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

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

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

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

(e - 1)^{2}

Decimal

2.95249 \dots

Solution steps

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

Take the constant out:

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

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

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

= (e - 1) (e - 1)

Simplify

(e - 1) (e - 1) : (e - 1)^{2}

= (e - 1)^{2}

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