Question

Evaluate the definite integral given as $$\underset{1}{\overset{3}{\int }}[|x-1|+|x-2|+|x-3|]dx$$.

Answer 1

Hint: We start solving the problems by recalling the properties of modulus function and we write functions for $|x-1|$, $|x-2|$ and $|x-3|$ in the intervals that were present as limits for definite integral. We divide the definite integral into two or more parts based on the modulus functions obtained. We substitute the functions obtained in the place of modulus functions in integral. Now, we do the integration and substitute the limits to get the required value.

Complete step-by-step solution:
According to the problem, we need to find the value of the definite integral $$\underset{1}{\overset{3}{\int }}[|x-1|+|x-2|+|x-3|]dx$$ ------(1).
Let us first learn about the properties of modulus function. We know that the value of $|x-a|$ varies as $$|x-a|=\{\begin{array}{c}(x-a)\text{, for x> a}\\ 0\text{, for x=a}\\ -(x-a)\text{, for x< a}\end{array}$$.
Using the properties of $|x-a|$, we check the variations of functions $|x-1|$, $|x-2|$ and $|x-3|$ in the intervals that were given in integration.
From the limits of the integral, we can see that the interval of x lies in between 1 and 3 i.e., $1<x<3$.
We find the variation in function for $|x-1|$ in the interval $1<x<3$.
So, the function $|x-1|$ is written as $$|x-1|=\{(x-1)\text{, for }1<x<3$$ ---------(2).
Now, we find the variation in function for $|x-2|$ in the interval $1<x<3$.
So, the function $|x-2|$ is written as $$|x-2|=\{\begin{array}{c}-(x-2),\text{ for }1<x<2\\ (x-2),\text{ for }2<x<3\end{array}$$ ------(3).
Now, we find the variation in function for $|x-3|$ in the interval $1<x<3$.
So, the function $|x-3|$ is written as $$|x-3|=\{-(x-3),\text{ for }1<x<3$$ ---(4).
We know that for $a< b Now, we divide the given definite integral into two intervals as shown below:
$$\underset{1}{\overset{3}{\int }}[|x-1|+|x-2|+|x-3|]dx=\underset{1}{\overset{2}{\int }}[|x-1|+|x-2|+|x-3|]dx+\underset{2}{\overset{3}{\int }}[|x-1|+|x-2|+|x-3|]dx$$.
Now, we use the functions that we have obtained in equation (2), (3), and (4) in the given definite integral.
$$\Rightarrow \underset{1}{\overset{3}{\int }}[|x-1|+|x-2|+|x-3|]dx=\underset{1}{\overset{2}{\int }}[(x-1)-(x-2)-(x-3)]dx+\underset{2}{\overset{3}{\int }}[(x-1)+(x-2)-(x-3)]dx$$.
$$\Rightarrow \underset{1}{\overset{3}{\int }}[|x-1|+|x-2|+|x-3|]dx=\underset{1}{\overset{2}{\int }}[x-1-x+2-x+3]dx+\underset{2}{\overset{3}{\int }}[x-1+x-2-x+3]dx$$.
$$\Rightarrow \underset{1}{\overset{3}{\int }}[|x-1|+|x-2|+|x-3|]dx=\underset{1}{\overset{2}{\int }}[-x+4]dx+\underset{2}{\overset{3}{\int }}\left[x\right]dx$$ --------(5).
We know that $\int {(a-x)}^{n}dx={\displaystyle \frac{-{(a-x)}^{n+1}}{n+1}}+c$, $\int x^{n}dx={\displaystyle \frac{x^{n+1}}{n+1}}+C$ and $\underset{a}{\overset{b}{\int }}{f}^{{}^{\prime }}\left(x\right)dx=f\left(b\right)-f\left(a\right)$. We use all these results in equation (5).
$$\Rightarrow \underset{1}{\overset{3}{\int }}[|x-1|+|x-2|+|x-3|]dx={{\displaystyle \frac{-{(4-x)}^2}{2}}|}_1^2+{{\displaystyle \frac{x^2}{2}}|}_2^3$$.
$$\Rightarrow \underset{1}{\overset{3}{\int }}[|x-1|+|x-2|+|x-3|]dx=\left({\displaystyle \frac{-{(4-2)}^2}{2}}\right)-\left({\displaystyle \frac{-{(4-1)}^2}{2}}\right)+\left({\displaystyle \frac{3^2}{2}}\right)-\left({\displaystyle \frac{2^2}{2}}\right)$$.
$$\Rightarrow \underset{1}{\overset{3}{\int }}[|x-1|+|x-2|+|x-3|]dx=\left({\displaystyle \frac{-4}{2}}\right)+\left({\displaystyle \frac{9}{2}}\right)+\left({\displaystyle \frac{9}{2}}\right)-\left({\displaystyle \frac{4}{2}}\right)$$.
$$\Rightarrow \underset{1}{\overset{3}{\int }}[|x-1|+|x-2|+|x-3|]dx=\left({\displaystyle \frac{-4+9+9-4}{2}}\right)$$.
$$\Rightarrow \underset{1}{\overset{3}{\int }}[|x-1|+|x-2|+|x-3|]dx=\left({\displaystyle \frac{10}{2}}\right)$$.
$$\Rightarrow \underset{1}{\overset{3}{\int }}[|x-1|+|x-2|+|x-3|]dx=5$$.
$\therefore$We have got the value of definite integral $$\underset{1}{\overset{3}{\int }}[|x-1|+|x-2|+|x-3|]dx$$ as 5.

Note: We should not take $\left[\right]$ as greatest integer function unless it is mentioned in the problem. We should not integrate taking limits directly from 1 to 3 as the functions are changing. We should not make mistakes or confuse with the sign changes that were present while integrating.