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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{e - 1}{2 e}

Decimal Notation

0.31606 \dots

Solution steps

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

Simplify

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

Apply u-substitution

: \int_{0}^{- 1} - \frac{e ^{u}}{2} d u

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

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

= - \int_{- 1}^{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_{- 1}^{0} e^{u} d u)

Use the common integral:

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

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

Simplify

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

Compute the boundaries:

1 - \frac{1}{e}

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

Simplify

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

\frac{d}{d x} (2 x^{7} - x^{5} + 5 x^{3} - 8 x + 4)

\frac{d}{d x} (\frac{( 5 x + 3 )}{3 e ^{x}})

\int sin (x) sec^{5} (x) d x

f (x) = x^{2} - 9, at x = 2

\frac{8}{x ^{3}}

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