What is the integral from-infinity to e of 16e^{-x} ?

The integral from-infinity to e of 16e^{-x} is diverges

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integral from-infinity to e of 16e^{-x}

\int_{- \infty}^{e} 16 e^{- x} d x

diverges

Solution steps

\int_{- \infty}^{e} 16 e^{- x} d x

Compute the indefinite integral:

\int 16 e^{- x} d x = - 16 e^{- x} + C

Compute the boundaries:

\int_{- \infty}^{e} 16 e^{- x} d x =

diverges

= diverges

\int_{- 1}^{0} x sec^{2} (x) d x

\int_{0}^{2} (\frac{x ^{2}}{2} + 2) d x

\int_{0}^{3} 0.5 e^{- 0.5 x} d x

\int_{1}^{5} \frac{x ^{5} - x}{3 x ^{2}} d x

\int_{0}^{\frac{π}{3}} sec^{5} (x) d x

What is the integral from-infinity to e of 16e^{-x} ?
The integral from-infinity to e of 16e^{-x} is diverges