What is the integral of 2e^{-x/4} ?

The integral of 2e^{-x/4} is -8e^{-x/4}+C

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integral of 2e^{-x/4}

\int 2 e^{- \frac{x}{4}} d x

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

Solution steps

\int 2 e^{- \frac{x}{4}} d x

Take the constant out:

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

= 2 \cdot \int e^{- \frac{x}{4}} d x

Apply u-substitution

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

= 2 \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

= 2 (- 4 \cdot \int e^{u} d u)

Use the common integral:

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

= 2 (- 4 e^{u})

Substitute back

u = - \frac{x}{4}

= 2 (- 4 e^{- \frac{x}{4}})

Simplify

= - 8 e^{- \frac{x}{4}}

Add a constant to the solution

= - 8 e^{- \frac{x}{4}} + C

x \to 9 + lim (6)

y^{^{'}} = y - x + 1

\int b^{4 x} d x

\frac{d}{d x} (\frac{2 x}{1 + x ^{2}})

\frac{\partial}{\partial x} ((x^{2} - y^{2})^{0.5})