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

The integral from 0 to 1 of xe^{xy} is e^x-1

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integral from 0 to 1 of xe^{xy}

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

e^{x} - 1

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Simplify

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

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

Compute the boundaries:

e^{x} - 1

= e^{x} - 1

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