Question

Find the value of $\int\limits_{ - 2}^2 {\min \left\{ {x - \left[ x \right], - x - \left[ { - x} \right]} \right\}dx} $ and choose the correct option from the given options below. (Where $\left[ . \right]$denotes the greatest integer function.)
A) $\dfrac{1}{2}$
B) $1$
C) $\dfrac{3}{2}$
D) $2$

Answer 1

Hint: To find the value of a given integral first we need to express the given function to integrate into the suitable form such that it is easy to integrate. Now we need to express the $x - \left[ x \right]$into $\left\{ x \right\}$ and then divide the given interval to integrate into the suitable intervals.

Formulas Used:
$\int\limits_{ - a}^a {f\left( x \right)dx} = 2\int\limits_o^a {f\left( x \right)dx} $ if $f\left( x \right) = f\left( { - x} \right)$,
$\int\limits_a^b {f\left( x \right)dx} = - \int\limits_b^a {f\left( x \right)dx} $,
$\int\limits_{ - a}^a {f\left( x \right)dx} = \int\limits_{ - a}^0 {f\left( x \right)dx} + \int\limits_o^a {f\left( x \right)dx} $

Complete step by step answer:
Let us take,
$f\left( x \right) = \min \left\{ {x - \left[ x \right], - x - \left[ { - x} \right]} \right\},g\left( x \right) = x - \left[ x \right]$
Observe that, $g\left( x \right) = \left\{ x \right\}$. Since $x - \left[ x \right] = \left\{ x \right\}$.
Consider,
$g\left( { - x} \right) = - x - \left[ { - x} \right] = \left\{ { - x} \right\}$.
We know that,
$$\left\{ x \right\} = x$$ in interval $\left( {0,1} \right)$, and
$$\left\{ { - x} \right\} = 1 - \left\{ x \right\}$$.
Now clearly,
$f\left( x \right) = \min \left\{ {x - \left[ x \right], - x - \left[ { - x} \right]} \right\} = \min \left\{ {\left\{ x \right\},\left\{ { - x} \right\}} \right\}$,
$ \Rightarrow f\left( x \right) = \min \left\{ {\left\{ x \right\},1 - \left\{ x \right\}} \right\}$
On further simplification, we get
$f= x, x \leqslant 0.5$
$f= 1-x, x$>$0.5$
Observe that, $f\left( x \right) = f\left( { - x} \right)$
$ \Rightarrow \int\limits_{ - 2}^2 {f\left( x \right)dx} = 2\int\limits_0^2 {f\left( x \right)dx} $
Now let's split the $\left( {0,2} \right)$into $\left( {0,1} \right) + \left( {1,2} \right)$
$ \Rightarrow \int\limits_{ - 2}^2 {f\left( x \right)dx} = 2\left\{ {\int\limits_0^1 {f\left( x \right)dx} + \int\limits_1^2 {f\left( x \right)dx} } \right\}$
$ \Rightarrow \int\limits_{ - 2}^2 {f\left( x \right)dx} = 2\left\{ {\int\limits_0^1 {\min \left\{ {\left\{ x \right\},1 - \left\{ x \right\}} \right\}dx} + \int\limits_1^2 {\min \left\{ {\left\{ x \right\},1 - \left\{ x \right\}} \right\}dx} } \right\}$
Observe, $$f\left( x \right) = f\left( {x + 1} \right)$$
$ \Rightarrow f\left( {\left( {0,1} \right)} \right) = f\left( {\left( {1,2} \right)} \right)$
$ \Rightarrow \int\limits_{ - 2}^2 {f\left( x \right)dx} = 2\left\{ {\int\limits_0^1 {\min \left\{ {\left\{ x \right\},1 - \left\{ x \right\}} \right\}dx} + \int\limits_0^1 {\min \left\{ {\left\{ x \right\},1 - \left\{ x \right\}} \right\}dx} } \right\}$
$ \Rightarrow \int\limits_{ - 2}^2 {f\left( x \right)dx} = 2\left\{ {2\left\{ {\int\limits_0^1 {\min \left\{ {\left\{ x \right\},1 - \left\{ x \right\}} \right\}dx} } \right\}} \right\}$
$ \Rightarrow \int\limits_{ - 2}^2 {f\left( x \right)dx} = 4\left\{ {\int\limits_0^1 {\min \left\{ {\left\{ x \right\},1 - \left\{ x \right\}} \right\}dx} } \right\}$
Now let’s split the interval $\left( {0,1} \right)$ to $\left( {0,0.5} \right) + \left( {0.5,1} \right)$
$ \Rightarrow \int\limits_{ - 2}^2 {f\left( x \right)dx} = 4\left\{ {\int\limits_0^{0.5} {\min \left\{ {\left\{ x \right\},1 - \left\{ x \right\}} \right\}dx} + \int\limits_{0.5}^1 {\min \left\{ {\left\{ x \right\},1 - \left\{ x \right\}} \right\}dx} } \right\}$
$ \Rightarrow \int\limits_{ - 2}^2 {f\left( x \right)dx} = 4\left\{ {\int\limits_0^{0.5} {\left\{ x \right\}dx} + \int\limits_{0.5}^1 {1 - \left\{ x \right\}dx} } \right\}$
We have, $$\left\{ x \right\} = x$$in the interval $\left( {0,1} \right)$
$ \Rightarrow \int\limits_{ - 2}^2 {f\left( x \right)dx} = 4\left\{ {\int\limits_0^{0.5} {xdx} + \int\limits_{0.5}^1 {1 - xdx} } \right\}$
$ \Rightarrow \int\limits_{ - 2}^2 {f\left( x \right)dx} = 4\left\{ {\left[ {\dfrac{{{x^2}}}{2}} \right]_0^{0.5} + \left[ {x - \dfrac{{{x^2}}}{2}} \right]_{0.5}^1} \right\}$
By applying the limits, we get
$ \Rightarrow \int\limits_{ - 2}^2 {f\left( x \right)dx} = 4\left\{ {\dfrac{{1 - \dfrac{1}{4}}}{2} + \left[ {\dfrac{1}{2} - 1 - \dfrac{{1 - \dfrac{1}{4}}}{2}} \right]} \right\}$$ \Rightarrow \int\limits_{ - 2}^2 {f\left( x \right)dx} = 4\left\{ {\dfrac{{\dfrac{1}{4} - 0}}{2} + \left[ {1 - \dfrac{1}{2} - \dfrac{{1 - \dfrac{1}{4}}}{2}} \right]} \right\}$
On further simplifications, we get
$ \Rightarrow \int\limits_{ - 2}^2 {f\left( x \right)dx} = 4\left\{ {\dfrac{1}{4}} \right\}$
$ \Rightarrow \int\limits_{ - 2}^2 {f\left( x \right)dx} = 1$
So, the value of $\int\limits_{ - 2}^2 {\min \left\{ {x - \left[ x \right], - x - \left[ { - x} \right]} \right\}dx} $ is 1.
Therefore, the correct option is (B).

