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

The integral from 0 to 1 of e^{x+y} is e^y(e-1)

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

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

e^{y} (e - 1)

Solution steps

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

e^{y + x} = e^{y} e^{x}

= \int_{0}^{1} e^{y} e^{x} d x

Take the constant out:

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

= e^{y} \cdot \int_{0}^{1} e^{x} d x

Use the common integral:

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

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

Compute the boundaries:

e - 1

= e^{y} (e - 1)

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