What is the integral from 4 to infinity of e^{-8x} ?

The integral from 4 to infinity of e^{-8x} is 1/(8e^{32)}

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integral from 4 to infinity of e^{-8x}

\int_{4}^{\infty} e^{- 8 x} d x

\frac{1}{8 e ^{32}}

Decimal

1.58302 E - 15

Solution steps

\int_{4}^{\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_{4}^{\infty} e^{- 8 x} d x = 0 - (- \frac{1}{8 e ^{32}})

= 0 - (- \frac{1}{8 e ^{32}})

Simplify

= \frac{1}{8 e ^{32}}

\frac{2 d y}{d x} + 10 y = 5

\frac{e ^{x} + e ^{- x}}{4}

y^{^{'}} + 2 y = x y^{3}

f (x) = 9 x + 2

t \to 2 lim ((t + 2)^{\frac{3}{2}} (2 t + 4)^{\frac{1}{3}})

What is the integral from 4 to infinity of e^{-8x} ?
The integral from 4 to infinity of e^{-8x} is 1/(8e^{32)}