What is the integral from 0 to xy of e^{z/y} ?

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

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integral from 0 to xy of e^{z/y}

\int_{0}^{x y} e^{\frac{z}{y}} d z

y (e^{x} - 1)

Solution steps

\int_{0}^{x y} e^{\frac{z}{y}} d z

Apply u-substitution

: \int_{0}^{x} e^{u} y d u

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

Take the constant out:

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

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

Use the common integral:

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

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

Compute the boundaries:

e^{x} - 1

= y (e^{x} - 1)

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

\int_{a}^{\infty} x^{n} d x

\int_{24}^{25} x x^{2} - 576 d x

\int_{- 1}^{3} 3 x^{2} - 6 x d x

\int_{0}^{3} \frac{x ^{2}}{3} d x

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