What is \(\mathop \smallint \limits_{ - 2}^2 {\rm{x\;dx}} - \mathop \smallint \limits_{ - 2}^2 \left[ {\rm{x}} \right]{\rm{dx}}\) equal to, where [⋅] is the greatest integer function?

Concept:

1. If f(x) is periodic function with period T then,

• \(\mathop \smallint \nolimits_0^{{\rm{nT}}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = {\rm{n}}\mathop \smallint \nolimits_0^{\rm{T}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}},{\rm{\;n}} \in {\rm{Z}}\)
• \(\mathop \smallint \nolimits_{\rm{a}}^{{\rm{a}} + {\rm{\;nT}}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = {\rm{n}}\mathop \smallint \nolimits_0^{\rm{T}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}},{\rm{\;n}} \in {\rm{Z}},{\rm{\;a}} \in {\rm{R}}\)

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 < 1
• Period 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.

\(\Rightarrow {\rm{I}} = \mathop \smallint \limits_{ - 2}^2 \left\{ {\rm{x}} \right\}{\rm{\;dx}} = 4 \times \mathop \smallint \limits_0^1 {\rm{x\;dx}} = 4 \times {\rm{\;}}\left[ {\frac{{{{\rm{x}}^2}}}{2}} \right]_0^1 = {\rm{\;}}4 \times \left( {\frac{1}{2} - 0} \right) = 2\)