\int\limits_{ - 1}^2 {\left| {{x^3} - x} \right|} dx

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}

Since {\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}

Since 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}

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