A calculus problem by A Former Brilliant Member

Calculus Level 3

Let I 1 = ∫ 0 1 e x 1 + x d x \displaystyle I_1 = \int _0^1 \frac {e^x}{1+ x } dx and I 2 = ∫ 0 1 x 2 e x 3 ( 2 − x 3 ) d x \displaystyle { I }_{ 2 }= \int _{ 0 }^{ 1 }{ \frac { { x }^{ 2 } }{ { e }^{ { x }^{ 3 } }\left( { 2 }-{ x }^{ 3 } \right) } } dx .

What is I 1 I 2 \dfrac { { I }_{ 1 } }{ { I }_{ 2 } } ?


The answer is 8.15484548537.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

Chew-Seong Cheong
Apr 26, 2018

I 2 = ∫ 0 1 x 2 e x 3 ( 2 − x 3 ) d x = ∫ 0 1 x 2 e − x 3 ( 2 − x 3 ) d x Let u = 1 − x 3 ⟹ d u = − 3 x 2 d x = − 1 3 ∫ 1 0 e u − 1 1 + u d u = 1 3 e ∫ 0 1 e u 1 + u d u = 1 3 e I 1 \begin{aligned} I_2 & = \int_0^1 \frac {x^2}{e^{x^3}(2-x^3)} dx \\ & = \int_0^1 \frac {x^2e^{-x^3}}{(2-x^3)} dx & \small \color{#3D99F6} \text{Let }u = 1 - x^3 \implies du = -3x^2 \ dx \\ & = - \frac 13 \int_1^0 \frac {e^{u-1}}{1+u} du \\ & = \frac 1{3e} \int_0^1 \frac {e^u}{1+u} du \\ & = \frac 1{3e} I_1 \end{aligned}

Therefore, I 1 I 2 = 3 e ≈ 8.155 \dfrac {I_1}{I_2} = 3e \approx \boxed{8.155} .

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