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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