What is the integral of e^{-0.02x} ?

The integral of e^{-0.02x} is -50e^{-0.02x}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of e^{-0.02x}

\int e^{- 0.02 x} d x

- 50 e^{- 0.02 x} + C

Solution steps

\int e^{- 0.02 x} d x

Apply u-substitution

: \int - \frac{1}{0.02} e^{u} d u

= \int - \frac{1}{0.02} e^{u} d u

Take the constant out:

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

= - \frac{1}{0.02} \cdot \int e^{u} d u

Simplify

= - 50 \cdot \int e^{u} d u

Use the common integral:

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

= - 50 e^{u}

Substitute back

u = - 0.02 x

= - 50 e^{- 0.02 x}

Add a constant to the solution

= - 50 e^{- 0.02 x} + C

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

t^{2} \frac{d y}{d t} - t = 1 + y + y t

y^{^{''}} = ey

\frac{d}{d x} (e^{x + 5})

y = ln (x - 1), y = x - 2, x = 3

What is the integral of e^{-0.02x} ?
The integral of e^{-0.02x} is -50e^{-0.02x}+C