What is the integral from 1 to infinity of 14e^{-7x} ?

The integral from 1 to infinity of 14e^{-7x} is 2/(e^7)

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integral from 1 to infinity of 14e^{-7x}

\int_{1}^{\infty} 14 e^{- 7 x} d x

\frac{2}{e ^{7}}

Decimal Notation

0.00182 \dots

Solution steps

\int_{1}^{\infty} 14 e^{- 7 x} d x

Compute the indefinite integral:

\int 14 e^{- 7 x} d x = - 2 e^{- 7 x} + C

Compute the boundaries:

\int_{1}^{\infty} 14 e^{- 7 x} d x = 0 - (- \frac{2}{e ^{7}})

= 0 - (- \frac{2}{e ^{7}})

Simplify

= \frac{2}{e ^{7}}

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\frac{\partial}{\partial x} (2 x^{2} y + 3 x y^{2})

n = 1 \sum \infty (\frac{- 1}{2})^{n}

\frac{\partial}{\partial x} ((4 x + 4 y) e^{y})

2 e^{- t} cos (2 t)

What is the integral from 1 to infinity of 14e^{-7x} ?
The integral from 1 to infinity of 14e^{-7x} is 2/(e^7)