The value of – 1 1 ( 1 + | x | – x ) e x + ( | x | – x ) e – x e x + e – x d x is equal to [2025]

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Solution

Q.
The value of $\int_{- 1}^{1} \frac{(1 + \sqrt{| x | - x}) e^{x} + (\sqrt{| x | - x}) e^{- x}}{e^{x} + e^{- x}} d x$ is equal to [2025]

1 $2 + \frac{2 \sqrt{2}}{3}$

2 $3 - \frac{2 \sqrt{2}}{3}$

3 $1 - \frac{2 \sqrt{2}}{3}$

4 $1 + \frac{2 \sqrt{2}}{3}$

Ans.
(4)

Consider, $I = \int_{- 1}^{1} \frac{(1 + \sqrt{| x | - x}) e^{x} + (\sqrt{| x | - x}) e^{- x}}{e^{x} + e^{- x}}$

$= \int_{- 1}^{1} \frac{e^{x} + e^{x} \sqrt{| x | - x} + e^{- x} \sqrt{| x - x |}}{e^{x} + e^{- x}} d x$

$= \int_{- 1}^{1} [\frac{e^{x}}{e^{x} + e^{- x}} + \frac{\sqrt{| x | - x} (e^{x} + e^{- x})}{e^{x} + e^{- x}}] d x$

$= \int_{- 1}^{1} \frac{e^{x}}{e^{x} + e^{- x}} d x + \int_{- 1}^{1} \sqrt{| x | - x} d x = I_{1} + I_{2}$

Now, $I_{1} = \int_{- 1}^{1} \frac{e^{x}}{e^{x} + e^{- x}} d x$ ... (i)

$\Rightarrow I_{1} = \int_{- 1}^{1} \frac{e^{- x}}{e^{- x} + e^{x}} d x$ ... (ii)

On adding (i) and (ii), we get

$\Rightarrow 2 I_{1} = \int_{- 1}^{1} \frac{e^{x} + e^{- x}}{e^{x} + e^{- x}} d x = \int_{- 1}^{1} 1 d x = 2 \Rightarrow I_{1} = 1$

and $I_{2} = \int_{- 1}^{1} \sqrt{| x | - x} d x = \int_{- 1}^{0} \sqrt{| x | - x} d x + \int_{0}^{1} \sqrt{| x | - x} d x$

$\Rightarrow I_{2} = \int_{- 1}^{0} \sqrt{- 2 x} d x + \int_{0}^{1} \sqrt{x - x} d x = \int_{- 1}^{0} \sqrt{- 2 x} d x$

Put $- 2 x = u \Rightarrow d x = - \frac{1}{2} d u$

$∴ I_{2} = \int_{2}^{0} \sqrt{u} (- \frac{1}{2} d u) = \frac{1}{2} \int_{0}^{2} \sqrt{u} d u = \frac{1}{2} {[\frac{u^{3 / 2}}{\frac{3}{2}}]}_{0}^{2} = \frac{2 \sqrt{2}}{3}$

$∴ I = I_{1} + I_{2} = 1 + \frac{2 \sqrt{2}}{3}$ .

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