What is the integral from 0 to 1 of x^2ye^{xy} ?

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

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integral from 0 to 1 of x^2ye^{xy}

\int_{0}^{1} x^{2} y e^{x y} d x

e^{y} + \frac{- 2 ( y e ^{y} - e ^{y} ) - 2}{y ^{2}}

Solution steps

\int_{0}^{1} x^{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}^{1} e^{y x} x^{2} d x

Apply Integration By Parts

: [e^{y x} \frac{1}{y} x^{2} - \int e^{y x} \frac{2}{y} x d x]_{0}^{1}

= y [e^{y x} \frac{1}{y} x^{2} - \int e^{y x} \frac{2}{y} x d x]_{0}^{1}

\int e^{y x} \frac{2}{y} x d x = \frac{2}{y} (e^{y x} \frac{1}{y} x - e^{y x} \frac{1}{y ^{2}})

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

Compute the boundaries:

e^{y} \frac{1}{y} - \frac{2}{y} (e^{y} \frac{1}{y} - e^{y} \frac{1}{y ^{2}}) - \frac{2}{y ^{3}}

= y (e^{y} \frac{1}{y} - \frac{2}{y} (e^{y} \frac{1}{y} - e^{y} \frac{1}{y ^{2}}) - \frac{2}{y ^{3}})

Simplify

= e^{y} + \frac{- 2 ( y e ^{y} - e ^{y} ) - 2}{y ^{2}}

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