The area of the figure bounded by the curves $y = e^{y}, y = e^{- x}$ and the straight line x = 1 is

e + \frac{1}{e}

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e - \frac{1}{e}

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e + \frac{1}{e} - 2

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e + \frac{1}{e} + 2

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Solution

The correct option is C $e + \frac{1}{e} - 2$
$A = \int_{0}^{1} e^{x} - e^{- x}$
$= e^{x} + e^{- x} \int_{0}^{1}$
$= (e + \frac{1}{e}) - (1 + 1)$
$= e + \frac{1}{e} - 2$

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