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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Compute the boundaries:

e^{2} - 1

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

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