JEE PYQ: Indefinite Integration Question 24
Question 24 - 2019 (10 Apr Shift 2)
If $\int x^5 e^{-x^2},dx = g(x)e^{-x^2} + c$, where $c$ is a constant of integration, then $g(-1)$ is equal to:
(1) $-1$
(2) 1
(3) $-\frac{5}{2}$
(4) $-\frac{1}{2}$
Show Answer
Answer: (3)
Solution
Put $-x^2 = t$, $-2x,dx = dt$. $I = \int x^4 e^{-x^2}\cdot x,dx = -\frac{1}{2}\int t^2 e^t,dt = -\frac{1}{2}e^t(t^2 - 2t + 2) + c = -\frac{1}{2}e^{-x^2}(x^4 + 2x^2 + 2) + c$. So $g(x) = -\frac{1}{2}(x^4 + 2x^2 + 2)$. $g(-1) = -\frac{1}{2}(1+2+2) = -\frac{5}{2}$.
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