What is the integral from 1 to infinity of 5e^{-5x} ?

The integral from 1 to infinity of 5e^{-5x} is 1/(e^5)

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integral from 1 to infinity of 5e^{-5x}

\int_{1}^{\infty} 5 e^{- 5 x} d x

\frac{1}{e ^{5}}

Decimal Notation

0.00673 \dots

Solution steps

\int_{1}^{\infty} 5 e^{- 5 x} d x

Compute the indefinite integral:

\int 5 e^{- 5 x} d x = - e^{- 5 x} + C

Compute the boundaries:

\int_{1}^{\infty} 5 e^{- 5 x} d x = 0 - (- \frac{1}{e ^{5}})

= 0 - (- \frac{1}{e ^{5}})

Simplify

= \frac{1}{e ^{5}}

\int e^{- x^{2}} d x

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

\frac{d}{d x} (x^{4} + 4 x^{3} - 2 x^{2} - 12 x)

\frac{d}{d x} (1 0^{- 2 x})

What is the integral from 1 to infinity of 5e^{-5x} ?
The integral from 1 to infinity of 5e^{-5x} is 1/(e^5)