What is the integral of e^{12-(w/2)} ?

The integral of e^{12-(w/2)} is -2e^{12-w/2}+C

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integral of e^{12-(w/2)}

\int e^{12 - (\frac{w}{2})} d w

- 2 e^{12 - \frac{w}{2}} + C

Solution steps

\int e^{12 - \frac{w}{2}} d w

e^{12 - \frac{w}{2}} = e^{12} e^{- \frac{w}{2}}

= \int e^{12} e^{- \frac{w}{2}} d w

Take the constant out:

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

= e^{12} \cdot \int e^{- \frac{w}{2}} d w

Apply u-substitution

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

= e^{12} \cdot \int - 2 e^{u} d u

Take the constant out:

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

= e^{12} (- 2 \cdot \int e^{u} d u)

Use the common integral:

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

= e^{12} (- 2 e^{u})

Substitute back

u = - \frac{w}{2}

= e^{12} (- 2 e^{- \frac{w}{2}})

Simplify

e^{12} (- 2 e^{- \frac{w}{2}}) : - 2 e^{12 - \frac{w}{2}}

= - 2 e^{12 - \frac{w}{2}}

Add a constant to the solution

= - 2 e^{12 - \frac{w}{2}} + C

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What is the integral of e^{12-(w/2)} ?
The integral of e^{12-(w/2)} is -2e^{12-w/2}+C