What is the integral from-1 to 2 of xe^{xy} ?

The integral from-1 to 2 of xe^{xy} is e^{2x}-e^{-x}

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integral from-1 to 2 of xe^{xy}

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

e^{2 x} - e^{- x}

Solution steps

\int_{- 1}^{2} 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_{- 1}^{2} e^{x y} d y

Apply u-substitution

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

= x \cdot \int_{- x}^{x \cdot 2} \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_{- x}^{x \cdot 2} e^{u} d u

Use the common integral:

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

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

Simplify

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

= [e^{u}]_{- x}^{2 x}

Compute the boundaries:

e^{2 x} - e^{- x}

= e^{2 x} - e^{- x}

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What is the integral from-1 to 2 of xe^{xy} ?
The integral from-1 to 2 of xe^{xy} is e^{2x}-e^{-x}