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

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

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

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

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

Decimal Notation

0.04978 \dots

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

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

Simplify

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

\int x^{3} \cdot e^{4 x} d x

\frac{d}{d y} (x^{2} y^{3})

\int - x e^{x^{2}} d x

x = sec (\frac{1}{y})

\int - 4 x^{3} d x

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