Let [t] denote the greatest integer $\le$ t. Then the value of
$8.\int\limits_{ - {1 \over 2}}^1 {([2x] + |x|)dx}$ is ___________.
Answer (integer)
5
Solution
$I = \,\int\limits_{ - {1 \over 2}}^1 {([2x] + |x|)dx}$<br><br>$= \int\limits_{ - 1/2}^1 {[2x]\,dx + \int\limits_{ - 1/2}^1 {|x|\,} dx}$<br><br>$= 0 + \int\limits_{ - 1/2}^0 {( - x)\,} dx + \int\limits_0^1 {x\,} dx$<br><br>$$ = \left( { - {{{x^2}} \over 2}} \right)_{ - 1/2}^0 + \left( {{{{x^2}} \over 2}} \right)_0^1$$<br><br>$= \left( {0 + {1 \over 8}} \right) + {1 \over 2}$<br><br>$= {5 \over 8}$<br><br>$\therefore$ 8I = 5
About this question
Subject: Mathematics · Chapter: Definite Integration · Topic: Properties of Definite Integrals
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