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

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

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

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

\frac{1}{2}

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{1}{2}

= \frac{1}{2}

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