What is the integral from-1 to 1 of e^{x+1} ?

The integral from-1 to 1 of e^{x+1} is e^2-1

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integral from-1 to 1 of e^{x+1}

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

e^{2} - 1

Decimal

6.38905 \dots

Solution steps

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

e^{x + 1} = e^{1} e^{x}

= \int_{- 1}^{1} e^{1} e^{x} d x

Take the constant out:

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

= e^{1} \cdot \int_{- 1}^{1} e^{x} d x

Simplify

= e \cdot \int_{- 1}^{1} e^{x} d x

Use the common integral:

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

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

Compute the boundaries:

e - \frac{1}{e}

= e (e - \frac{1}{e})

Simplify

= e^{2} - 1

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What is the integral from-1 to 1 of e^{x+1} ?
The integral from-1 to 1 of e^{x+1} is e^2-1