What is the integral from 4 to infinity of e^{-9x} ?

The integral from 4 to infinity of e^{-9x} is 1/(9e^{36)}

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integral from 4 to infinity of e^{-9x}

\int_{4}^{\infty} e^{- 9 x} d x

\frac{1}{9 e ^{36}}

Decimal

2.57725 E - 17

Solution steps

\int_{4}^{\infty} e^{- 9 x} d x

Compute the indefinite integral:

\int e^{- 9 x} d x = - \frac{1}{9} e^{- 9 x} + C

Compute the boundaries:

\int_{4}^{\infty} e^{- 9 x} d x = 0 - (- \frac{1}{9 e ^{36}})

= 0 - (- \frac{1}{9 e ^{36}})

Simplify

= \frac{1}{9 e ^{36}}

\int 16 - 9 x^{2} d x

\frac{\partial}{\partial x} (3 x + y^{2})

\frac{d}{d x} (\frac{x}{a ^{2} - x ^{2}})

\frac{d}{d x} (x^{3} - 3 x^{2} + 3 x)

f (x) = x + e^{- 2 x}

What is the integral from 4 to infinity of e^{-9x} ?
The integral from 4 to infinity of e^{-9x} is 1/(9e^{36)}