Evaluate $\int_{-1}^{2}\left|x^{3}-x\right| d x$

CBSE Class 12 · 2016 · Definite IntegralsLong Answer

Evaluate

\int_{- 1}^{2} x^{3} - x d x

Answer: $\frac{11}{4}$

x^{3} - x = x (x - 1) (x + 1)

changes sign at

- 1, 0, 1

. On

[- 1, 0]

it is

\geq 0

, on

[0, 1]

it is

\leq 0

, on

[1, 2]

it is

\geq 0

\int_{- 1}^{0} (x^{3} - x) d x = [\frac{x ^{4}}{4} - \frac{x ^{2}}{2}]_{- 1}^{0} = 0 - (\frac{1}{4} - \frac{1}{2}) = \frac{1}{4}

\int_{0}^{1} (x - x^{3}) d x = [\frac{x ^{2}}{2} - \frac{x ^{4}}{4}]_{0}^{1} = \frac{1}{4}

\int_{1}^{2} (x^{3} - x) d x = [\frac{x ^{4}}{4} - \frac{x ^{2}}{2}]_{1}^{2} = 2 - (- \frac{1}{4}) = \frac{9}{4}

I = \frac{1}{4} + \frac{1}{4} + \frac{9}{4} = \frac{11}{4}

CBSE Class 12 Board · 2016 · Subjective