What is the integral from 0 to 1 of y+xe^y ?

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

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integral from 0 to 1 of y+xe^y

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

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

Solution steps

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

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

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

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

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