What is the integral from 0 to x of 4e^{-4x} ?

The integral from 0 to x of 4e^{-4x} is -e^{-4x}+1

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integral from 0 to x of 4e^{-4x}

\int_{0}^{x} 4 e^{- 4 x} d x

- e^{- 4 x} + 1

Solution steps

\int_{0}^{x} 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_{0}^{x} e^{- 4 x} d x

Apply u-substitution

: \int_{0}^{- 4 x} - \frac{1}{4} e^{u} d u

= 4 \cdot \int_{0}^{- 4 x} - \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_{0}^{- 4 x} e^{u} d u)

Use the common integral:

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

= 4 (- \frac{1}{4} [e^{u}]_{0}^{- 4 x})

Simplify

4 (- \frac{1}{4} [e^{u}]_{0}^{- 4 x}) : - [e^{u}]_{0}^{- 4 x}

= - [e^{u}]_{0}^{- 4 x}

Compute the boundaries:

e^{- 4 x} - 1

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

Simplify

= - e^{- 4 x} + 1

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What is the integral from 0 to x of 4e^{-4x} ?
The integral from 0 to x of 4e^{-4x} is -e^{-4x}+1