What is the integral of 4e^{-0.1x} ?

The integral of 4e^{-0.1x} is -40e^{-0.1x}+C

Popular >

integral of 4e^{-0.1x}

\int 4 e^{- 0.1 x} d x

- 40 e^{- 0.1 x} + C

Solution steps

\int 4 e^{- 0.1 x} d x

Take the constant out:

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

= 4 \cdot \int e^{- 0.1 x} d x

Apply u-substitution

: \int - \frac{1}{0.1} e^{u} d u

= 4 \cdot \int - \frac{1}{0.1} e^{u} d u

Take the constant out:

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

= 4 (- \frac{1}{0.1} \cdot \int e^{u} d u)

Simplify

= 4 (- 10 \cdot \int e^{u} d u)

Use the common integral:

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

= 4 (- 10 e^{u})

Substitute back

u = - 0.1 x

= 4 (- 10 e^{- 0.1 x})

Simplify

= - 40 e^{- 0.1 x}

Add a constant to the solution

= - 40 e^{- 0.1 x} + C

3 sin (2 x), - 2 π, π

x \to 0 lim (\frac{( 3 arccos ( \frac{1}{2} + x ) - 3 arccos ( \frac{1}{2} ) )}{x})

\frac{d}{d x} (e^{- x^{7}})

(ln (u))^{^{'}}

\frac{x}{1 + x}

What is the integral of 4e^{-0.1x} ?
The integral of 4e^{-0.1x} is -40e^{-0.1x}+C