What is the integral from 0 to y of 6e^{-x}e^{-2y} ?

The integral from 0 to y of 6e^{-x}e^{-2y} is -6e^{-2y}(e^{-y}-1)

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integral from 0 to y of 6e^{-x}e^{-2y}

\int_{0}^{y} 6 e^{- x} e^{- 2 y} d x

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

Solution steps

\int_{0}^{y} 6 e^{- x} e^{- 2 y} d x

Take the constant out:

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

= 6 e^{- 2 y} \cdot \int_{0}^{y} e^{- x} d x

Apply u-substitution

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

= 6 e^{- 2 y} \cdot \int_{0}^{- y} - e^{u} d u

Take the constant out:

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

= 6 e^{- 2 y} (- \int_{0}^{- y} e^{u} d u)

Use the common integral:

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

= 6 e^{- 2 y} (- [e^{u}]_{0}^{- y})

Simplify

= - 6 e^{- 2 y} [e^{u}]_{0}^{- y}

Compute the boundaries:

e^{- y} - 1

= - 6 e^{- 2 y} (e^{- y} - 1)

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What is the integral from 0 to y of 6e^{-x}e^{-2y} ?
The integral from 0 to y of 6e^{-x}e^{-2y} is -6e^{-2y}(e^{-y}-1)