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

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

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integral from 0 to 2 of ye^{-xy}

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

- e^{- 2 y} + 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

Apply u-substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Simplify

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

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