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

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

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

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

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

Solution steps

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

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

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

Use the common integral:

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

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

Compute the boundaries:

e^{y} - 1

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

\int_{- 2}^{2} (8 - ∣ x ∣) d x

\int_{- 2}^{0} e^{x} d x

\int_{0}^{8} x^{2} - 8 x d x

\int_{2}^{4} \frac{1 - ( \frac{x}{2} ) ^{2}}{x} d x

\int_{1}^{27} (3 x) d x

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