JEE MAIN - Mathematics (2021 - 24th February Morning Shift - No. 20)
If $$\int\limits_{ - a}^a {\left( {\left| x \right| + \left| {x - 2} \right|} \right)} dx = 22$$, (a > 2) and [x] denotes the greatest integer $$ \le $$ x, then$$\int\limits_{ - a}^a {\left( {x + \left[ x \right]} \right)} dx$$ is equal to _________.
Answer
3
Explanation
$$\int\limits_{ - a}^a {\left( {\left| x \right| + \left| {x - 2} \right|} \right)} dx = 22$$
$$ \Rightarrow $$ $$\int\limits_{ - a}^0 {( - 2x + 2)dx} + \int\limits_0^2 {(x + 2 - x)dx} + \int\limits_2^a {(2x - 2)dx} = 22$$
$$ \Rightarrow $$ $${x^2} - 2x|_0^{ - a} + 2x|_0^2 + {x^2} - 2x|_2^a = 22$$
$$ \Rightarrow $$ $${a^2} + 2a + 4 + {a^2} - 2a - (4 - 4) = 22$$
$$ \Rightarrow $$ $$2{a^2} = 18 \Rightarrow a = 3$$
$$\int\limits_3^{ - 3} {(x + [x])dx} = - \left( {\int\limits_{ - 3}^3 {(x + [x])dx} } \right) = - \left( {\int\limits_{ - 3}^3 {[x]dx} } \right)$$
= -($$\int\limits_{ - 3}^{ - 2} {\left[ x \right]dx} + \int\limits_{ - 2}^{ - 1} {\left[ x \right]dx} + \int\limits_{ - 1}^0 {\left[ x \right]dx} $$
+ $$\int\limits_0^1 {\left[ x \right]dx} + \int\limits_1^2 {\left[ x \right]dx} + \int\limits_2^3 {\left[ x \right]dx} $$)
$$ = - ( - 3 - 2 - 1 + 0 + 1 + 2) = 3$$
$$ \Rightarrow $$ $$\int\limits_{ - a}^0 {( - 2x + 2)dx} + \int\limits_0^2 {(x + 2 - x)dx} + \int\limits_2^a {(2x - 2)dx} = 22$$
$$ \Rightarrow $$ $${x^2} - 2x|_0^{ - a} + 2x|_0^2 + {x^2} - 2x|_2^a = 22$$
$$ \Rightarrow $$ $${a^2} + 2a + 4 + {a^2} - 2a - (4 - 4) = 22$$
$$ \Rightarrow $$ $$2{a^2} = 18 \Rightarrow a = 3$$
$$\int\limits_3^{ - 3} {(x + [x])dx} = - \left( {\int\limits_{ - 3}^3 {(x + [x])dx} } \right) = - \left( {\int\limits_{ - 3}^3 {[x]dx} } \right)$$
= -($$\int\limits_{ - 3}^{ - 2} {\left[ x \right]dx} + \int\limits_{ - 2}^{ - 1} {\left[ x \right]dx} + \int\limits_{ - 1}^0 {\left[ x \right]dx} $$
+ $$\int\limits_0^1 {\left[ x \right]dx} + \int\limits_1^2 {\left[ x \right]dx} + \int\limits_2^3 {\left[ x \right]dx} $$)
$$ = - ( - 3 - 2 - 1 + 0 + 1 + 2) = 3$$
Comments (0)
