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

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

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

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

\frac{1}{3 e ^{3}}

Decimal Notation

0.01659 \dots

Solution steps

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

Compute the indefinite integral:

\int e^{- 3 x} d x = - \frac{1}{3} e^{- 3 x} + C

Compute the boundaries:

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

= 0 - (- \frac{1}{3 e ^{3}})

Simplify

= \frac{1}{3 e ^{3}}

\int_{3}^{\infty} \frac{1}{x - 2} d x

\int x^{2} + 2 x + 3 d x

\frac{d}{d x} ((x^{2} + y^{2})^{6})

\frac{\partial}{\partial v} (\frac{u + v}{2})

y^{^{''''}} - 9 y^{^{''}} + 20 y = 0

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