Integral Calculus Question 445
Question: $ \int_{{}}^{{}}{\frac{t}{{e^{3t^{2}}}}\ dt=} $
[MP PET 1997]
Options:
A) $ \frac{1}{6}{e^{3t^{2}}}+c $
B) $ -\frac{1}{6}{e^{3t^{2}}}+c $
C) $ \frac{1}{6}{e^{-3t^{2}}}+c $
D) $ -\frac{1}{6}{e^{-3t^{2}}}+c $
Show Answer
Answer:
Correct Answer: D
Solution:
$ I=\int_{{}}^{{}}{t,.,{e^{-3t^{2}}}dt} $ Put $ -3t^{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^{-3t^{2}}}}{6}+c. $
BETA
