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

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

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 2pix(2e^{-x^2})

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

2 π (1 - \frac{1}{e})

Decimal Notation

3.97173 \dots

Solution steps

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

Take the constant out:

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

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

Simplify

= 4 π \cdot \int_{0}^{1} x e^{- x^{2}} d x

Apply u-substitution

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

= 4 π \cdot \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

= 4 π (- \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

= 4 π (- (- \frac{1}{2} \cdot \int_{- 1}^{0} e^{u} d u))

Use the common integral:

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

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

Simplify

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

= 2 π [e^{u}]_{- 1}^{0}

Compute the boundaries:

1 - \frac{1}{e}

= 2 π (1 - \frac{1}{e})

\int_{0}^{π} t cos (2 t) d t

\int \frac{( x - 2 ) ( x + 3 )}{x x} d x

\int ln (x^{7}) d x

(e^{x} + e^{- x}) \frac{d y}{d x} = y^{2}

\int (2 a x + b) (a x^{2} + b x)^{7} d x

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