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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y (e - 1)

Solution steps

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

Apply u-substitution

: \int_{0}^{1} e^{u} y d u

= \int_{0}^{1} 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}^{1} e^{u} d u

Use the common integral:

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

= y [e^{u}]_{0}^{1}

Compute the boundaries:

e - 1

= y (e - 1)

\int_{\frac{π}{4}}^{\frac{π}{2}} cot (y) d y

\int_{0}^{1} 6 x^{2} y d x

\int_{0}^{0.5} e^{x^{2}} d x

\int_{π}^{\frac{3 π}{2}} cos (x) d x

\int_{0}^{1} (x^{2} + 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 y(e-1)