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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Simplify

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

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

Compute the boundaries:

e^{x} - 1

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

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