What is the integral from 0 to 1 of xye^{xy^2} ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to 1 of xye^{xy^2}

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

\frac{1}{2} (e^{x} - 1)

Solution steps

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

Take the constant out:

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

= x \cdot \int_{0}^{1} e^{x y^{2}} y d y

Apply u-substitution

= x \cdot \int_{0}^{x} \frac{e ^{u}}{2 x} d u

Take the constant out:

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

= x \frac{1}{2 x} \cdot \int_{0}^{x} e^{u} d u

Use the common integral:

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

= x \frac{1}{2 x} [e^{u}]_{0}^{x}

Simplify

x \frac{1}{2 x} [e^{u}]_{0}^{x} : \frac{1}{2} [e^{u}]_{0}^{x}

= \frac{1}{2} [e^{u}]_{0}^{x}

Compute the boundaries:

e^{x} - 1

= \frac{1}{2} (e^{x} - 1)

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