Question

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

Answer 1

${\int }_{-1}^2|x^3-x|dx$

It is clear that

$x^3-x⩾0on[-1,0]$

$x^3-x⩽0on[0,1]$

$x^3-x⩾0on[1,2]$

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