JEE MAIN - Mathematics (2019 - 10th April Evening Slot - No. 6)
If $$\int {{x^5}} {e^{ - {x^2}}}dx = g\left( x \right){e^{ - {x^2}}} + c$$, where c is a constant of integration, then $$g$$(–1) is equal to :
1
- 1
$$ - {5 \over 2}$$
$$ - {1 \over 2}$$
Explanation
Let x2 = t
$$ \Rightarrow {1 \over 2}\int {{t^2}{e^{ - t}}dt} $$
$$ \Rightarrow {1 \over 2}\left[ { - {t^2}{e^{ - t}} + \int {2t{e^{ - t}}dt} } \right]$$
$$ \Rightarrow {{ - {t^2}{e^{ - t}}} \over 2} - t{e^{ - t}} - {e^{ - t}}$$
$$ \Rightarrow \left( { - {{{x^4}} \over 2} - {x^2} - 1} \right){e^{ - {x^2}}} + c$$
Then $$g(x) = - {{{x^4}} \over 2} - {x^2} - 1$$
$$ \Rightarrow g( - 1) = - {1 \over 2} - 1 - 1$$
$$ \Rightarrow - {5 \over 2}$$
$$ \Rightarrow {1 \over 2}\int {{t^2}{e^{ - t}}dt} $$
$$ \Rightarrow {1 \over 2}\left[ { - {t^2}{e^{ - t}} + \int {2t{e^{ - t}}dt} } \right]$$
$$ \Rightarrow {{ - {t^2}{e^{ - t}}} \over 2} - t{e^{ - t}} - {e^{ - t}}$$
$$ \Rightarrow \left( { - {{{x^4}} \over 2} - {x^2} - 1} \right){e^{ - {x^2}}} + c$$
Then $$g(x) = - {{{x^4}} \over 2} - {x^2} - 1$$
$$ \Rightarrow g( - 1) = - {1 \over 2} - 1 - 1$$
$$ \Rightarrow - {5 \over 2}$$
Comments (0)
