What is the integral from-infinity to-1 of e^{-38t} ?

The integral from-infinity to-1 of e^{-38t} is diverges

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integral from-infinity to-1 of e^{-38t}

\int_{- \infty}^{- 1} e^{- 38 t} d t

diverges

Solution steps

\int_{- \infty}^{- 1} e^{- 38 t} d t

Compute the indefinite integral:

\int e^{- 38 t} d t = - \frac{1}{38} e^{- 38 t} + C

Compute the boundaries:

\int_{- \infty}^{- 1} e^{- 38 t} d t =

diverges

= diverges

n = 1 \sum \infty \frac{5}{n ^{2} + 16}

s^{2} + 6 s + 7

x \to \frac{π}{2} lim (x)

y = \frac{7 ln ( x )}{3 x + 4}

x \to \frac{1}{3} lim (3 x^{3} + 2 x^{2})

What is the integral from-infinity to-1 of e^{-38t} ?
The integral from-infinity to-1 of e^{-38t} is diverges