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STD 12 - 7.2 definite integral — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 7.2 definite integral4 Marks
MCQ
Let ${I_1} = \int\limits_0^1 {\frac{{{e^x}}}{{\left( {1 + x} \right)}}} \,dx$ and ${I_2} = \int\limits_0^1 {\frac{{{x^2}}}{{{e^{{x^3}}}\left( {2 - {x^3}} \right)}}} \,dx$ then value of $\frac{{{I_1}}}{{{I_2}}}$ is equal to
  • A
    $\frac{3}{e}$
  • B
    $\frac{e}{3}$
  • ✓
    $3e$
  • D
    $\frac{1}{3e}$
✓

Answer

Correct option: C.
$3e$
c
$I_{2}=\frac{1}{3} \int_{0}^{1} \frac{3 x^{2} d x}{e^{x^{3}}\left(2-x^{3}\right)},$ substituting $t=x^{3}$

$\Rightarrow I_{2}=\frac{1}{3} \int_{0}^{1} \frac{d t}{e^{t}(2-t)} \Rightarrow$ Substituting $t=1-z$

then $I_{2}=\frac{1}{3} \int_{0}^{1} \frac{d z}{e^{1-z}(1+z)}=\frac{1}{3 e} \int_{0}^{1} \frac{e^{z} d z}{(z+1)}$

$\Rightarrow \frac{I_{1}}{I_{2}}=3 e$

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