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Question: \(\int_{- 2}^{2}{\min\{|x|,\lbrack x\rbrack\} dx}\)equals:...

∫−22min⁡{∣x∣,[x]}dx\int_{- 2}^{2}{\min\{|x|,\lbrack x\rbrack\} dx}equals:

A

–2

B

0

C

1

D

–1

Answer

–2

Explanation

Solution

Clearly min. {|x|, [x]} = [x]

\∫−22[x]dx\int_{- 2}^{2}{\lbrack x\rbrack dx}= ∫−2−1[x]dx\int_{- 2}^{- 1}{\lbrack x\rbrack dx}+∫−10[x]dx\int_{- 1}^{0}{\lbrack x\rbrack dx}+ ∫01[x]dx\int_{0}^{1}{\lbrack x\rbrack dx}+

∫12[x]dx\int_{1}^{2}{\lbrack x\rbrack dx}

= – 2 ∫−2−1dx\int_{- 2}^{- 1}{dx}– ∫−10dx\int_{- 1}^{0}{dx}+ 0 + ∫12dx\int_{1}^{2}{dx}

= –2 –1 + 1 = –2