Question

Evaluate : \[\int\limits_{- 2}^1 \left| x^3 - x \right|dx\] .

Answer 1

Let \[I = \int\limits_{- 2}^1 \left| x^3 - x \right|dx \text { and } f\left( x \right) = x^3 - x\]

Clearly,

\[f\left( x \right) = x^3 - x = x\left( x - 1 \right)\left( x + 1 \right)\]

The signs of f(x) for different values of x are shown in the figure below.
We observe that:

\[f\left( x \right) > 0\text { for all }x \in \left( - 1, 0 \right) \text { and } , f\left( x \right) < 0 \text { for all } x \in \left( - 2, - 1 \right) \cup \left( 0, 1 \right)\]

\[\left| f\left( x \right) \right| = \binom{f\left( x \right), x \in \left( - 1, 0 \right)}{ - f\left( x \right), x \in \left( - 2, - 1 \right) \cup \left( 0, 1 \right)}\]

\[ \Rightarrow \left| x^3 - x \right| = \binom{ x^3 - x, x \in \left( - 1, 0 \right)}{ - \left( x^3 - x \right), x \in \left( - 2, - 1 \right) \cup \left( 0, 1 \right)}\]

\[ \Rightarrow I = \int_{- 2}^{- 1} \left| x^3 - x \right|dx + \int_{- 1}^0 \left| x^3 - x \right|dx + \int_0^1 \left| x^3 - x \right|dx\]

\[\Rightarrow I = \int_{- 2}^{- 1} - \left( x^3 - x \right)dx + \int_{- 1}^0 \left( x^3 - x \right)dx + \int_0^1 - \left( x^3 - x \right)dx\]

\[ \Rightarrow I = \left[ - \frac{x^4}{4} + \frac{x^2}{2} \right]_{- 2}^{- 1} + \left[ \frac{x^4}{4} - \frac{x^2}{2} \right]_{- 1}^0 + \left[ - \frac{x^4}{4} + \frac{x^2}{2} \right]_0^1 \]

\[ \Rightarrow I = \left[ \left( - \frac{1}{4} + \frac{1}{2} \right) - \left( - \frac{16}{4} + \frac{4}{2} \right) \right] + \left[ 0 - \left( \frac{1}{4} - \frac{1}{2} \right) \right] + \left[ \left( - \frac{1}{4} + \frac{1}{2} \right) - 0 \right]\]

\[ \Rightarrow I = \left[ \frac{1}{4} + 2 \right] + \left[ 0 + \frac{1}{4} \right] + \left[ \frac{1}{4} - 0 \right]\]

\[ \Rightarrow I = \frac{11}{4}\]