What is the integral from 2 to infinity of e^{-8x} ?

The integral from 2 to infinity of e^{-8x} is 1/(8e^{16)}

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integral from 2 to infinity of e^{-8x}

\int_{2}^{\infty} e^{- 8 x} d x

\frac{1}{8 e ^{16}}

Decimal Notation

1.40669 E - 8

Solution steps

\int_{2}^{\infty} e^{- 8 x} d x

Compute the indefinite integral:

\int e^{- 8 x} d x = - \frac{1}{8} e^{- 8 x} + C

Compute the boundaries:

\int_{2}^{\infty} e^{- 8 x} d x = 0 - (- \frac{1}{8 e ^{16}})

= 0 - (- \frac{1}{8 e ^{16}})

Simplify

= \frac{1}{8 e ^{16}}

y^{^{'}} = e^{x} y

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

x \to - \infty lim (x - e^{x})

\int e^{- x} d x

y = 3 x^{2} + 5 x

What is the integral from 2 to infinity of e^{-8x} ?
The integral from 2 to infinity of e^{-8x} is 1/(8e^{16)}