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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals4 Marks
Question
Evaluate $\int _ { - 1 } ^ { 2 } \left| x ^ { 3 } - x \right| d x$.
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Answer

According to the question , $I =\int _ { - 1 } ^ { 2 } \left| x ^ { 3 } - x \right| d x$
We can observe that,
$\left| x ^ { 3 } - x \right| = \left\{ \begin{array} { c } { \left( x ^ { 3 } - x \right) , \text { when } - 1 < x < 0 } \\ { - \left( x ^ { 3 } - x \right) , \text { when } 0 \leq x < 1 } \\ { \left( x ^ { 3 } - x \right) , \text { when } 1 \leq x < 2 } \end{array} \right.$
By Splitting the intervals , we get  
$I =\int _ { - 1 } ^ { 0 } \left| x ^ { 3 } - x \right| d x + \int _ { 0 } ^ { 1 } \left| x ^ { 3 } - x \right| d x +\int _ { 1} ^ { 2 } \left| x ^ { 3 } - x \right| d x$
 $\therefore I = \int _ { - 1 } ^ { 0 } \left( x ^ { 3 } - x \right) d x + \int _ { 0 } ^ { 1 } - \left( x ^ { 3 } - x \right) d x + \int _ { 1 } ^ { 2 } \left( x ^ { 3 } - x \right) d x$
$= \int _ { - 1 } ^ { 0 } \left( x ^ { 3 } - x \right) d x - \int _ { 0 } ^ { 1 } \left( x ^ { 3 } - x \right) d x + \int _ { 1 } ^ { 2 } \left( x ^ { 3 } - x \right) d x$
$= \left[ \frac { x ^ { 4 } } { 4 } - \frac { x ^ { 2 } } { 2 } \right] _ { - 1 } ^ { 0 } - \left[ \frac { x ^ { 4 } } { 4 } - \frac { x ^ { 2 } } { 2 } \right] _ { 0 } ^ { 1 } + \left[ \frac { x ^ { 4 } } { 4 } - \frac { x ^ { 2 } } { 2 } \right] _ { 1 } ^ { 2 }$
$= \left[ 0 - \left( \frac { 1 } { 4 } - \frac { 1 } { 2 } \right) \right] - \left[ \left( \frac { 1 } { 4 } - \frac { 1 } { 2 } \right) - 0 \right]$$+ \left[ \left( \frac { 16 } { 4 } - \frac { 4 } { 2 } \right) - \left( \frac { 1 } { 4 } - \frac { 1 } { 2 } \right) \right]$
$= - \frac { 1 } { 4 } + \frac { 1 } { 2 } - \frac { 1 } { 4 } + \frac { 1 } { 2 } + 4 - 2 - \frac { 1 } { 4 } + \frac { 1 } { 2 }$
$= - \frac { 3 } { 4 } + \frac { 3 } { 2 } + 2$
$ = \frac { - 3 + 6 + 8 } { 4 }$
$ = \frac { 11 } { 4 }$
$\therefore I = \frac { 11 } { 4 }$ sq units.

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