Answer

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Hint: Pay attention to breaking the modulus, according to the interval given to us, which is (-1,2), we need to break the interval according to the values of $x$ where the expression inside the modulus changes sign. The interval where it gives a negative value will have the expression integrated only after multiplication with -1 otherwise it will be integrated as it is, on opening the modulus. Thus, first find the values of $x$ where the expression changes sign, and then break the interval, and integrate separately the functions, after breaking the modulus.

This question is a bit different from usual questions of definite integration. First, let’s see what the expression inside the integral will become on opening the modulus.

First case, let’s assume that we open the modulus using a minus sign, or the value that we get on evaluating ${{x}^{3}}-x$ is negative. Only in that case, would we multiply ${{x}^{3}}-x$ with $-1$, so that we get a positive value on evaluating the modulus.

Let’s find the limiting case first, or find the boundary, on the left hand side of which the expression ${{x}^{3}}-x$ will be negative, and on the right side of which, the expression ${{x}^{3}}-x$ will be positive.

To do so, let’s find the value of $x$ at which the expression evaluates to a negative value.

$\begin{align}

& \Rightarrow {{x}^{3}}-x<0 \\

& \Rightarrow x({{x}^{2}}-1)<0 \\

& \Rightarrow (x-0)(x-1)(x+1)<0 \\

& \Rightarrow x\in (-\infty ,-1)\cup (0,1) \\

\end{align}$

Therefore for every value of $x$ less than $-1$, or between 0 and 1, ${{x}^{3}}-x$ will be negative, and we can open the modulus with a negative sign.

This also automatically means that at all the remaining values of $x$, the expression ${{x}^{3}}-x$ will evaluate to positive, and hence, we can open the modulus with a plus or a positive sign.

Next step, let’s think of the upper and lower limits of the integral.

The lower limit of the integral is -1, and the upper limit is 2.

However, to evaluate the integral we have to take it out of the modulus first.

We know that in the interval (-1,0), the expression ${{x}^{3}}-x$ will be positive, in the interval (0,1), the expression ${{x}^{3}}-x$ will be negative, and in the interval (1,2), the expression ${{x}^{3}}-x$ will be positive.

Thus, we can break the given integral into three different definite integrals, and our answer would be their sum, due to the property that :

\[\int\limits_{a}^{c}{f(x)dx=}\int\limits_{a}^{b}{f(x)dx+\int\limits_{b}^{c}{f(x)dx}}\] provided that $b$ lies between $a$ and $c$.

Thus, breaking the above integral we get :

$\int\limits_{-1}^{2}{|{{x}^{3}}-x|dx=\int\limits_{-1}^{0}{|{{x}^{3}}-x|dx+\int\limits_{0}^{1}{|{{x}^{3}}-x|dx+\int\limits_{1}^{2}{|{{x}^{3}}-x|dx}}}}$

Now, let’s open the modulus for each interval, according to the value we get for the expression ${{x}^{3}}-x$ in each interval.

Doing so, we get :

$\begin{align}

& \int\limits_{-1}^{2}{|{{x}^{3}}-x|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}}}} \\

& =\int\limits_{-1}^{0}{({{x}^{3}}-x)dx+\int\limits_{0}^{1}{-({{x}^{3}}-x)dx+\int\limits_{1}^{2}{({{x}^{3}}-x)dx}}} \\

& =\int\limits_{-1}^{0}{({{x}^{3}}-x)dx+\int\limits_{0}^{1}{(x-{{x}^{3}})dx+\int\limits_{1}^{2}{({{x}^{3}}-x)dx}}} \\

\end{align}$

Now, integrating the expressions we get :

\[\begin{align}

& =(\dfrac{{{x}^{4}}}{4}-{{\dfrac{x}{2}}^{2}})_{-1}^{0}+({{\dfrac{x}{2}}^{2}}-\dfrac{{{x}^{4}}}{4})_{0}^{1}+(\dfrac{{{x}^{4}}}{4}-{{\dfrac{x}{2}}^{2}})_{1}^{2} \\

& =[(0-0)-(\dfrac{1}{4}-\dfrac{1}{2})]+[(\dfrac{1}{2}-\dfrac{1}{4})-(0-0)]+[(\dfrac{16}{4}-\dfrac{4}{2})-(\dfrac{1}{4}-\dfrac{1}{2})] \\

& =\dfrac{1}{4}+\dfrac{1}{4}+(4-2)+\dfrac{1}{4} \\

& =\dfrac{3}{4}+2 \\

& =\dfrac{11}{4} \\

\end{align}\]

Hence, the integral evaluates to $\dfrac{11}{4}$ is the answer.

Note: It is very important to analyse the intervals the integral is told to be solved between. In case of modulus functions, they are always opened with either a plus or a minus sign, depending what the expression within the modulus evaluates to. This is because a modulus always returns the absolute value, so even if the answer we get on evaluating the expression is negative, the modulus will give us its absolute, or positive value only. Therefore, when the expression gives a positive value, we don’t need to multiply the expression with a -1, since on opening the modulus, we’ll get the original value of the expression only. But, if the expression gives a negative value, then multiplying the value with -1 will give us a positive value, or the actual value we’ll get on opening the modulus.

