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

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

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

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

9

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

= 0 - (- 9)

Simplify

= 9

\frac{\partial}{\partial x} ((x - 1)^{2} + x y + (y - 1)^{2})

\int \frac{2 x ^{2} - 3 x - 36}{( 2 x - 1 ) ( x ^{2} + 9 )} d x

\frac{d}{d x} (1 - ln (x))

\int \frac{1}{3 + x ^{2}} d x

x \to 3 lim (\frac{x + 1}{x - 2})

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