What is the integral from 0 to sqrt(2 of)xe^{-x^2} ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to sqrt(2 of)xe^{-x^2}

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

\frac{e ^{2} - 1}{2 e ^{2}}

Decimal Notation

0.43233 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{- 2} - \frac{e ^{u}}{2} d u

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Take the constant out:

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

= - (- \frac{1}{2} \cdot \int_{- 2}^{0} e^{u} d u)

Use the common integral:

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

= - (- \frac{1}{2} [e^{u}]_{- 2}^{0})

Simplify

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

Compute the boundaries:

1 - \frac{1}{e ^{2}}

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

Simplify

= \frac{e ^{2} - 1}{2 e ^{2}}

\int ∣ x ∣ d x

\frac{\partial}{\partial x} (a^{x} - b^{x})

\int t^{2} e^{- 22 t} d t

\int 1 - 7 x^{2} d x

n = 2 \sum \infty \frac{5}{2 ^{n}}

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