Question

How do you find the integral of $\left| x \right|dx$ on the interval $\left[ -2,1 \right]$?

Answer 1

We start solving the problem by equating the variable to the given definite integral. We then recall the properties of modulus function as \left| x \right|=\left\\{ \begin{matrix} x,\text{ for }x > 0 \\\ -x,\text{ for }x < 0 \\\ 0,\text{ for }x=0 \\\ \end{matrix} \right.. We then apply this property between different values of x in the given definite integral to proceed through the problem. We then make use of the facts that $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+C$, $\int\limits_{a}^{b}{{{f}^{'}}\left( x \right)dx}=\left[ f\left( x \right) \right]_{a}^{b}=f\left( b \right)-f\left( a \right)$ to proceed further through the problem. We then make the necessary calculations to get the required value of the given definite integral.

Complete step-by-step answer:
According to the problem, we are asked to find the integral of $\left| x \right|dx$ on the interval $\left[ -2,1 \right]$.
Let us assume $I=\int\limits_{-2}^{1}{\left| x \right|dx}$ ---(1).
We know that modulus function is defined as \left| x \right|=\left\\{ \begin{matrix} x,\text{ for }x > 0 \\\ -x,\text{ for }x < 0 \\\ 0,\text{ for }x=0 \\\ \end{matrix} \right.. Let us use this property in equation (1).
$\Rightarrow I=\int\limits_{-2}^{0}{\left( -x \right)dx}+\int\limits_{0}^{1}{\left( x \right)dx}$.
$\Rightarrow I=-\int\limits_{-2}^{0}{xdx}+\int\limits_{0}^{1}{xdx}$ ---(2).
We know that $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+C$, $\int\limits_{a}^{b}{{{f}^{'}}\left( x \right)dx}=\left[ f\left( x \right) \right]_{a}^{b}=f\left( b \right)-f\left( a \right)$. Let us use these results in equation (2).
$\Rightarrow I=-\left[ \dfrac{{{x}^{2}}}{2} \right]_{-2}^{0}+\left[ \dfrac{{{x}^{2}}}{2} \right]_{0}^{1}$.
$\Rightarrow I=-\left( \left( \dfrac{{{0}^{2}}}{2} \right)-\left( \dfrac{{{\left( -2 \right)}^{2}}}{2} \right) \right)+\left( \left( \dfrac{{{1}^{2}}}{2} \right)-\left( \dfrac{{{0}^{2}}}{2} \right) \right)$.
$\Rightarrow I=-\left( \left( \dfrac{0}{2} \right)-\left( \dfrac{4}{2} \right) \right)+\left( \left( \dfrac{1}{2} \right)-\left( \dfrac{0}{2} \right) \right)$.
$\Rightarrow I=-\left( 0-2 \right)+\left( 1-0 \right)$.
$\Rightarrow I=-\left( -2 \right)+\left( 1 \right)$.
$\Rightarrow I=2+1$.
$\Rightarrow I=3$.
So, we have found the value of the integral of $\left| x \right|dx$ on the interval $\left[ -2,1 \right]$ as 3.
$\therefore$ The value of the integral of $\left| x \right|dx$ on the interval $\left[ -2,1 \right]$ is 3.

Note: We should perform each step carefully in order to avoid confusion and calculation mistakes while solving this problem. We can also solve this problem by plotting the given function and then finding the area by using the formula of area of the triangle. Whenever we get this type of problem, we divide the interval into two or more sub-intervals to use the properties of modulus, greatest integer functions. Similarly, we can expect problems to find the value of the integral of $\left[ 3x \right]dx$ on the interval $\left[ -3,0 \right]$, where $\left[ . \right]$ is the Greatest Integer Function.