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

The integral from 0 to 1 of e^x(x+1) is e

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

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

e

Decimal Notation

2.71828 \dots

Solution steps

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

Apply Integration By Parts

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

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

\int e^{x} d x = e^{x}

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

Expand

e^{x} (x + 1) - e^{x} : e^{x} x

= [e^{x} x]_{0}^{1}

Compute the boundaries:

e

= e

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