JEE MAIN - Mathematics (2021 - 20th July Evening Shift - No. 3)
If [x] denotes the greatest integer less than or equal to x, then the value of the integral $$\int_{ - \pi /2}^{\pi /2} {[[x] - \sin x]dx} $$ is equal to :
$$-$$ $$\pi$$
$$\pi$$
0
1
Explanation
$$I = \int_{ - \pi /2}^{\pi /2} {[[x] - \sin x]dx} $$
$$ = \int_{ - \pi /2}^{\pi /2} {\left( {[x] + [ - \sin x]} \right)dx} $$
$$ = \int_0^{\pi /2} {\left( {[x] + [ - \sin x] + [ - x] + [\sin x]} \right)} dx$$
$$ = \int_0^{\pi /2} {( - 2)dx} $$
$$ = - \pi $$
$$ = \int_{ - \pi /2}^{\pi /2} {\left( {[x] + [ - \sin x]} \right)dx} $$
$$ = \int_0^{\pi /2} {\left( {[x] + [ - \sin x] + [ - x] + [\sin x]} \right)} dx$$
$$ = \int_0^{\pi /2} {( - 2)dx} $$
$$ = - \pi $$
Comments (0)
