Question

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

Answer 1

Hint: First we have to define what the terms we need to solve the problem are.
Integral value will be in a modulus so it may be positive or negative term.
If the integral is dislocated into three parts the first and third part must be positive and the second part must be negative terms.

Complete step by step answer:
It is given that to evaluate the integral $\int\limits_{ - 1}^2 {\left| {{x^3} - x} \right|} dx$
Since it is in a modulus value it may be positive term or negative term so we need to dislocate this integral into several integrals after then we approach the integral formula;
Let $\int\limits_{ - 1}^2 {\left| {{x^3} - x} \right|} dx$in this the integral values is $ - 1$ to $2$, now breaking this integral into three parts which are $ - 1$ to $0$, $0$ to $1$ and finally $1$ to $2$
That is$\int\limits_{ - 1}^0 {{x^3} - x} dx$ , $\int\limits_0^1 {{x^3} - x} dx$ and $\int\limits_1^2 {{x^3} - x} dx$ so these are the integrals with dislocated integral part
Thus applying in formula; $\int\limits_{ - 1}^2 {\left| {{x^3} - x} \right|} dx$= $\int\limits_{ - 1}^0 {{x^3} - x} dx$$ - $($\int\limits_0^1 {{x^3} - x} dx$)$ + $$\int\limits_1^2 {{x^3} - x} dx$
Now further proceeding to $ \Rightarrow $$\int\limits_{ - 1}^0 {{x^3} - x} dx$$ - $($\int\limits_0^1 {{x^3} - x} dx$)$ + $$\int\limits_1^2 {{x^3} - x} dx$
Multiply the negative terms into the integral values we get
$ \Rightarrow $$\int\limits_{ - 1}^0 {{x^3} - x} dx$ $ + $ $\int\limits_0^1 { - {x^3} + x} dx$$ + $$\int\limits_1^2 {{x^3} - x} dx$ $(1)$
This will be integrals after we dislocated with only positive values of integrals from the general integral part with modulus may be positive or negative.
Now further proceeding with these three integrals,
Let us take the first integral value which is$\int\limits_{ - 1}^0 {{x^3} - x} dx$, now applying integral we get
$\int\limits_{ - 1}^0 {{x^3} - x} dx$=$\mathop {\left( {\dfrac{{{x^4}}}{4} - {{\dfrac{x}{2}}^2}} \right)}\nolimits_{ - 1}^0 $= $0 - \left( {{{\dfrac{{( - 1)}}{4}}^4} - {{\dfrac{{( - 1)}}{2}}^2}} \right)$= $ - (\dfrac{1}{4} - \dfrac{1}{2}) = - ( - \dfrac{1}{4}) = \dfrac{1}{4}$
This will be the integrated value of the first integral and proceeding the same for others,

 $\int\limits_0^1 { - {x^3} + x} dx$=$\mathop {\left( { - \dfrac{{{x^4}}}{4} + {{\dfrac{x}{2}}^2}} \right)}\nolimits_0^1 $= $\left( { - {{\dfrac{{( - 1)}}{4}}^4} + {{\dfrac{{( - 1)}}{2}}^2}} \right) - 0$= $( - \dfrac{1}{4} + \dfrac{1}{2}) = ( - \dfrac{1}{4} + \dfrac{2}{4}) = \dfrac{1}{4}$
Similarly for the final integral part we get
$\int\limits_1^2 {{x^3} - x} dx$= $\mathop {\left( {\dfrac{{{x^4}}}{4} - {{\dfrac{x}{2}}^2}} \right)}\nolimits_1^2 $= $\left( {{{\dfrac{2}{4}}^4} - \dfrac{{{1^4}}}{2}} \right) - \left( {{{\dfrac{2}{4}}^2} - {{\dfrac{1}{2}}^2}} \right) = \dfrac{{14}}{4} - \dfrac{1}{2} = \dfrac{12}{4}=3$
Hence from the above three dislocated integrals part we can able find the original integral of this equation,
Thus adding all three integrals we get
$\int\limits_{ - 1}^2 {\left| {{x^3} - x} \right|} dx$= $\int\limits_{ - 1}^0 {{x^3} - x} dx$$ - $($\int\limits_0^1 {{x^3} - x} dx$)$ + $$\int\limits_1^2 {{x^3} - x} dx$
                     = $\dfrac{1}{4} + \dfrac{1}{4} + 3$
                     = $\dfrac{{7}}{2}$
Hence after finding every integral and adding every values we get
$\int\limits_{ - 1}^2 {\left| {{x^3} - x} \right|} dx$= $\dfrac{{7}}{2}$

Note: If the integral is dislocated into three parts the first and third part must be positive and the second part must be negative terms.
Also for an integration first apply the upper limit and negative sign of the lower limit.