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

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

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^{-y^2}

\int_{0}^{\infty} e^{- y^{2}} d y

\frac{π}{2}

Decimal

0.88622 \dots

Solution steps

\int_{0}^{\infty} e^{- y^{2}} d y

Compute the indefinite integral:

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

Compute the boundaries:

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

= \frac{π}{2} - 0

Simplify

= \frac{π}{2}

\int_{0}^{49} \frac{8}{x} d x

\int_{0}^{1} sin (x y)

\int_{0}^{1} (10 e^{2 t} - 7) d t

\int_{1}^{2} x - \frac{1}{x ^{2}} d x

\int_{a}^{b} (C x + D)^{p} d x

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