What is the integral of e^{(-r)/a} ?

The integral of e^{(-r)/a} is -ae^{-r/a}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of e^{(-r)/a}

\int e^{\frac{- r}{a}} d r

- a e^{- \frac{r}{a}} + C

Solution steps

\int e^{\frac{- r}{a}} d r

Simplify

= \int e^{- \frac{r}{a}} d r

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

= - a e^{u}

Substitute back

u = - \frac{r}{a}

= - a e^{- \frac{r}{a}}

Add a constant to the solution

= - a e^{- \frac{r}{a}} + C

\int \frac{1}{8 - 2 x} d x

\int y^{2} + y d y

\int (tan (\frac{x}{3}) + cot (\frac{x}{3}))^{2} d x

\int \frac{8}{x 1 + 2 x} d x

\int \frac{2 x - 1}{( x - 1 ) ^{2}} d x