What is the integral of e^{-bx} ?

The integral of e^{-bx} is -1/b e^{-bx}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of e^{-bx}

\int e^{- b x} d x

- \frac{1}{b} e^{- b x} + C

Solution steps

\int e^{- b x} d x

Apply u-substitution

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

= \int - \frac{e ^{u}}{b} d u

Take the constant out:

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

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

Use the common integral:

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

= - \frac{1}{b} e^{u}

Substitute back

u = - b x

= - \frac{1}{b} e^{- b x}

Add a constant to the solution

= - \frac{1}{b} e^{- b x} + C

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

\int \frac{5 x ^{2} + 2 x - 5}{x ^{3} - x} d x

f (x) = lo g_{10} (1 + x^{2})

y^{^{''}} + 2 y^{^{'}} + 2 y = sin (3 x)

\frac{\partial}{\partial x} (6 y - 5 x^{2} e^{5 y})