Question

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

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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, $|{x^3} - x| = {x^3} - x$
Therefore, on applying integration, we get,
$\Rightarrow \int\limits_{ - 1}^2 {{x^3} - x}$
$\Rightarrow {\left[ {\dfrac{{{x^4}}}{4} - \dfrac{{{x^2}}}{2}} \right]^2}_{ - 1}\left[ {\because \int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} } \right]$
Applying the upper limit and lower limit values in the above equation, we get
$\Rightarrow \left[ {\dfrac{{{2^4}}}{4} - \dfrac{{{2^2}}}{2}} \right] - \left[ {\dfrac{{{1^4}}}{4} - \dfrac{{{1^2}}}{2}} \right]$
$\Rightarrow 2 + \dfrac{1}{4} = \dfrac{9}{4}$
Hence the value of $I = \int\limits_{ - 1}^2 {|{x^3} - x|dx}$ is $\dfrac{9}{4}$.

Note: Make sure to find the appropriate formula to solve the given integral and need to be very careful while solving the modulus functions. At first we need to evaluate the modulus and then we have to apply the required formulae.