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

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\\} = xin 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.