Evaluate: 
$\int\limits_{ – 1}^2 {\left| {{x^3} – x} \right|dx}$

[CBSE 2012]

Solution

$\int\limits_{ – 1}^2 {\left| {{x^3} – x} \right|dx} = I = \int\limits_{ – 1}^1 {\left| {{x^3} – x} \right|dx} + \int\limits_1^2 {\left| {{x^3} – x} \right|dx}$

${\left| {{x^3} – x} \right|}$ is an even function,

$\int\limits_{ – 1}^1 {\left| {{x^3} – x} \right|dx} = 2\int\limits_0^1 {\left| {{x^3} – x} \right|dx}$

$I = 2\int\limits_0^1 {\left| {{x^3} – x} \right|dx} + \int\limits_1^2 {\left| {{x^3} – x} \right|dx}$

$x > {x^3}$ when x lies between 0 to 1 and ${x^3} > x$ when x lies between 1 and 2,

$I = 2\int\limits_0^1 {\left( {x – {x^3}} \right)dx} + \int\limits_1^2 {\left( {{x^3} – x} \right)dx}$

$= 2\left. {\left( {\frac{{{x^2}}}{2} – \frac{{{x^4}}}{4}} \right)} \right|_0^1 + \left. {\left( {\frac{{{x^4}}}{4} – \frac{{{x^2}}}{2}} \right)} \right|_1^2$

$= 2\left[ {\left( {\frac{1}{2} – \frac{1}{4}} \right) – 0} \right] + \left[ {\left( {\frac{{16}}{4} – \frac{4}{2}} \right) – \left( {\frac{1}{4} – \frac{1}{2}} \right)} \right]$

$= 3\left( {\frac{1}{2} – \frac{1}{4}} \right) + (4 – 2)$

$= \frac{3}{4} + 2 = \frac{{11}}{4}$