What is the integral from 0 to 1 of xye^{x^2y} ?

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

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integral from 0 to 1 of xye^{x^2y}

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

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

Solution steps

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

Apply u-substitution

: \int_{0}^{y} \frac{e ^{u}}{2 y} d u

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

Take the constant out:

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

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

Use the common integral:

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

= y \frac{1}{2 y} [e^{u}]_{0}^{y}

Simplify

y \frac{1}{2 y} [e^{u}]_{0}^{y} : \frac{1}{2} [e^{u}]_{0}^{y}

= \frac{1}{2} [e^{u}]_{0}^{y}

Compute the boundaries:

e^{y} - 1

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

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