\(\mathop \smallint \limits_{ - 1}^1 {\rm{x}}\left| {\rm{x}} \right|{\rm{dx}}\) is equal to

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  1. 0
  2. \(\frac{2}{3}\)
  3. 2
  4. -2

Answer (Detailed Solution Below)

Option 1 : 0
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Detailed Solution

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Concept:

1. \(\mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right) = \mathop \smallint \nolimits_{\rm{a}}^{\rm{c}} {\rm{f}}\left( {\rm{x}} \right) + \mathop \smallint \nolimits_{\rm{c}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right)\)

2. \({\rm{f}}\left( {\rm{x}} \right) = \left| {\rm{x}} \right| = \left\{ {\begin{array}{*{20}{c}} { - x,\;x < 0}\\ {x,\;x \ge 0} \end{array}} \right.\)

 

Calculation:

Let, \({\rm{I}} = \mathop \smallint \limits_{ - 1}^1 {\rm{x}}\left| {\rm{x}} \right|{\rm{dx}}\)

\( = \mathop \smallint \nolimits_{ - 1}^1 {\rm{x}}\left| {\rm{x}} \right|{\rm{dx}}\)

\(\because \mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right) = \mathop \smallint \nolimits_{\rm{a}}^{\rm{c}} {\rm{f}}\left( {\rm{x}} \right) + \mathop \smallint \nolimits_{\rm{c}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right)\)

\(= \mathop \smallint \nolimits_{ - 1}^0 {\rm{x}}\left| {\rm{x}} \right|{\rm{dx}} + \mathop \smallint \nolimits_0^1 {\rm{x}}\left| {\rm{x}} \right|{\rm{dx}}\) 

For, -1 to 0 |x| = -x

For, 0 to 1 |x| = x         

\( \Rightarrow \mathop \smallint \nolimits_{ - 1}^0 {\rm{x}}\left( { - {\rm{x}}} \right){\rm{dx}} + \mathop \smallint \nolimits_0^1 {\rm{x}}\left( {\rm{x}} \right){\rm{dx}}\)

\( \Rightarrow \mathop \smallint \nolimits_{ - 1}^0 - {{\rm{x}}^2}{\rm{dx}} + \mathop \smallint \nolimits_0^1 {{\rm{x}}^2}{\rm{dx}}\)

\( \Rightarrow - \left[ {\frac{{{{\rm{x}}^3}}}{3}} \right]_{ - 1}^0 + \left[ {\frac{{{{\rm{x}}^3}}}{3}} \right]_0^1\)

\( \Rightarrow - \left[ {0 - \left( { - \frac{1}{3}} \right)} \right] + \left[ {\frac{1}{3}} \right]\)

\(\Rightarrow - \frac{1}{3} + \frac{1}{3}\)

⇒ 0

Hence, option (1) is correct
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