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

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

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integral from 0 to 1 of e^{-(x+y)}

\int_{0}^{1} e^{- (x + y)} d x

- e^{- y - 1} + e^{- y}

Solution steps

\int_{0}^{1} e^{- (x + y)} d x

Apply u-substitution

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

= \int_{- y}^{- y - 1} - e^{u} d u

Take the constant out:

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

= - \int_{- y}^{- y - 1} e^{u} d u

Use the common integral:

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

= - [e^{u}]_{- y}^{- y - 1}

Compute the boundaries:

e^{- y - 1} - e^{- y}

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

Simplify

= - e^{- y - 1} + e^{- y}

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