 \right ]\end{bmatrix}$ & $\begin{bmatrix} \left [ -1 \leq x < 0 \Rightarrow [x] = -1 \right] \\ \left [ -2 \leq x < -1 \Rightarrow [x] = -2 \right ] \\ \left [ -3 \leq x < -2 \Rightarrow [x] = -3 \right ]\end{bmatrix}$
So we can plot f(x) = [x] as
Now here in $Q^n$,
f(x) = x- [x]
so I $=\int_{0.25}^{1.25} f(x)dx$
$=\int_{0.25}^{1.25} (x-[x])dx$
$=\int_{0.25}^{1.25} xdx$ $-$ $\int_{0.25}^{1.25} [x]dx$
$=\int_{0.25}^{1.25} xdx$ - $\left [ \int_{0.25}^{1} [x]dx + \int_{1}^{1.25} [x]dx \right ]$ $\begin{Bmatrix}\because \int_{a}^{c} f(x)dx = \int_{a}^{b} f(x)dx + \int_{b}^{c} f(x)dx \\ if \ a< b<c\end{Bmatrix}$
since integral of a function gives area under curve so here,
$\int_{0.25}^{1} [x]dx = 0$ {from the above graph}
and $\int_{1}^{1.25} [x]dx = 1 * (1.25 - 1) $
$\because$ in the above graph
so overall ,
I $ = \int_{0}^{1.25} xdx - [0 + 0.25]$
$=\frac{1}{2}[(1.25)^2 - (0.25)^2] - 0.25$
$=\frac{1}{2}[(1.25+ 0.25) (1.25 - 0.25)] - 0.25$
$=\frac{1}{2}*1.50*1 - 0.25$
$= 0.75 - 0.25$
$=0.50$