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

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

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integral of 2e^{4-4x}

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

- \frac{e ^{4}}{2} e^{- 4 x} + C

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

Simplify

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

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

Add a constant to the solution

= - \frac{e ^{4}}{2} e^{- 4 x} + C

What is the integral of 2e^{4-4x} ?
The integral of 2e^{4-4x} is -(e^4)/2 e^{-4x}+C