What is the integral from 0 to infinity of e^{-9x} ?

The integral from 0 to infinity of e^{-9x} is 1/9

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integral from 0 to infinity of e^{-9x}

\int_{0}^{\infty} e^{- 9 x} d x

\frac{1}{9}

Decimal Notation

0.11111 \dots

Solution steps

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

= 0 - (- \frac{1}{9})

Simplify

= \frac{1}{9}

(\frac{t ^{3}}{t ^{6} + 5})^{2}

\frac{d}{d x} (\frac{e ^{2 x}}{1 + sin ( π x )})

x \to 3 - lim (- \frac{1}{x - 3})

\int_{1}^{3} (x^{2} + 1) d x

y^{^{''}} + 3 y^{^{'}} = 6 e^{- 3 t}, y (0) = 8, y^{^{'}} (0) = - 4

What is the integral from 0 to infinity of e^{-9x} ?
The integral from 0 to infinity of e^{-9x} is 1/9