What is the integral from 0 to 2 of e^{x+1} ?

The integral from 0 to 2 of e^{x+1} is e^3-e

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integral from 0 to 2 of e^{x+1}

\int_{0}^{2} e^{x + 1} d x

e^{3} - e

Decimal Notation

17.36725 \dots

Solution steps

\int_{0}^{2} e^{x + 1} d x

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

= \int_{0}^{2} 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_{0}^{2} e^{x} d x

Simplify

= e \cdot \int_{0}^{2} e^{x} d x

Use the common integral:

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

= e [e^{x}]_{0}^{2}

Compute the boundaries:

e^{2} - 1

= e (e^{2} - 1)

Simplify

= e^{3} - e

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What is the integral from 0 to 2 of e^{x+1} ?
The integral from 0 to 2 of e^{x+1} is e^3-e