What is the integral from 4 to infinity of e^{-3x} ?

The integral from 4 to infinity of e^{-3x} is 1/(3e^{12)}

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integral from 4 to infinity of e^{-3x}

\int_{4}^{\infty} e^{- 3 x} d x

\frac{1}{3 e ^{12}}

Decimal Notation

2.04807 E - 6

Solution steps

\int_{4}^{\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_{4}^{\infty} e^{- 3 x} d x = 0 - (- \frac{1}{3 e ^{12}})

= 0 - (- \frac{1}{3 e ^{12}})

Simplify

= \frac{1}{3 e ^{12}}

\frac{\partial}{\partial x} (\frac{x}{4})

\frac{5}{s ( s + 2 )}

\frac{d}{d x} (524.8 x^{0.75})

(x^{3} - 2 x + 4), 2

\int (\frac{1}{2 x ^{2} - 5 x + 7} + (2 x + 3) e^{2 x}) d x

What is the integral from 4 to infinity of e^{-3x} ?
The integral from 4 to infinity of e^{-3x} is 1/(3e^{12)}