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

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

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

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

\frac{1}{2}

Decimal

0.5

Solution steps

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

Apply Integration By Parts

: [y (e^{y} - y) - \int e^{y} - y d y]_{0}^{1}

= [y (e^{y} - y) - \int e^{y} - y d y]_{0}^{1}

\int e^{y} - y d y = e^{y} - \frac{y ^{2}}{2}

= [y (e^{y} - y) - (e^{y} - \frac{y ^{2}}{2})]_{0}^{1}

Simplify

[y (e^{y} - y) - (e^{y} - \frac{y ^{2}}{2})]_{0}^{1} : [\frac{1}{2} (- y^{2} + 2 e^{y} y - 2 e^{y})]_{0}^{1}

= [\frac{1}{2} (- y^{2} + 2 e^{y} y - 2 e^{y})]_{0}^{1}

Compute the boundaries:

\frac{1}{2}

= \frac{1}{2}

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