What is the integral from 1 to infinity of 2e^{-x} ?

The integral from 1 to infinity of 2e^{-x} is 2/e

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integral from 1 to infinity of 2e^{-x}

\int_{1}^{\infty} 2 e^{- x} d x

\frac{2}{e}

Decimal Notation

0.73575 \dots

Solution steps

\int_{1}^{\infty} 2 e^{- x} d x

Compute the indefinite integral:

\int 2 e^{- x} d x = - 2 e^{- x} + C

Compute the boundaries:

\int_{1}^{\infty} 2 e^{- x} d x = 0 - (- \frac{2}{e})

= 0 - (- \frac{2}{e})

Simplify

= \frac{2}{e}

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\frac{\partial}{\partial y} (yz e^{x})

\int ln (1 + x^{2}) d x

\frac{\partial}{\partial x} ((2 x + 8) ln (x y))

What is the integral from 1 to infinity of 2e^{-x} ?
The integral from 1 to infinity of 2e^{-x} is 2/e