What is the integral of 5e^{-0.2x} ?

The integral of 5e^{-0.2x} is -25e^{-0.2x}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 5e^{-0.2x}

\int 5 e^{- 0.2 x} d x

- 25 e^{- 0.2 x} + C

Solution steps

\int 5 e^{- 0.2 x} d x

Take the constant out:

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

= 5 \cdot \int e^{- 0.2 x} d x

Apply u-substitution

: \int - \frac{1}{0.2} e^{u} d u

= 5 \cdot \int - \frac{1}{0.2} e^{u} d u

Take the constant out:

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

= 5 (- \frac{1}{0.2} \cdot \int e^{u} d u)

Simplify

= 5 (- 5 \cdot \int e^{u} d u)

Use the common integral:

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

= 5 (- 5 e^{u})

Substitute back

u = - 0.2 x

= 5 (- 5 e^{- 0.2 x})

Simplify

= - 25 e^{- 0.2 x}

Add a constant to the solution

= - 25 e^{- 0.2 x} + C

\frac{\partial}{\partial x} (ln (x - e^{x} y))

y^{^{''}} + 16 y = 8 sin (4 t)

\int \frac{1}{x ^{2} + 4 x + 9} d x

y (t + 5) (t + 2) y^{^{'}} + t + 3 = 0

y = (x - 10)^{- 3}

What is the integral of 5e^{-0.2x} ?
The integral of 5e^{-0.2x} is -25e^{-0.2x}+C