What is the integral of (e^{5-(3y)/7}) ?

The integral of (e^{5-(3y)/7}) is -7/3 e^{5-(3y)/7}+C

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integral of (e^{5-(3y)/7})

\int (e^{5 - \frac{3 y}{7}}) d y

- \frac{7}{3} e^{5 - \frac{3 y}{7}} + C

Solution steps

\int e^{5 - \frac{3 y}{7}} d y

e^{5 - \frac{3 y}{7}} = e^{5} e^{- \frac{3 y}{7}}

= \int e^{5} e^{- \frac{3 y}{7}} d y

Take the constant out:

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

= e^{5} \cdot \int e^{- \frac{3 y}{7}} d y

Apply u-substitution

= e^{5} \cdot \int - \frac{7}{3} e^{u} d u

Take the constant out:

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

= e^{5} (- \frac{7}{3} \cdot \int e^{u} d u)

Use the common integral:

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

= e^{5} (- \frac{7}{3} e^{u})

Substitute back

u = - \frac{3 y}{7}

= e^{5} (- \frac{7}{3} e^{- \frac{3 y}{7}})

Simplify

e^{5} (- \frac{7}{3} e^{- \frac{3 y}{7}}) : - \frac{7}{3} e^{5 - \frac{3 y}{7}}

= - \frac{7}{3} e^{5 - \frac{3 y}{7}}

Add a constant to the solution

= - \frac{7}{3} e^{5 - \frac{3 y}{7}} + C

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What is the integral of (e^{5-(3y)/7}) ?
The integral of (e^{5-(3y)/7}) is -7/3 e^{5-(3y)/7}+C