_ ^ t e^ 3 t^2 ;dt =

MCQ

$\int_{}^{} {\frac{t}{{{e^{3{t^2}}}}}\;dt = } $

A
$\frac{1}{6}{e^{3{t^2}}} + c$
B
$ - \frac{1}{6}{e^{3{t^2}}} + c$
C
$\frac{1}{6}{e^{ - 3{t^2}}} + c$
✓
$ - \frac{1}{6}{e^{ - 3{t^2}}} + c$

✓

Answer

Correct option: D.

$ - \frac{1}{6}{e^{ - 3{t^2}}} + c$

d
(d) $I = \int_{}^{} {t\,.\,{e^{ - 3{t^2}}}dt} $
Put $ - 3{t^2} = z \Rightarrow - 6t\,dt = dz \Rightarrow t\,dt = \frac{{ - 1}}{6}\,dz$
$\therefore \,\,\,I = - \frac{1}{6}\int_{}^{} {{e^z}dt} = \frac{{ - {e^z}}}{6} + c = - \frac{{{e^{ - 3{t^2}}}}}{6} + c.$

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