If I1=∫−1011dx(5+2x−2x2)(1+e2−4x) and I2=∫−1011 dx5+2x−2x2 then I1I2=

Use ∫abf(x)dx=∫abf(a+b−x)dx I1=∫−1011dx(5+2x−2x2)(1+e2−4x) I1=∫−1011e2−4x(5+2x−2x2)(e2−4x+1) dx 2I1=∫−1011dx5+2x−x2=I2 I1I2=12

If I1=∫−1011dx(5+2x−2x2)(1+e2−4x) and I2=∫−1011 dx5+2x−2x2 then I1I2=

Use ∫abf(x)dx=∫abf(a+b−x)dx I1=∫−1011dx(5+2x−2x2)(1+e2−4x) I1=∫−1011e2−4x(5+2x−2x2)(e2−4x+1) dx 2I1=∫−1011dx5+2x−x2=I2 I1I2=12

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If $I_{1} = \int_{- 10}^{11} \frac{d x}{(5 + 2 x - 2 x^{2}) (1 + e^{2 - 4 x})}$ and $I_{2} = \int_{- 10}^{11} \frac{d x}{5 + 2 x - 2 x^{2}}$ then $\frac{I_{1}}{I_{2}} =$

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$2$

$\frac{1}{2}$

$1$

$\frac{- 1}{2}$

answer is B.

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Detailed Solution

Use $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$
$I_{1} = \int_{- 10}^{11} \frac{d x}{(5 + 2 x - 2 x^{2}) (1 + e^{2 - 4 x})}$
$I_{1} = \int_{- 10}^{11} \frac{e^{2 - 4 x}}{(5 + 2 x - 2 x^{2}) (e^{2 - 4 x} + 1)} d x$
$2 I_{1} = \int_{- 10}^{11} \frac{d x}{5 + 2 x - x^{2}} = I_{2}$
$\frac{I_{1}}{I_{2}} = \frac{1}{2}$