​PLEASE SOLVE AND GIVE CORRECT ANSWER
Question 62 of 90
${\int }_{-2}^{2}min\left(x-\left[x\right], -x-\left[-x\right]\right) dx$, where [.] is the greatest integer function =
1. 0
2. 1
3. 2
4. None of these.

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{I}={\int }_{-2}^2 \min \left(\mathrm{x}-\left[\mathrm{x}\right],-\mathrm{x}-\left[-\mathrm{x}\right]\right)\mathrm{dx} \\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{x}-\left[\mathrm{x}\right]=\left\{\mathrm{x}\right\} \\ -\mathrm{x}-\left[-\mathrm{x}\right]=\left\{-\mathrm{x}\right\} \\ ={\int }_{-2}^2 \min \left(\left\{\mathrm{x}\right\},\left\{-\mathrm{x}\right\}\right)\mathrm{dx} \\ \mathrm{Now} \mathrm{we} \mathrm{solve} \mathrm{graphically}. \end{gathered}$$
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{smaller} \mathrm{triangle}=\frac{1}{2}\times 1\times 1=\frac{1}{2} \\ \mathrm{Hence} \mathrm{area}=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=2 \left[\mathrm{Answer}\right] \end{gathered}$$

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