Evaluate $\infty \int 0 [2 e^{- x}] d x$ , where $[x]$ represents the greatest integer function.

I n 2

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I n 4

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I n 3

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I n 8

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Solution

The correct option is B $I n 2$
$\int_{0}^{\infty} 2 e^{- x} d x$
$\int 2 e^{- x} d x = - 2 e^{- x} + C$
$\int 2 e^{- x} d x = 2 \int e^{- x} d x$
Substitute $u = - x$
$= 2 \int - e^{u} d u$
$= - 2 \int e^{u} d u$
$= - 2 e^{u}$
$u = - x$
$= - 2 e^{- x} + C$
Computing the boundaries
$\int_{0}^{\infty} 2 e^{- x} d x$
$= 2 [e^{- \infty} - e^{- 0}]$
$= 2 [0 - (- 1)]$
$= 0 + 2 = 2$
$\int_{0}^{\infty} 2 e^{- x} d x = 2$ .
(A) $l n 2$ is the correct answer.

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