QUESTION

# Solve: $I = \int\limits_{ - 1}^2 {|{x^3} - x|dx}$.

Hint: Here, we need to find the sign of the integral as the modulus is present and then we need to apply the required integration formulae and at last we need to substitute the integral values in the place of x to find the value of integral.

For $x = - 1$ to $2$ ${x^3} > x$
Therefore, $|{x^3} - x| = {x^3} - x$
$\Rightarrow \int\limits_{ - 1}^2 {{x^3} - x}$
$\Rightarrow {\left[ {\dfrac{{{x^4}}}{4} - \dfrac{{{x^2}}}{2}} \right]^2}_{ - 1}\left[ {\because \int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} } \right]$
$\Rightarrow \left[ {\dfrac{{{2^4}}}{4} - \dfrac{{{2^2}}}{2}} \right] - \left[ {\dfrac{{{1^4}}}{4} - \dfrac{{{1^2}}}{2}} \right]$
$\Rightarrow 2 + \dfrac{1}{4} = \dfrac{9}{4}$
Hence the value of $I = \int\limits_{ - 1}^2 {|{x^3} - x|dx}$ is $\dfrac{9}{4}$.