What is the integral of (e^{5-2t}) ?

The integral of (e^{5-2t}) is -1/2 e^{5-2t}+C

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integral of (e^{5-2t})

\int (e^{5 - 2 t}) d t

- \frac{1}{2} e^{5 - 2 t} + C

Solution steps

\int e^{5 - 2 t} d t

Apply u-substitution

: \int - \frac{1}{2} e^{u} d u

= \int - \frac{1}{2} e^{u} d u

Take the constant out:

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

= - \frac{1}{2} \cdot \int e^{u} d u

Use the common integral:

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

= - \frac{1}{2} e^{u}

Substitute back

u = 5 - 2 t

= - \frac{1}{2} e^{5 - 2 t}

Add a constant to the solution

= - \frac{1}{2} e^{5 - 2 t} + C

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What is the integral of (e^{5-2t}) ?
The integral of (e^{5-2t}) is -1/2 e^{5-2t}+C