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

The integral from-infinity to-1 of e^{-14t} is diverges

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

\int_{- \infty}^{- 1} e^{- 14 t} d t

diverges

Solution steps

\int_{- \infty}^{- 1} e^{- 14 t} d t

Compute the indefinite integral:

\int e^{- 14 t} d t = - \frac{1}{14} e^{- 14 t} + C

Compute the boundaries:

\int_{- \infty}^{- 1} e^{- 14 t} d t =

diverges

= diverges

\int 6 x^{2} - 5 d x

y = \frac{7}{x}

\int_{9}^{15} 2 cos (\frac{π x}{6}) d x

\frac{d t}{d t}

y = x^{2} - 43, y = 5 - 2 x

What is the integral from-infinity to-1 of e^{-14t} ?
The integral from-infinity to-1 of e^{-14t} is diverges