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

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

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

\int_{0}^{1} (- y - \frac{1}{2} e^{y}) d y

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

Solution steps

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

Apply the Sum Rule:

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

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

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

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

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

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