Question

Evaluate the definite integral $\int\limits_{-1}^{2}{\left| {{x}^{3}}-x \right|}\cdot dx$

Answer 1

Hint: We can see that the integrand is not differentiable everywhere as modulus function is used here. So we need to break the interval limits of the integrand into differentiable intervals. Proceed by simple modulus arithmetic of multiplication. Then use the formula $\int_{a}^{b}{f\left( x \right)=}\int_{a}^{c}{f\left( x \right)+\int_{c}^{b}{f\left( x \right)}}$ where $f$ is a real differentiable function and $a < c < b$.

Complete step by step answer:
As we cannot integrate directly because the modulus function is not differentiable everywhere in $R$. Hence we first need to break the limits of the integrand into smaller intervals.
First using the property of modulus $\left| a\cdot b \right|=\left| a \right|\left| b \right|$,
\[\left| {{x}^{3}}-x \right|=\left| x\left( {{x}^{2}}-1 \right) \right|=\left| x\left( x-1 \right)\left( x+1 \right) \right|=\left| x \right|\left| x-1 \right|\left| x+1 \right|\]
Now we know that the t integrand is not differentiable at $x=0,x=1$ and $x=-1$. So we break the interval at $x=0,x=1$ and $x=-1$. Now the integral transforms to ,
\[ \int\limits_{ - 1}^2 {\left| {{x^3} - x} \right|} \cdot dx = \int\limits_{ - 1}^2 {\left| x \right|\left| {x - 1} \right|\left| {x + 1} \right|} \\
   = \int\limits_{ - 1}^0 {\left| x \right|\left| {x - 1} \right|\left| {x + 1} \right| + \int\limits_{ - 1}^1 {\left| x \right|\left| {x - 1} \right|\left| {x + 1} \right| + \int\limits_1^2 {\left| x \right|\left| {x - 1} \right|\left| {x + 1} \right|} } } \\
\]
The values of the modulus will be different at different intervals. So we check modulus at each of the new intervals and put proper modulus values using the definition of modulus function. We get,
\[\left| {{x}^{3}}-x \right|=\left\{ \begin{matrix}
   \left( -x \right)\left( -\left( x-1 \right) \right)\left( x+1 \right)={{x}^{3}}-x & \text{if }-1 < x< 0 \\
   x\left( -\left( x-1 \right) \right)\left( x+1 \right)=x-{{x}^{3}} & \text{if } 0 < x <1 \\
   \left( x \right)\left( x-1 \right)\left( x+1 \right)={{x}^{3}}-x & \text{if } 1 < x < 2 \\
\end{matrix} \right.\]
Substituting the modulus values in the transformed integral,
\[\begin{align}
  & \int\limits_{-1}^{2}{\left| {{x}^{3}}-x \right|}\cdot dx \\
 & =\int\limits_{-1}^{0}{\left| x \right|\left| x-1 \right|\left| x+1 \right|+\int\limits_{0}^{1}{\left| x \right|\left| x-1 \right|\left| x+1 \right|+\int\limits_{1}^{2}{\left| x \right|\left| x-1 \right|\left| x+1 \right|}}} \\
 & =\int\limits_{-1}^{0}{{{x}^{3}}-x}+\int\limits_{0}^{1}{x-{{x}^{3}}}+\int\limits_{1}^{2}{{{x}^{3}}-x} \\
\end{align}\]
Using the formula $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}+c$ for some real constant $c$ and assigning the integral limits accordingly we get,
.\[
   = \left[ {\dfrac{{{x^4}}}{4}} \right]_{ - 1}^0 - \left[ {\dfrac{{{x^2}}}{2}} \right]_{ - 1}^0 + \left[ {\dfrac{{{x^2}}}{2}} \right]_0^1 - \left[ {\dfrac{{{x^4}}}{4}} \right]_0^1 + \left[ {\dfrac{{{x^4}}}{4}} \right]_1^2 - \left[ {\dfrac{{{x^2}}}{2}} \right]_1^2 \\
   = - \dfrac{1}{4} + \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{4} + \dfrac{{15}}{4} - \dfrac{3}{2} = \dfrac{{11}}{4} \\
\]

Note: The question tests two concepts: one the definition of modulus function and splitting of integral limit while calculating definite integral. We need to be careful of positive and negative signs while calculating the modulus value. In the case of definite integral, the mistakes occur at the placement and computation of value during substitution. That is another point to take note of.