What is the integral from 1 to infinity of 12e^{-4x} ?

The integral from 1 to infinity of 12e^{-4x} is 3/(e^4)

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integral from 1 to infinity of 12e^{-4x}

\int_{1}^{\infty} 12 e^{- 4 x} d x

\frac{3}{e ^{4}}

Solution steps

\int_{1}^{\infty} 12 e^{- 4 x} d x

Compute the indefinite integral:

\int 12 e^{- 4 x} d x = - 3 e^{- 4 x} + C

Compute the boundaries:

\int_{1}^{\infty} 12 e^{- 4 x} d x = 0 - (- \frac{3}{e ^{4}})

= 0 - (- \frac{3}{e ^{4}})

Simplify

= \frac{3}{e ^{4}}

(\frac{t ^{3}}{t ^{8} + 7})^{2}

\int (x + 1) ln (2 x + 4) d x

\frac{d}{d x} ((\frac{1}{x ^{2}} - \frac{3}{x ^{4}}) (x + 7 x^{3}))

\frac{d y}{d x} = x (9 - y)

\frac{d}{d x} (\frac{1 + 3 x}{1 - 3 x})

What is the integral from 1 to infinity of 12e^{-4x} ?
The integral from 1 to infinity of 12e^{-4x} is 3/(e^4)