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

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

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integral of 4e^{-4x}

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

- e^{- 4 x} + C

Solution steps

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

Apply u-substitution

: \int - \frac{1}{4} e^{u} d u

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

Use the common integral:

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

= 4 (- \frac{1}{4} e^{u})

Substitute back

u = - 4 x

= 4 (- \frac{1}{4} e^{- 4 x})

Simplify

4 (- \frac{1}{4} e^{- 4 x}) : - e^{- 4 x}

= - e^{- 4 x}

Add a constant to the solution

= - e^{- 4 x} + C

\frac{x}{1 - ln ( x - 5 )}

\int_{0}^{\frac{5 π}{4}} 3 tan (\frac{x}{5}) d x

n = 0 \sum \infty \frac{1}{n ^{2} + 1}

\frac{d y}{d x}, y = 7^{x}

n = 0 \sum \infty \frac{1 ^{n}}{4 ^{n}}