What is the integral of e^{(-t)/2} ?

The integral of e^{(-t)/2} is -2e^{-t/2}+C

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integral of e^{(-t)/2}

\int e^{\frac{- t}{2}} d t

- 2 e^{- \frac{t}{2}} + C

Solution steps

\int e^{\frac{- t}{2}} d t

Simplify

= \int e^{- \frac{t}{2}} d t

Apply u-substitution

: \int - 2 e^{u} d u

= \int - 2 e^{u} d u

Take the constant out:

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

= - 2 \cdot \int e^{u} d u

Use the common integral:

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

= - 2 e^{u}

Substitute back

u = - \frac{t}{2}

= - 2 e^{- \frac{t}{2}}

Add a constant to the solution

= - 2 e^{- \frac{t}{2}} + C

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