Mathematics · 2019
JEE Main · 10 April 2019, Shift 2 · Q74
If ∫ x^5 e^-x^2 d x= g (x) e^-x^2+ c, where c is a constant of integration, then g (-1) is equal to:
If $\displaystyle \int x^5 \mathrm{e}^{-x^2} \mathrm{~d} x=\mathrm{g}(x) \mathrm{e}^{-x^2}+\mathrm{c}$, where c is a constant of integration, then $\displaystyle \mathrm{g}(-1)$ is equal to :
Official answer
(1)
$\displaystyle -\frac{5}{2}$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.