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

The integral from 0 to 2 of 2e^{x^2-2x} is (sqrt(pi)\erfi(1)-sqrt(pi)\erfi(-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 2 of 2e^{x^2-2x}

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

\frac{π erfi ( 1 ) - π erfi ( - 1 )}{e}

Solution steps

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

Take the constant out:

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

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

Complete the square

x^{2} - 2 x : (x - 1)^{2} - 1

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

Apply u-substitution

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

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

Simplify

e^{u^{2} - 1} : e^{u^{2}} e^{- 1}

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

Take the constant out:

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

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

Apply exponent rule:

a^{- 1} = \frac{1}{a}

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

Use the nonelementary integral:

\int e^{u^{2}} d u = \frac{π}{2} erfi (u)

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

Simplify

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

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

Compute the boundaries:

\frac{π erfi ( 1 ) - π erfi ( - 1 )}{2}

= \frac{2}{e} \cdot \frac{π erfi ( 1 ) - π erfi ( - 1 )}{2}

Simplify

= \frac{π erfi ( 1 ) - π erfi ( - 1 )}{e}

\int_{0}^{\frac{π}{4}} t cos (2 t) d t

\int_{0}^{2} y 2 + 4 y^{2} d y

\int_{8}^{16} (x + \frac{1}{x}) d x

\int_{6}^{7} x^{2} d x

\int_{0}^{1} 19 x e^{x} d x

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