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

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

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

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

e + \frac{1}{e}

Decimal

3.08616 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

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

Expand

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

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

Compute the boundaries:

e + \frac{1}{e}

= e + \frac{1}{e}

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