Area of the region bounded by the curves, $y = e^{x}, y = e^{- x}$ and the straight line $x = 1$ is given by

(e - e^{- 1} + 2)

sq.unit

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(e - e^{- 1} - 2)

sq.unit

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(e + e^{- 1} - 2)

sq.unit

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None of the above

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Solution

The correct option is C $(e + e^{- 1} - 2)$ sq.unit
$∴$ Required Area $= \int_{0}^{1} (e^{x} - e^{- x}) d x$
$= {[e^{x} - e^{- x}]}_{0}^{x}$
$= (e + e^{- 1}) - (e^{0} + e^{- 0})$
$= (e + e^{- 1} - 2)$ .sq.unit.

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