What is the integral from-1 to 0 of (xe^y-y) ?

The integral from-1 to 0 of (xe^y-y) is -y-1/2 e^y

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integral from-1 to 0 of (xe^y-y)

\int_{- 1}^{0} (x e^{y} - y) d x

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

Solution steps

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{- 1}^{0} y d x + \int_{- 1}^{0} e^{y} x d x

\int_{- 1}^{0} y d x = y

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

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

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What is the integral from-1 to 0 of (xe^y-y) ?
The integral from-1 to 0 of (xe^y-y) is -y-1/2 e^y