- 1 1 e | x | ? d x =

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Solution

Q.
$\int_{- 1}^{1} e^{| x |} d x =$

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

2 $2 (e + 1)$

3 $e - 1$

4 $2 (e - 1)$

Ans.
(4)

Ans.    $2 (e - 1)$

Explanation :

$I = \int_{- 1}^{1} e^{| x |} d x$

$= \int_{0}^{1} e^{x} d x + \int_{- 1}^{0} e^{- x} d x$

   $= {[e^{x}]}_{0}^{1} - {[e^{- x}]}_{- 1}^{0}$

   $= (e - 1) - (1 - e)$

   $= 2 e - 2$

$= 2 (e - 1)$

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