Mathematics

# Evaluate: $\displaystyle\int^2_{-1}|x^3-x|dx$.

##### SOLUTION
$\int_{-1}^{2} |x^{3}-x|dx$
It is clear that
$x^{3} - x \geqslant 0 on [-1,0]$
$x^{3} - x \leqslant 0 on [0,1]$
$x^{3} - x \geqslant 0 on [1,2]$
Hence the interval of the integral can be subdivided as
$\int_{-1}^{2} |x^{3} - x |dx = \int_{0}^{-1} (x^{3}-x)dx + \int_{0}^{1} -(x^{3}-x)dx + \int_{1}^{2} (x^{3}-x)dx$
$= \int_{-1}^{0} (x^{3} - x ) dx + \int_{0}^{1} (x-x^{3}) dx + \int_{1}^{2}(x^{3}-x)dx$
$= [\frac{x^{4}}{4} - \frac{x^{2}}{2}]_{-1}^{0} + [\frac{x^{2}}{2}-\frac{x^{4}}{4}]_{0}^{1} + [\frac{x^{4}}{4}-\frac{x^{2}}{2}]_{1}^{2}$
$[\int x^{n} dx = \frac{x^{n+1}}{n+}]$
$= -(\frac{1}{4} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{4}) + (4-2) - (\frac{1}{4}-\frac{1}{2})$
$= \frac{-1}{4} + \frac{1}{2} + \frac{1}{2} - \frac{1}{4} + 2 - \frac{1}{4} + \frac{1}{2}$
$= \frac{3}{2} - \frac{3}{4} + 2$
$= \frac{11}{4}$ Ans

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
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