∫_0^∞ e^ xdx = [e^ x/ (1)] _0^∞ = [ e^ x] _0^∞ =[ e^ ∞ ( e^ 0)] = 0+e^0 = 1 #160; [∵∫ e^x.dx = e^x] ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Property 5

Evaluate: ∫0...

Question

Evaluate:
$\int_{0}^{\infty} e^{- x} d x$

Open in App

Solution

$\int_{0}^{\infty} e^{- x} d x = [\frac{e^{- x}}{- (1)}]_{0}^{\infty} = [- e^{- x}]_{0}^{\infty} = [- e^{- \infty} - (- e^{- 0})]$
$= 0 + e^{0}$
$= 1$
$[∵ \int e^{x} . d x = e^{x}]$

Suggest Corrections