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

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

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

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

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

Solution steps

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

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Compute the boundaries:

e^{x} - 1

= e^{x} (e^{x} - 1)

\frac{x + 1}{x + 2}

\frac{d}{d x} (- 64 sin (2 x))

y = 6 e^{x}, at x = 1

\int \frac{- 8 e ^{2 x}}{e ^{2 x} + 3 e ^{x} + 2} d x

\int cos^{2} (θ) sin^{4} (θ) d θ

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