JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Indefinite Integrals

\int 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

Answer: (C)

Solution

Substitute

t = x^{2}

, so

d t = 2 x d x

and

\int x^{5} e^{- x^{2}} d x = \frac{1}{2} \int t^{2} e^{- t} d t

. Integrating by parts twice,

\frac{1}{2} \int t^{2} e^{- t} d t = - (\frac{t ^{2}}{2} + t + 1) e^{- t}

, and restoring

t = x^{2}

identifies

g (x) = - (\frac{x ^{4}}{2} + x^{2} + 1)

. Hence

g (- 1) = - (\frac{1}{2} + 1 + 1) = - \frac{5}{2}

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Single correct option

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