What is the integral from 0 to 2 of x^1e^{-x^2} ?

The integral from 0 to 2 of x^1e^{-x^2} is (e^4-1)/(2e^4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 2 of x^1e^{-x^2}

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

\frac{e ^{4} - 1}{2 e ^{4}}

Decimal

0.49084 \dots

Solution steps

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

Simplify

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

Apply u-substitution

: \int_{0}^{- 4} - \frac{e ^{u}}{2} d u

= \int_{0}^{- 4} - \frac{e ^{u}}{2} d u

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

= - \int_{- 4}^{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_{- 4}^{0} e^{u} d u)

Use the common integral:

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

= - (- \frac{1}{2} [e^{u}]_{- 4}^{0})

Simplify

= \frac{1}{2} [e^{u}]_{- 4}^{0}

Compute the boundaries:

1 - \frac{1}{e ^{4}}

= \frac{1}{2} (1 - \frac{1}{e ^{4}})

Simplify

= \frac{e ^{4} - 1}{2 e ^{4}}

\int_{- 1}^{2} (x^{3} - x^{2} - 2 x) d x

\int_{- 6}^{6} (2 x - 5) d x

\int_{0}^{2} e^{- 2 x + 1} d x

\int_{1}^{2} π (e^{y})^{2} d y

\int_{- 2}^{3} e^{x + 2} d x

What is the integral from 0 to 2 of x^1e^{-x^2} ?
The integral from 0 to 2 of x^1e^{-x^2} is (e^4-1)/(2e^4)