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

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

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

\int_{0}^{l n (y)} e^{x + y} d x

e^{y} (y - 1)

Solution steps

\int_{0}^{l n (y)} e^{x + y} d x

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

= \int_{0}^{l n (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}^{l n (y)} e^{x} d x

Use the common integral:

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

= e^{y} [e^{x}]_{0}^{l n (y)}

Compute the boundaries:

y - 1

= e^{y} (y - 1)

\int_{- 1}^{1} 7 d x

\int_{0}^{2} 2 t 1 + 4 t^{2} d t

\int_{1}^{4} (x^{2} - 2 x + 5) d x

\int_{0}^{2} 8 x^{3} - x^{6} d x

\int_{2}^{6} \frac{1}{4} x d x

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