Correct Answer - Option 4 : 7/3
Concept:
|x| = x x ≥ 0
|x| = -x x < 0
Calculation:
Given:
\(\mathop \smallint \nolimits_{ - 1}^2 x\left| x \right|dx\)
\( = \mathop \smallint \nolimits_{ - 1}^0 - {x^2}dx + \mathop \smallint \nolimits_0^2 {x^2}dx\)
\( = \left[ {\frac{{ - {x^3}}}{3}} \right]_{ - 3}^0 + \left[ {\frac{{{x^3}}}{3}} \right]_0^2\)
\( = \frac{{ - 1}}{3} + \frac{8}{3} = \frac{7}{3}\)
