What is the integral from 0 to infinity of e^{-8x} ?

The integral from 0 to infinity of e^{-8x} is 1/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to infinity of e^{-8x}

\int_{0}^{\infty} e^{- 8 x} d x

\frac{1}{8}

Decimal

0.125

Solution steps

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

= 0 - (- \frac{1}{8})

Simplify

= \frac{1}{8}

y^{^{''}} - 6 y^{^{'}} + 9 y = t^{- 5} e^{3 t}

\frac{d}{d x} (ln (sin (x + y)))

f (x) = 3 x + 6

\frac{d}{d A} (\frac{1 + cot ( A )}{csc ( A )})

f (x) = x^{4} - 3 x^{2} + 5, at x = - 1

What is the integral from 0 to infinity of e^{-8x} ?
The integral from 0 to infinity of e^{-8x} is 1/8