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Question: Let I<sub>1</sub> = \(\int_{0}^{1}\frac{e^{x}dx}{x + 1}\) and I<sub>2</sub> = \(\int_{0}^{1}\frac{x^...

Let I1 = ∫01exdxx+1\int_{0}^{1}\frac{e^{x}dx}{x + 1} and I2 = ∫01x2dxex3(2−x3)\int_{0}^{1}\frac{x^{2}dx}{e^{x^{3}}(2 - x^{3})}, then I1I2\frac{I_{1}}{I_{2}} is equal to

A

3e\frac{3}{e}

B

e3\frac{e}{3}

C

3e

D

13e\frac{1}{3e}

Answer

3e

Explanation

Solution

I2 = ∫01x2dxex3(2−x3)\int_{0}^{1}\frac{x^{2}dx}{e^{x^{3}}(2 - x^{3})}

{let 1 – x3 = t Ž x2dx = −dt3\frac{- dt}{3}}

= −13∫10dte1−t(1+t)- \frac{1}{3}\int_{1}^{0}\frac{dt}{e^{1 - t}(1 + t)} = 13e∫01et1+tdt\frac{1}{3e}\int_{0}^{1}{\frac{e^{t}}{1 + t}dt}

I2 = 13eI1\frac{1}{3e}I_{1}Ž I1I2\frac{I_{1}}{I_{2}} = 3e