Let I1 - -01ex dx1 - x and I2 - -01x2 dxex32 - x3- then I1I2 is equal

I_1 = ∫_0^1e^x dx1 + x , I_2 = ∫_0^1x^2 dxe^x^3(2 x^3) In I_2, put 1 x^3 = t ⇒ I_2 = 13∫_1^0 dte^1 t(1 + t) = 13e∫_0^1e^t dt1 + t = 13e I_1 ⇒I_1I_2 = 3e ...

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Standard XII

Mathematics

Sandwich Theorem

Let I1 = ∫01e...

Question

Let $I_{1} = 1 \int 0 \frac{e^{x} d x}{1 + x}$ and $I_{2} = 1 \int 0 \frac{x^{2} d x}{e^{x^{3}} (2 - x^{3})}$ , then $\frac{I_{1}}{I_{2}}$ is equal

\frac{3}{e}

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\frac{e}{3}

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3 e

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\frac{1}{3 e}

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Solution

The correct option is C $3 e$
$I_{1} = 1 \int 0 \frac{e^{x} d x}{1 + x}, I_{2} = 1 \int 0 \frac{x^{2} d x}{e^{x^{3}} (2 - x^{3})}$
In $I_{2}$ , put $1 - x^{3} = t$
$\Rightarrow I_{2} = \frac{1}{3} 0 \int 1 \frac{- d t}{e^{1 - t} (1 + t)}$
$= \frac{1}{3 e} 1 \int 0 \frac{e^{t} d t}{1 + t} = \frac{1}{3 e} I_{1}$
$\Rightarrow \frac{I_{1}}{I_{2}} = 3 e$

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Sandwich Theorem

Standard XII Mathematics

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Let I1=1∫0exdx1+x and I2=1∫0x2dxex3(2−x3), then I1I2 is equal

The correct option is C 3eI1=1∫0exdx1+x,I2=1∫0x2dxex3(2−x3) In I2, put 1−x3=t ⇒I2=130∫1−dte1−t(1+t) =13e1∫0etdt1+t=13eI1 ⇒I1I2=3e

Let $I_{1} = 1 \int 0 \frac{e^{x} d x}{1 + x}$ and $I_{2} = 1 \int 0 \frac{x^{2} d x}{e^{x^{3}} (2 - x^{3})}$ , then $\frac{I_{1}}{I_{2}}$ is equal