SMALLINT \nolimits_0^{{\rm{nT}}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = {\rm{n}}\mathop \smallint \nolimits_0^{\rm{T}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}},{\rm{\;n}} \in {\rm{Z}}\)\(\mathop \smallint \nolimits_{\rm{a}}^{{\rm{a}} + {\rm{\;nT}}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = {\rm{n}}\mathop \smallint \nolimits_0^{\rm{T}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}},{\rm{\;n}} \in {\rm{Z}},{\rm{\;a}} \in {\rm{R}}\)2. Fractional PART of x: This is the difference between x and its greatest integer part, [x]The fractional part of x is given by: {x} = x – [x]{x} = x for 0 ≤ x < 1Period of the fractional part of x is one.Graph of the fractional part of x:Calculation:We have to find the VALUE of \(\mathop \smallint \limits_{ - 2}^2 {\rm{x\;dx}} - \mathop \smallint \limits_{ - 2}^2 \left[ {\rm{x}} \right]{\rm{dx}}\)\({\rm{LET\;I\;}} = \mathop \smallint \limits_{ - 2}^2 {\rm{x\;dx}} - \mathop \smallint \limits_{ - 2}^2 \left[ {\rm{x}} \right]{\rm{dx}} = {\rm{\;}}\mathop \smallint \limits_{ - 2}^2 {\rm{x}} - {\rm{\;}}\left[ {\rm{x}} \right]{\rm{\;dx}}\)\(\Rightarrow {\rm{I}} = \mathop \smallint \limits_{ - 2}^2 \left\{ {\rm{x}} \right\}{\rm{\;dx}}\) (∵ x – [x] = {x})As we know that period of {x} is one.\(\Rightarrow {\rm{I}} = \mathop \smallint \limits_{ - 2}^2 \left\{ {\rm{x}} \right\}{\rm{\;dx}} = 4 \times \mathop \smallint \limits_0^1 {\rm{x\;dx}} = 4 \times {\rm{\;}}\left[ {\frac{{{{\rm{x}}^2}}}{2}} \right]_0^1 = {\rm{\;}}4 \times \left( {\frac{1}{2} - 0} \right) = 2\)