Question

The value of \[\int_{-2}^{2}{\min .\left\{ |x|,\left[ x \right] \right\}}dx\] , where [.] denotes the greatest integer function, is
(a) -2
(b) -1
(c) 0
(d) 1

Answer 1

Hint: Here, we will form a piecewise function of f(x) in differential intervals and then we will find the value of the given integration.

Complete step-by-step answer:
Since, we have to integrate the function from -2 to 2.
We can divide it into 2 intervals.
The first interval can be $x\in \left( -2,0 \right)$ and the second interval can be $x\in \left( 0,2 \right)$.
We know that fractional part of x, that is {x} is given as:
$\begin{align}
  & \left\{ x \right\}=x-\left[ x \right] \\
 & \left[ x \right]=x-\left\{ x \right\} \\
\end{align}$
When we are talking about positive integers, that is , when $x\in \left( 0,2 \right)$, |x| always gives a result x.
Since, $\left\{ x \right\}\in \left( 0,1 \right)$, so $x-\left\{ x \right\}$, is always less than x.
This implies that in the interval (0, 2):
  $x>x-\left\{ x \right\}$
So, $\min \left\{ |x|,\left[ x \right] \right\}$ is [x].
 Now, when we are talking about negative integers, that is when $x\in \left( -2,0 \right)$. We know that |x| is always greater than zero.
Since, all the x in the interval (-2, 0) are negative, we can write that $\left[ x \right]<0$.
So, in the interval $x\in \left( -2,0 \right)$, $\min \left\{ |x|,\left[ x \right] \right\}$ is [x].
Therefore, we can write:
\[\int_{-2}^{2}{\min .\left\{ |x|,\left[ x \right] \right\}}dx=\int_{-2}^{-1}{\left[ x \right]dx+\int_{-1}^{0}{\left[ x \right]dx+\int_{0}^{1}{\left[ x \right].dx}+\int_{1}^{2}{\left[ x \right]dx}...........\left( 1 \right)}}\]
We know that the greatest integer function of x always returns an integer which is less than or equal to x. Therefore, we have:
$\int_{-2}^{-1}{\left[ x \right]dx=\int_{-2}^{-1}{-2dx}}$
$\int_{-1}^{0}{\left[ x \right]dx=\int_{-1}^{0}{-1dx}}$
$\int_{0}^{1}{\left[ x \right]dx=\int_{0}^{1}{0dx}}$
$\int_{1}^{2}{\left[ x \right]dx=\int_{1}^{2}{1dx}}$
On substituting these values in equation (1), we get:
\[\begin{align}
  & \int_{-2}^{2}{\min .\left\{ |x|,\left[ x \right] \right\}}dx=\int_{-2}^{-1}{-2dx+\int_{-1}^{0}{-1dx+\int_{0}^{1}{0dx}+\int_{1}^{2}{1dx}}} \\
 & \Rightarrow \int_{-2}^{2}{\min .\left\{ |x|,\left[ x \right] \right\}}dx=-2\int_{-2}^{-1}{dx+-1\int_{-1}^{0}{dx+0+\int_{1}^{2}{dx}}} \\
 & \Rightarrow \int_{-2}^{2}{\min .\left\{ |x|,\left[ x \right] \right\}}dx=-2\left[ x \right]_{-2}^{-1}-1\left[ x \right]_{-1}^{0}+\left[ x \right]_{1}^{2} \\
 & \Rightarrow \int_{-2}^{2}{\min .\left\{ |x|,\left[ x \right] \right\}}dx=-2\left\{ -1-\left( -2 \right) \right\}-1\left\{ 0-\left( -1 \right) \right\}+\left( 2-1 \right) \\
 & \Rightarrow \int_{-2}^{2}{\min .\left\{ |x|,\left[ x \right] \right\}}dx=-2\left( -1+2 \right)-1\left( 0+1 \right)+1 \\
 & \Rightarrow \int_{-2}^{2}{\min .\left\{ |x|,\left[ x \right] \right\}}dx=-2\times 1-1+1 \\
 & \Rightarrow \int_{-2}^{2}{\min .\left\{ |x|,\left[ x \right] \right\}}dx=-2 \\
\end{align}\]
So, the value of given integration is -2.
Hence, option (a) is the correct answer.

Note: Students should note here that it is necessary to form piecewise functions here because the function behaves differently in different intervals. It is important to remember that $\left[ x \right]=x-\left\{ x \right\}$ and $\left\{ x \right\}\in \left( 0,1 \right)$.