What is the integral from 1 to infinity of 9e^{-3x} ?

The integral from 1 to infinity of 9e^{-3x} is 3/(e^3)

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integral from 1 to infinity of 9e^{-3x}

\int_{1}^{\infty} 9 e^{- 3 x} d x

\frac{3}{e ^{3}}

Decimal

0.14936 \dots

Solution steps

\int_{1}^{\infty} 9 e^{- 3 x} d x

Compute the indefinite integral:

\int 9 e^{- 3 x} d x = - 3 e^{- 3 x} + C

Compute the boundaries:

\int_{1}^{\infty} 9 e^{- 3 x} d x = 0 - (- \frac{3}{e ^{3}})

= 0 - (- \frac{3}{e ^{3}})

Simplify

= \frac{3}{e ^{3}}

What is the integral from 1 to infinity of 9e^{-3x} ?
The integral from 1 to infinity of 9e^{-3x} is 3/(e^3)