What is the integral from-infinity to-1 of e^{-28t} ?

The integral from-infinity to-1 of e^{-28t} is diverges

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integral from-infinity to-1 of e^{-28t}

\int_{- \infty}^{- 1} e^{- 28 t} d t

diverges

Solution steps

\int_{- \infty}^{- 1} e^{- 28 t} d t

Compute the indefinite integral:

\int e^{- 28 t} d t = - \frac{1}{28} e^{- 28 t} + C

Compute the boundaries:

\int_{- \infty}^{- 1} e^{- 28 t} d t =

diverges

= diverges

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y = \frac{1}{( x + 1 ) ^{2}}

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What is the integral from-infinity to-1 of e^{-28t} ?
The integral from-infinity to-1 of e^{-28t} is diverges