A) \[4\]
B) \[4e\]
C) \[3e\]
D) \[2e\]
Correct Answer: B
Solution :
[b]We have, \[{{I}_{1}}=\int_{0}^{1}{\frac{{{e}^{x}}}{x+3}dx}\] |
\[{{I}_{2}}=\int_{0}^{1}{\frac{{{x}^{3}}}{{{e}^{{{x}^{4}}}}(4-{{x}^{4}})}dx}\] |
In \[{{I}_{2}}\]put \[{{x}^{4}}=t\] \[\Rightarrow \] \[4{{x}^{3}}\,\,\,dx=dt\] |
\[\therefore \] \[{{I}_{2}}=\frac{1}{4}\int_{0}^{1}{\frac{dt}{{{e}^{t}}(4-t)}}\] |
\[{{I}_{2}}=\frac{1}{4}\int_{0}^{1}{\frac{dt}{{{e}^{1-t}}(4-(1-t)}}\] \[\left[ \because \int_{0}^{a}{f(x)dx=\int_{0}^{a}{f(a-x)dx}} \right]\] |
\[{{I}_{2}}=\frac{1}{4}\int_{0}^{1}{\frac{{{e}^{t}}}{e\,(3+t)}dt}\] |
\[{{I}_{2}}=\frac{1}{4e}{{I}_{1}}\]\[\Rightarrow \]\[\frac{{{I}_{1}}}{{{I}_{2}}}=4e\] |
You need to login to perform this action.
You will be redirected in
3 sec