What is the integral of ae^{-ax} ?

The integral of ae^{-ax} is -e^{-ax}+C

Popular >

integral of ae^{-ax}

\int a e^{- a x} d x

- e^{- a x} + C

Solution steps

\int a e^{- a x} d x

Take the constant out:

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

= a \cdot \int e^{- a x} d x

Apply u-substitution

: \int - \frac{e ^{u}}{a} d u

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = - a x

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

Simplify

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

= - e^{- a x}

Add a constant to the solution

= - e^{- a x} + C

(x^{3} cos (x))^{^{'}}

\int \frac{1}{16 x ^{4} - 8 x ^{2} + 1} d x

n = 0 \sum \infty \frac{4 n}{8 n + 5}

\int_{1}^{2} \frac{ln ( x )}{x ^{3}} d x

\int \frac{1}{( \frac{1}{e ^{x}} + 1 )} d x