Floor o'er the fractional part !

Or you can also use the following trivial fact (that follows from definition of fractional part):

$\int\limits_p^{p+1} \{x\}\,\mathrm dx=\int\limits_0^1 x\,\mathrm dx\quad\forall~p\in\Bbb{Z_0^+}$

The solution to this problem can further be generalized using the above mentioned trivial identity.

Consider the following integral $\mathcal{I}_{p,q}$ where $q\geq p~\land~p,q\in\Bbb{Z_0^+}$ :

$\mathcal{I}_{p,q}=\int\limits_p^q \{x\}^{\lfloor x\rfloor}\,\mathrm dx=\int\limits_0^1\left(\sum_{k=p}^{q-1} x^k\right)\,\mathrm dx=\sum_{k=p}^{q-1}\int\limits_0^1 x^k\,\mathrm dx\\ \implies \mathcal{I}_{p,q}=\sum_{k=p}^{q-1}\left(\frac{x^{k+1}}{k+1}\bigg|_0^1\right)=\sum_{k=p}^{q-1}\frac{1}{k+1}=\sum_{k=p+1}^q\frac{1}{k}$

We can easily conclude our result as:

$\mathcal{I}_{p,q}=\begin{cases}\begin{array}{ccc}H_q-H_p&\textrm{when}&p\gt 0\\ H_q&\textrm{when}&p=0~\land~q\neq 0\end{array}\end{cases}$

where $H_x$ is the $x^{\textrm{th}}$ harmonic number, i.e., $\displaystyle H_x=\sum_{k=1}^x\frac 1k~\forall~x\in\Bbb{Z_{\geq 1}}$ .

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The answer to this problem is simply the value of $~\mathcal{I}_{0,3}=H_3=\dfrac{11}{6}=1.8\overline 3$

Cool question!! I couldn't even think of that

great....such questions can surely come in $JEE\quad$ exam..

Really, nice one!

great question .... i did it by using the graph of the given function.