What is the integral from y to 1 of (ye^x-y) ?

The integral from y to 1 of (ye^x-y) is -y+y^2+ey-ye^y

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integral from y to 1 of (ye^x-y)

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

- y + y^{2} + ey - y e^{y}

Solution steps

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

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

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

= - (y - y^{2}) + y (e - e^{y})

Simplify

- (y - y^{2}) + y (e - e^{y}) : - y + y^{2} + ey - y e^{y}

= - y + y^{2} + ey - y e^{y}

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What is the integral from y to 1 of (ye^x-y) ?
The integral from y to 1 of (ye^x-y) is -y+y^2+ey-ye^y