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

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

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

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

\frac{1}{e ^{2}}

Decimal

0.13533 \dots

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

= 0 - (- \frac{1}{e ^{2}})

Simplify

= \frac{1}{e ^{2}}

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