What is the integral from 0 to y of 2e^{-2x} ?

The integral from 0 to y of 2e^{-2x} is -e^{-2y}+1

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integral from 0 to y of 2e^{-2x}

\int_{0}^{y} 2 e^{- 2 x} d x

- e^{- 2 y} + 1

Solution steps

\int_{0}^{y} 2 e^{- 2 x} d x

Take the constant out:

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

= 2 \cdot \int_{0}^{y} e^{- 2 x} d x

Apply u-substitution

: \int_{0}^{- 2 y} - \frac{1}{2} e^{u} d u

= 2 \cdot \int_{0}^{- 2 y} - \frac{1}{2} e^{u} d u

Take the constant out:

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

= 2 (- \frac{1}{2} \cdot \int_{0}^{- 2 y} e^{u} d u)

Use the common integral:

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

= 2 (- \frac{1}{2} [e^{u}]_{0}^{- 2 y})

Simplify

2 (- \frac{1}{2} [e^{u}]_{0}^{- 2 y}) : - [e^{u}]_{0}^{- 2 y}

= - [e^{u}]_{0}^{- 2 y}

Compute the boundaries:

e^{- 2 y} - 1

= - (e^{- 2 y} - 1)

Simplify

= - e^{- 2 y} + 1

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What is the integral from 0 to y of 2e^{-2x} ?
The integral from 0 to y of 2e^{-2x} is -e^{-2y}+1