What is the integral of 8e^{-t} ?

The integral of 8e^{-t} is -8e^{-t}+C

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integral of 8e^{-t}

\int 8 e^{- t} d t

- 8 e^{- t} + C

Solution steps

\int 8 e^{- t} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \int e^{- t} d t

Apply u-substitution

= 8 \cdot \int - e^{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 (- \int e^{u} d u)

Use the common integral:

\int e^{u} d u = e^{u}

= 8 (- e^{u})

Substitute back

u = - t

= 8 (- e^{- t})

Simplify

= - 8 e^{- t}

Add a constant to the solution

= - 8 e^{- t} + C

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