If I1- -100 101 dx-5-2x-2x21-e2-4x and I2- -100 101 dx-5-2x-2x2 then I1I2 is

The correct option is B 1/2 I_1= nbsp; nbsp;∫_ 100^101dx/(5+2x 2x^2)(1+e^(2 4x)) = nbsp;∫_ 100^101dx/(5+2(1 x) 2(1 x)^2)(1+e^(2 4(1 x))) ...

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Byju's Answer

Standard XII

Mathematics

Property 1

If I1=∫ -10...

Question

If $I_{1} = \int_{- 100}^{101} \frac{d x}{(5 + 2 x - 2 x^{2}) (1 + e^{(2 - 4 x)})}$ and $I_{2} = \int_{- 100}^{101} \frac{d x}{(5 + 2 x - 2 x^{2})}$ then
$\frac{I_{1}}{I_{2}}$ is

2

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\frac{1}{2}

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1

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- \frac{1}{2}

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Solution

The correct option is B $\frac{1}{2}$
$I_{1} = \int_{- 100}^{101} \frac{d x}{(5 + 2 x - 2 x^{2}) (1 + e^{(2 - 4 x)})}$
$= \int_{- 100}^{101} \frac{d x}{(5 + 2 (1 - x) - 2 (1 - x)^{2}) (1 + e^{(2 - 4 (1 - x))})}$
$= \int_{- 100}^{101} \frac{d x}{(5 + 2 x - 2 x^{2}) (1 + e^{- (2 - 4 x)})}$
$= \int_{- 100}^{101} \frac{e^{(2 - 4 x)} d x}{(5 + 2 x - 2 x^{2}) (e^{(2 - 4 x)} + 1)}$
$2 I_{1} = \int_{- 100}^{101} \frac{1 + e^{(2 - 4 x)} d x}{(5 + 2 x - 2 x^{2}) (e^{(2 - 4 x)} + 1)}$
or $2 I_{1} = \int_{- 100}^{101} \frac{d x}{5 + 2 x - 2 x^{2}} = I_{2}$
or $\frac{I_{1}}{I_{2}} = \frac{1}{2}$

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If I1=∫101−100dx(5+2x−2x2)(1+e(2−4x))and I2=∫101−100dx(5+2x−2x2) thenI1I2 is

If $I_{1} = \int_{- 100}^{101} \frac{d x}{(5 + 2 x - 2 x^{2}) (1 + e^{(2 - 4 x)})}$ and $I_{2} = \int_{- 100}^{101} \frac{d x}{(5 + 2 x - 2 x^{2})}$ then
$\frac{I_{1}}{I_{2}}$ is