JEE MAIN - Mathematics (2022 - 27th June Morning Shift - No. 8)
The value of the integral
$$\int\limits_{ - 2}^2 {{{|{x^3} + x|} \over {({e^{x|x|}} + 1)}}dx} $$ is equal to :
$$\int\limits_{ - 2}^2 {{{|{x^3} + x|} \over {({e^{x|x|}} + 1)}}dx} $$ is equal to :
5e2
3e$$-$$2
4
6
Explanation
$$I = \int\limits_{ - 2}^2 {{{|{x^3} + x|} \over {{e^{x|x|}} + 1}}dx} $$ ..... (i)
$$I = \int\limits_{ - 2}^2 {{{|{x^3} + x|} \over {{e^{ - x|x|}} + 1}}dx} $$ ..... (ii)
$$2I = \int\limits_{ - 2}^2 {|{x^3} + x|dx} $$
$$2I =2 \int\limits_0^2 {({x^3} + x)dx} $$
$$I = \int\limits_0^2 {({x^3} + x)dx} $$
$$ = \left. {{{{x^4}} \over 4} + {{{x^2}} \over 2}} \right]_0^2$$
$$ = \left( {{{16} \over 4} + {4 \over 2}} \right) - 0$$
$$ = 4 + 2 = 6$$
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