What is the integral of e^{-1/3 t} ?

The integral of e^{-1/3 t} is -3e^{-t/3}+C

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integral of e^{-1/3 t}

\int e^{- \frac{1}{3} t} d t

- 3 e^{- \frac{t}{3}} + C

Solution steps

\int e^{- \frac{1}{3} t} d t

Simplify

- \frac{1}{3} t : - \frac{t}{3}

= \int e^{- \frac{t}{3}} d t

Apply u-substitution

: \int - 3 e^{u} d u

= \int - 3 e^{u} d u

Take the constant out:

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

= - 3 \cdot \int e^{u} d u

Use the common integral:

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

= - 3 e^{u}

Substitute back

u = - \frac{t}{3}

= - 3 e^{- \frac{t}{3}}

Add a constant to the solution

= - 3 e^{- \frac{t}{3}} + C

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