• # question_answer If ${{I}_{1}}=\int_{0}^{1}{\frac{{{e}^{x}}}{x+3}dx}$ and ${{I}_{2}}=\int_{0}^{1}{\frac{{{x}^{3}}}{{{e}^{{{x}^{4}}}}(4-{{x}^{4}})}dx,}$ then $\frac{{{I}_{1}}}{{{I}_{2}}}$ is A) $4$                              B) $4e$ C) $3e$                             D) $2e$

[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$