What is the integral from-infinity to-1 of e^{-18t} ?

The integral from-infinity to-1 of e^{-18t} is diverges

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integral from-infinity to-1 of e^{-18t}

\int_{- \infty}^{- 1} e^{- 18 t} d t

diverges

Solution steps

\int_{- \infty}^{- 1} e^{- 18 t} d t

Compute the indefinite integral:

\int e^{- 18 t} d t = - \frac{1}{18} e^{- 18 t} + C

Compute the boundaries:

\int_{- \infty}^{- 1} e^{- 18 t} d t =

diverges

= diverges

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x \to \frac{π}{4} lim (sin (x))

What is the integral from-infinity to-1 of e^{-18t} ?
The integral from-infinity to-1 of e^{-18t} is diverges