This question is a bit different from usual questions of definite integration. First, let’s see what the expression inside the integral will become on opening the modulus.

First case, let’s assume that we open the modulus using a minus sign, or the value that we get on evaluating ${{x}^{3}}-x$ is negative. Only in that case, would we multiply ${{x}^{3}}-x$ with $-1$, so that we get a positive value on evaluating the modulus.

Let’s find the limiting case first, or find the boundary, on the left hand side of which the expression ${{x}^{3}}-x$ will be negative, and on the right side of which, the expression ${{x}^{3}}-x$ will be positive.

To do so, let’s find the value of $x$ at which the expression evaluates to a negative value.

$\begin{align}

& \Rightarrow {{x}^{3}}-x<0 \\

& \Rightarrow x({{x}^{2}}-1)<0 \\

& \Rightarrow (x-0)(x-1)(x+1)<0 \\

& \Rightarrow x\in (-\infty ,-1)\cup (0,1) \\

\end{align}$

Therefore for every value of $x$ less than $-1$, or between 0 and 1, ${{x}^{3}}-x$ will be negative, and we can open the modulus with a negative sign.

This also automatically means that at all the remaining values of $x$, the expression ${{x}^{3}}-x$ will evaluate to positive, and hence, we can open the modulus with a plus or a positive sign.

Next step, let’s think of the upper and lower limits of the integral.

The lower limit of the integral is -1, and the upper limit is 2.

However, to evaluate the integral we have to take it out of the modulus first.

We know that in the interval (-1,0), the expression ${{x}^{3}}-x$ will be positive, in the interval (0,1), the expression ${{x}^{3}}-x$ will be negative, and in the interval (1,2), the expression ${{x}^{3}}-x$ will be positive.

Thus, we can break the given integral into three different definite integrals, and our answer would be their sum, due to the property that :

\[\int\limits_{a}^{c}{f(x)dx=}\int\limits_{a}^{b}{f(x)dx+\int\limits_{b}^{c}{f(x)dx}}\] provided that $b$ lies between $a$ and $c$.

Thus, breaking the above integral we get :

$\int\limits_{-1}^{2}{|{{x}^{3}}-x|dx=\int\limits_{-1}^{0}{|{{x}^{3}}-x|dx+\int\limits_{0}^{1}{|{{x}^{3}}-x|dx+\int\limits_{1}^{2}{|{{x}^{3}}-x|dx}}}}$

Now, let’s open the modulus for each interval, according to the value we get for the expression ${{x}^{3}}-x$ in each interval.

Doing so, we get :

$\begin{align}

& \int\limits_{-1}^{2}{|{{x}^{3}}-x|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}}}} \\

& =\int\limits_{-1}^{0}{({{x}^{3}}-x)dx+\int\limits_{0}^{1}{-({{x}^{3}}-x)dx+\int\limits_{1}^{2}{({{x}^{3}}-x)dx}}} \\

& =\int\limits_{-1}^{0}{({{x}^{3}}-x)dx+\int\limits_{0}^{1}{(x-{{x}^{3}})dx+\int\limits_{1}^{2}{({{x}^{3}}-x)dx}}} \\

\end{align}$

Now, integrating the expressions we get :

\[\begin{align}

& =(\dfrac{{{x}^{4}}}{4}-{{\dfrac{x}{2}}^{2}})_{-1}^{0}+({{\dfrac{x}{2}}^{2}}-\dfrac{{{x}^{4}}}{4})_{0}^{1}+(\dfrac{{{x}^{4}}}{4}-{{\dfrac{x}{2}}^{2}})_{1}^{2} \\

& =[(0-0)-(\dfrac{1}{4}-\dfrac{1}{2})]+[(\dfrac{1}{2}-\dfrac{1}{4})-(0-0)]+[(\dfrac{16}{4}-\dfrac{4}{2})-(\dfrac{1}{4}-\dfrac{1}{2})] \\

& =\dfrac{1}{4}+\dfrac{1}{4}+(4-2)+\dfrac{1}{4} \\

& =\dfrac{3}{4}+2 \\

& =\dfrac{11}{4} \\

\end{align}\]

Hence, the integral evaluates to $\dfrac{11}{4}$ is the answer.

Note: It is very important to analyse the intervals the integral is told to be solved between. In case of modulus functions, they are always opened with either a plus or a minus sign, depending what the expression within the modulus evaluates to. This is because a modulus always returns the absolute value, so even if the answer we get on evaluating the expression is negative, the modulus will give us its absolute, or positive value only. Therefore, when the expression gives a positive value, we don’t need to multiply the expression with a -1, since on opening the modulus, we’ll get the original value of the expression only. But, if the expression gives a negative value, then multiplying the value with -1 will give us a positive value, or the actual value we’ll get on opening the modulus.

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