Additional Information:
Proof of $\int\limits_{ - a}^a {f\left( x \right)dx} = 2\int\limits_0^a {f\left( x \right)dx} $ if $f\left( x \right) = f\left( { - x} \right)$
We have,
$\int\limits_{ - a}^a {f\left( x \right)dx} = \int\limits_{ - a}^0 {f\left( x \right)dx} + \int\limits_o^a {f\left( x \right)dx} $
Consider,$\int\limits_{ - a}^0 {f\left( x \right)dx} $
Replace $$x$$ with $$ - x$$ in $\int\limits_{ - a}^0 {f\left( x \right)dx} $.
We get,
$\int\limits_{ - a}^0 {f\left( x \right)dx} = - \int\limits_a^0 {f\left( { - x} \right)dx} $
$ \Rightarrow \int\limits_{ - a}^0 {f\left( x \right)dx} = - \int\limits_a^0 {f\left( x \right)dx} $. Since $f\left( x \right) = f\left( { - x} \right)$
$ \Rightarrow \int\limits_{ - a}^0 {f\left( x \right)dx} = \int\limits_0^a {f\left( x \right)dx} $. Since $\int\limits_a^b {f\left( x \right)dx} = - \int\limits_b^a {f\left( x \right)dx} $
Therefore, $\int\limits_{ - a}^a {f\left( x \right)dx} = \int\limits_0^a {f\left( x \right)dx} + \int\limits_o^a {f\left( x \right)dx} $
$\int\limits_{ - a}^a {f\left( x \right)dx} = 2\int\limits_o^a {f\left( x \right)dx} $
Hence Proved.

Note:
We can solve this problem in many ways. We have to be careful while solving this type of sums because when we divide the interval there is a high possibility that we are making mistakes and we have to be careful while calculating the greatest integer functions and fractional part functions. Here also there is a high chance of making mistakes. We have to be careful while calculating. Also by making mistakes in the calculations also we can get the wrong answer.