What is the integral from 0 to 1 of ye^{xy} ?

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

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integral from 0 to 1 of ye^{xy}

\int_{0}^{1} y e^{x y} d x

e^{y} - 1

Solution steps

\int_{0}^{1} 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}^{1} e^{y x} d x

Apply u-substitution

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

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

Use the common integral:

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

= y \frac{1}{y} [e^{u}]_{0}^{y}

Simplify

y \frac{1}{y} [e^{u}]_{0}^{y} : [e^{u}]_{0}^{y}

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

Compute the boundaries:

e^{y} - 1

= e^{y} - 1

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