What is the integral of+4e^{-7x} ?

The integral of+4e^{-7x} is -4/7 e^{-7x}+C

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integral of+4e^{-7x}

\int + 4 e^{- 7 x} d x

- \frac{4}{7} e^{- 7 x} + C

Solution steps

\int 4 e^{- 7 x} d x

Take the constant out:

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

= 4 \cdot \int e^{- 7 x} d x

Apply u-substitution

: \int - \frac{1}{7} e^{u} d u

= 4 \cdot \int - \frac{1}{7} e^{u} d u

Take the constant out:

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

= 4 (- \frac{1}{7} \cdot \int e^{u} d u)

Use the common integral:

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

= 4 (- \frac{1}{7} e^{u})

Substitute back

u = - 7 x

= 4 (- \frac{1}{7} e^{- 7 x})

Simplify

4 (- \frac{1}{7} e^{- 7 x}) : - \frac{4}{7} e^{- 7 x}

= - \frac{4}{7} e^{- 7 x}

Add a constant to the solution

= - \frac{4}{7} e^{- 7 x} + C

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