Please solve Ques no. 33

Dear Student,
Please find below the solution to the asked query:

We know that : |X| = \{\begin{cases} X & if X > 0 \\ - X & if X \leq 0 \end{cases}; (i) We have : I = \int_{- 1}^{2} |x^{3} - x| dx = \int_{- 1}^{1} |x^{3} - x| dx + \int_{1}^{2} |x^{3} - x| dx We know that : \int_{- a}^{a} f (x) dx = 2 \int_{0}^{a} f (x) dx if f (x) is an even function . [f (- x) = f (x)] As f (x) = |x^{3} - x| is an even function, hence we get : I = 2 \int_{0}^{1} |x^{3} - x| dx + \int_{1}^{2} |x^{3} - x| dx Now x^{3} - x = x (x^{2} - 1) = x (x - 1) (x + 1) By method of interval . we get : x^{3} - x = x (x - 1) (x + 1) < 0 when x \in (0, 1) and x^{3} - x = x (x - 1) (x + 1) > 0 if x \geq 1 Hence using (i), we get : I = - 2 \int_{0}^{1} (x^{3} - x) dx + \int_{1}^{2} (x^{3} - x) dx = - 2 {[\frac{x^{4}}{4} - \frac{x^{2}}{2}]}_{0}^{1} + {[\frac{x^{4}}{4} - \frac{x^{2}}{2}]}_{1}^{2} = - 2 [\frac{1}{4} - \frac{1}{2} - 0] + [(\frac{16}{4} - \frac{4}{2}) - (\frac{1}{4} - \frac{1}{2})] = - 2 (- \frac{1}{4}) + [2 + \frac{1}{4}] = \frac{1}{2} + 2 + \frac{1}{4} = \frac{11}{4} = 2.75 Hence \int_{- 1}^{2} |x^{3} - x| dx = 2.75

Regards