Question

3. Evaluate S' [2 – [x]] dx limits from -1 to 1 integrate

Answer 1

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$\large\underline\purple{\bold{Solution :- }}$

$\tt \:  Let \: f(x) \: = [2 - [x]] \: where \: - 1 \leqslant x \leqslant 1$

$\tt \:  \begin{gathered}\begin{gathered}\tt \: So, \: f(x) \: = \begin{cases} &\sf{[2 - ( - 1)]} \: when \: - 1 \leqslant x \ < 0 \\ &\sf{[2 - 0]} \: when \: 0 \leqslant x \leqslant 1\end{cases}\end{gathered}\end{gathered}$

$\tt \:  \begin{gathered}\begin{gathered}\tt \: So, \: f(x) \: = \begin{cases} &\sf{[3]} \: when \: - 1 \leqslant x < 0 \\ &\sf{[2]} \: when \: 0 \leqslant x \leqslant 1\end{cases}\end{gathered}\end{gathered}$

$\tt \:  \begin{gathered}\begin{gathered}\tt \: So, \: f(x) \: = \begin{cases} &\sf{3} \: when \: - 1 \leqslant x \ < 0 \\ &\sf{2} \: when \: 0 \leqslant x \leqslant 1\end{cases}\end{gathered}\end{gathered}$

$\tt \:  Consider \: \int_{ - 1}^1[2 - [x]]dx$

$\tt \:  = \: \int_{ - 1}^0[2 - [x]]dx + \int_{ 0}^1[2 - [x]]dx$

$\tt \:  = \: \int_{ - 1}^03 \: dx \: + \int_{ 0}^1 \: 2 \: dx$

$\tt \:  = [3x]_{ - 1}^0 \: + \: [2x]_0^1$

$\tt \:  = ( 0 - (- 3)) + (2 - 0)$

$\tt \:  = \: \: 5$

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