A) \[-\frac{5}{2}\]
B) \[1\]
C) \[-\frac{1}{2}\]
D) \[-1\]
Correct Answer: A
Solution :
Let \[{{x}^{2}}=t\] \[2xdx=dt\] \[\Rightarrow \frac{1}{2}\int_{{}}^{{}}{{{t}^{2}}.{{e}^{-1}}}dt=\frac{1}{2}\left[ -{{t}^{2}}.{{e}^{-t}}+\int_{{}}^{{}}{2t.{{e}^{-t}}dt} \right]\] \[=\frac{1}{2}\left( -{{t}^{2}}.{{e}^{-t}} \right)+\left( -t{{e}^{-t}}+\int_{{}}^{{}}{1.{{e}^{-t}}.dt} \right)\] \[=-\frac{{{t}^{2}}{{e}^{-t}}}{2}-t\,{{e}^{-t}}-\,{{e}^{-t}}=\left( -\frac{{{t}^{2}}}{2}-t-1 \right){{e}^{-t}}\] \[=\left( -\frac{{{x}^{4}}}{2}-{{x}^{2}}-1 \right){{e}^{-{{x}^{2}}}}+C\] \[g(x)=-1-{{x}^{2}}-\frac{{{x}^{4}}}{2}+k{{e}^{{{x}^{2}}}}\]for \[k=0\] \[g(-1)=-1-1-\frac{1}{2}=-\frac{5}{2}\]
You need to login to perform this action.
You will be redirected in
3 sec
