A) \[\frac{1}{6}{{e}^{3{{t}^{2}}}}+c\]
B) \[-\frac{1}{6}{{e}^{3{{t}^{2}}}}+c\]
C) \[\frac{1}{6}{{e}^{-3{{t}^{2}}}}+c\]
D) \[-\frac{1}{6}{{e}^{-3{{t}^{2}}}}+c\]
Correct Answer: D
Solution :
\[I=\int_{{}}^{{}}{t\,.\,{{e}^{-3{{t}^{2}}}}dt}\] 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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