What is the integral from 1 to infinity of 8e^{-4x} ?

The integral from 1 to infinity of 8e^{-4x} is 2/(e^4)

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integral from 1 to infinity of 8e^{-4x}

\int_{1}^{\infty} 8 e^{- 4 x} d x

\frac{2}{e ^{4}}

Decimal

0.03663 \dots

Solution steps

\int_{1}^{\infty} 8 e^{- 4 x} d x

Compute the indefinite integral:

\int 8 e^{- 4 x} d x = - 2 e^{- 4 x} + C

Compute the boundaries:

\int_{1}^{\infty} 8 e^{- 4 x} d x = 0 - (- \frac{2}{e ^{4}})

= 0 - (- \frac{2}{e ^{4}})

Simplify

= \frac{2}{e ^{4}}

y = cot^{2} (5 x)

f (x) = \frac{4}{x ^{3}}

(x^{11} - 1)^{3}

y = 6 e^{x} cos (x)

\int 1^{- x} d x

What is the integral from 1 to infinity of 8e^{-4x} ?
The integral from 1 to infinity of 8e^{-4x} is 2/(e^4)