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

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

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

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

1

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

= 0 - (- 1)

Simplify

= 1

\frac{\partial}{\partial x} (1 - \frac{3}{y} + x)

\int_{1}^{8} 1 + 3 x d x

\frac{d y}{d x} = ka e^{- k x} - b y

\frac{s}{( s ^{2} - 1 ^{2} ) ^{2}}

n = 1 \sum \infty \frac{1}{n ^{3} - n}

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