Question

The value of $\int\limits_{ - 2}^2 {\min \left\{ {x - [x], - x - [ - x]} \right\}dx} $ is equal to (where [ ] denotes the greatest integer function).
 $\eqalign{
  & A)\,\,\,\,\dfrac{1}{2} \cr
  & B)\,\,\,\,1 \cr
  & C)\,\,\,\,\dfrac{3}{2} \cr
  & D)\,\,\,\,2 \cr} $.

Answer 1

Hint:
[x] denotes the greatest integer which is less than or equal to x.As an example [2.5]=2 and [-2.5]=(-3) .We have to split the whole integral w.r.t. their limits and find the integrand there for solving it easily.

Complete step by step solution:
Step1:
                  $$\eqalign{
  & For\,\,\, - 2 < x < - 1\,\, \Rightarrow [x] = - 2 \Rightarrow x - [x] = x + 2 \cr
  & Then,\,\,1 < - x < 2\, \Rightarrow [ - x] = 1 \Rightarrow - x - [ - x] = - x - 1 \cr
  & \min \left\{ {x - [x], - x - [ - x]} \right\} \cr
  & = \min \left\{ {x + 2, - x - 1} \right\} \cr
  & = x + 2,\,\, - 2 < x \leqslant - \dfrac{3}{2} \cr
  & = - x - 1,\,\, - \dfrac{3}{2} < x < - 1 \cr} $$
Step2:
                 $$\eqalign{
  & For\,\,\, - 1 < x < 0\,\, \Rightarrow [x] = - 1 \Rightarrow x - [x] = x + 1 \cr
  & Then,\,\,0 < - x < 1\, \Rightarrow [ - x] = 0 \Rightarrow - x - [ - x] = - x \cr
  & \min \left\{ {x - [x], - x - [ - x]} \right\} \cr
  & = \min \left\{ {x + 1, - x} \right\} \cr
  & = x + 1,\,\, - 1 < x \leqslant - \dfrac{1}{2} \cr
  & = - x,\,\, - \dfrac{1}{2} < x < 0 \cr} $$
Step3:
                 $$\eqalign{
  & For\,\,\,0 < x < 1\,\, \Rightarrow [x] = 0 \Rightarrow x - [x] = x \cr
  & Then,\,\, - 1 < - x < 0\, \Rightarrow [ - x] = - 1 \Rightarrow - x - [ - x] = - x + 1 \cr
  & \min \left\{ {x - [x], - x - [ - x]} \right\} \cr
  & = \min \left\{ {x,1 - x} \right\} \cr
  & = x,\,\,0 < x \leqslant \dfrac{1}{2} \cr
  & = 1 - x,\,\,\dfrac{1}{2} < x < 1 \cr} $$
Step4:
                $$\eqalign{
  & For\,\,\,1 < x < 2\,\, \Rightarrow [x] = 1 \Rightarrow x - [x] = x - 1 \cr
  & Then,\,\, - 2 < - 1 < 0\, \Rightarrow [ - x] = - 2 \Rightarrow - x - [ - x] = - x + 2 \cr
  & \min \left\{ {x - [x], - x - [ - x]} \right\} \cr
  & = \min \left\{ {x - 1,2 - x} \right\} \cr
  & = x - 1,\,\,1 < x \leqslant \dfrac{3}{2} \cr
  & = 2 - x,\,\,\dfrac{3}{2} < x < 2 \cr} $$
Step5:Using the above value of the integrand in different range and the property of definite integral,
               $\eqalign{
  & \int\limits_{ - 2}^2 {\min \left\{ {x - [x], - x - [ - x]} \right\}dx} \cr
  & = \int\limits_{ - 2}^{\,\dfrac{{ - 3}}{2}} {(x + 2)dx} \, + \,\int\limits_{\,\dfrac{{ - 3}}{2}}^{\, - 1} {( - x - 1)dx} \, + \int\limits_{ - 1}^{\,\dfrac{{ - 1}}{2}} {(x + 1)dx} \, + \int\limits_{\dfrac{{ - 1}}{2}}^{\,0} {( - x)dx} \, + \int\limits_0^{\,\dfrac{1}{2}} {xdx} \, + \int\limits_{\dfrac{1}{2}}^{\,1} {(1 - x)dx} \, + \int\limits_1^{\,\dfrac{3}{2}} {(x - 1)dx} \, + \int\limits_{\dfrac{3}{2}}^{\,2} {(2 - x)dx} \cr
  & = \dfrac{1}{2}\left\{ {\left[ {{{(x + 2)}^2}} \right]_{ - 2}^{ - 1.5} - \left[ {{{(x + 1)}^2}} \right]_{ - 1.5}^{ - 1} + \left[ {{{(x + 1)}^2}} \right]_{ - 1}^{ - 0.5} - \left[ {{x^2}} \right]_{ - 0.5}^0 + \left[ {{x^2}} \right]_0^{0.5} - \left[ {{{(1 - x)}^2}} \right]_{0.5}^1 + \left[ {{{(x - 1)}^2}} \right]_1^{1.5} - \left[ {{{(2 - x)}^2}} \right]_{1.5}^2} \right\} \cr
  & = \dfrac{1}{2}\left\{ {\dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4}} \right\} \cr
  & = \dfrac{1}{2} \times 8 \times \dfrac{1}{4} \cr
  & = 1 \cr} $

Hence, the option B) is correct here.

Note:
In most of the questions on greatest integer function, we have to split into intervals. In each of the intervals, we have to define the function. This will be easy for solving these questions.