What is the integral of 4e^{-4t} ?

The integral of 4e^{-4t} is -e^{-4t}+C

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integral of 4e^{-4t}

\int 4 e^{- 4 t} d t

- e^{- 4 t} + C

Solution steps

\int 4 e^{- 4 t} d t

Take the constant out:

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

= 4 \cdot \int e^{- 4 t} d t

Apply u-substitution

: \int - \frac{1}{4} e^{u} d u

= 4 \cdot \int - \frac{1}{4} e^{u} d u

Take the constant out:

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

= 4 (- \frac{1}{4} \cdot \int e^{u} d u)

Use the common integral:

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

= 4 (- \frac{1}{4} e^{u})

Substitute back

u = - 4 t

= 4 (- \frac{1}{4} e^{- 4 t})

Simplify

4 (- \frac{1}{4} e^{- 4 t}) : - e^{- 4 t}

= - e^{- 4 t}

Add a constant to the solution

= - e^{- 4 t} + C

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