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

The integral from-infinity to-1 of e^{-8t} is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int_{- \infty}^{- 1} e^{- 8 t} d t

diverges

Solution steps

\int_{- \infty}^{- 1} e^{- 8 t} d t

Compute the indefinite integral:

\int e^{- 8 t} d t = - \frac{1}{8} e^{- 8 t} + C

Compute the boundaries:

\int_{- \infty}^{- 1} e^{- 8 t} d t =

diverges

= diverges

y = x^{2} - 3 x, x = 0, x = 7, y = 4 x

\frac{\partial}{\partial y} (x y + yz + x z)

x \to (\frac{π}{2}) - lim (tan (x))

\frac{\partial}{\partial x} (x^{m} y^{n})

\int \frac{x}{3 + x} d x

What is the integral from-infinity to-1 of e^{-8t} ?
The integral from-infinity to-1 of e^{-8t} is diverges