What is the integral from 0 to infinity of e^{-7x^2} ?

The integral from 0 to infinity of e^{-7x^2} is (sqrt(pi))/(2sqrt(7))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to infinity of e^{-7x^2}

\int_{0}^{\infty} e^{- 7 x^{2}} d x

\frac{π}{2 7}

Decimal Notation

0.33496 \dots

Solution steps

\int_{0}^{\infty} e^{- 7 x^{2}} d x

Compute the indefinite integral:

\int e^{- 7 x^{2}} d x = \frac{π}{2 7} erf (7 x) + C

Compute the boundaries:

\int_{0}^{\infty} e^{- 7 x^{2}} d x = \frac{π}{2 7} - 0

= \frac{π}{2 7} - 0

Simplify

= \frac{π}{2 7}

\int_{0}^{2} (2 x + 2) d x

\int_{- 1}^{1} e^{y} sin (y) d y

\int_{0}^{5} 2 e^{- 2 x} d x

\int_{0}^{1} ln (x) d x

\int_{- 3}^{3} - x^{2} + 5 d x

What is the integral from 0 to infinity of e^{-7x^2} ?
The integral from 0 to infinity of e^{-7x^2} is (sqrt(pi))/(2sqrt(7))