Let $I_1=\int\limits_0^1 \frac{e^x}{1+x} d x$ and $I_2=\int\limits_0^1 \frac{x^2}{e^{x^3}\left(2-x^3\right)} d x$. Then, $\frac{I_1}{I_2}$ is equal to
Answer & explanation
Correct answer: option 3
We have,
$I_1 =\int\limits_0^1 \frac{e^x}{1+x} d x$ and $I_2=\int\limits_0^1 \frac{x^2}{e^{x^3}\left(2-x^3\right)} d x$
∴ $I_2 =\frac{1}{3} \int\limits_0^1 \frac{1}{e^{x^3}\left(2-x^3\right)} 3 x^2 d x$
$\Rightarrow I_2 =\frac{1}{3} \int\limits_0^1 \frac{1}{e^t(2-t)} d t$, where $t=x^3$
$\Rightarrow I_2=\frac{1}{3} \int\limits_0^1 \frac{1}{e^{1-t}(1+t)} d t$ $\left[∵ \int\limits_0^a f(x) d x=\int\limits_0^a f(a-x)\right]$
$\Rightarrow I_2=\frac{1}{3 e} \int\limits_0^1 \frac{e^t}{1+t} d t \Rightarrow I_2=\frac{1}{3 e} I_1 \Rightarrow \frac{I_1}{I_2}=3 e$