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

The integral of-4e^{-x/4} is 16e^{-x/4}+C

Popular >

integral of-4e^{-x/4}

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

16 e^{- \frac{x}{4}} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int - 4 e^{u} d u

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

Take the constant out:

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

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

Use the common integral:

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

= - 4 (- 4 e^{u})

Substitute back

u = - \frac{x}{4}

= - 4 (- 4 e^{- \frac{x}{4}})

Simplify

= 16 e^{- \frac{x}{4}}

Add a constant to the solution

= 16 e^{- \frac{x}{4}} + C

(x, y) \to (2, 1) lim (\frac{x - 2 y}{x ^{3} - 8 y ^{3}})

\frac{\partial}{\partial y} (y^{4} sin (6 x))

\int \frac{1}{u ( u + 2 )} d u

2 x^{2} + x - 1

y^{^{''}} + 4 y = 4 cos (2 t)