JEE MAIN - Mathematics (2021 - 31st August Morning Shift - No. 16)
Let [t] denote the greatest integer $$\le$$ t. Then the value of
$$8.\int\limits_{ - {1 \over 2}}^1 {([2x] + |x|)dx} $$ is ___________.
$$8.\int\limits_{ - {1 \over 2}}^1 {([2x] + |x|)dx} $$ is ___________.
Answer
5
Explanation
$$I = \,\int\limits_{ - {1 \over 2}}^1 {([2x] + |x|)dx} $$
$$ = \int\limits_{ - 1/2}^1 {[2x]\,dx + \int\limits_{ - 1/2}^1 {|x|\,} dx} $$
$$ = 0 + \int\limits_{ - 1/2}^0 {( - x)\,} dx + \int\limits_0^1 {x\,} dx$$
$$ = \left( { - {{{x^2}} \over 2}} \right)_{ - 1/2}^0 + \left( {{{{x^2}} \over 2}} \right)_0^1$$
$$ = \left( {0 + {1 \over 8}} \right) + {1 \over 2}$$
$$ = {5 \over 8}$$
$$ \therefore $$ 8I = 5
$$ = \int\limits_{ - 1/2}^1 {[2x]\,dx + \int\limits_{ - 1/2}^1 {|x|\,} dx} $$
$$ = 0 + \int\limits_{ - 1/2}^0 {( - x)\,} dx + \int\limits_0^1 {x\,} dx$$
$$ = \left( { - {{{x^2}} \over 2}} \right)_{ - 1/2}^0 + \left( {{{{x^2}} \over 2}} \right)_0^1$$
$$ = \left( {0 + {1 \over 8}} \right) + {1 \over 2}$$
$$ = {5 \over 8}$$
$$ \therefore $$ 8I = 5
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