What is the integral from 1 to infinity of e^{-x} ?

The integral from 1 to infinity of e^{-x} is 1/e

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integral from 1 to infinity of e^{-x}

\int_{1}^{\infty} e^{- x} d x

\frac{1}{e}

Decimal Notation

0.36787 \dots

Solution steps

\int_{1}^{\infty} e^{- x} d x

Compute the indefinite integral:

\int e^{- x} d x = - e^{- x} + C

Compute the boundaries:

\int_{1}^{\infty} e^{- x} d x = 0 - (- \frac{1}{e})

= 0 - (- \frac{1}{e})

Simplify

= \frac{1}{e}

(x^{x^{x^{x}}})^{^{'}}

(7 x + 6 y) d x + (6 x - 8 y 3) d y = 0

x \to 0 lim (\frac{sin ( 7 x )}{x})

\frac{d}{d x} (1 - tan (x))

\frac{d ^{2}}{d x ^{2}} (cos^{2} (4 π x))

What is the integral from 1 to infinity of e^{-x} ?
The integral from 1 to infinity of e^{-x} is 1/e