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

The integral from y to 1 of y(e^x-1) is y(-1+y+e-e^y)

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

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

y (- 1 + y + e - e^{y})

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

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

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

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

= y (- (1 - y) + e - e^{y})

Simplify

= y (- 1 + y + e - e^{y})

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