What is the integral from 0 to y^2 of e^{x/y} ?

The integral from 0 to y^2 of e^{x/y} is y(e^y-1)

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integral from 0 to y^2 of e^{x/y}

\int_{0}^{y^{2}} e^{\frac{x}{y}} d x

y (e^{y} - 1)

Solution steps

\int_{0}^{y^{2}} e^{\frac{x}{y}} d x

Apply u-substitution

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

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

Use the common integral:

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

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

Compute the boundaries:

e^{y} - 1

= y (e^{y} - 1)

\int_{0}^{1} (- x + 1) d x

\int_{0}^{5} t^{2} - t d t

\int_{4.6}^{7} (12 x^{2} - 6 x + 4) d x

\int_{0}^{1} 2 (1 - y) d y

What is the integral from 0 to y^2 of e^{x/y} ?
The integral from 0 to y^2 of e^{x/y} is y(e^y-1)