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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to y of e^{-t^2}

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

\frac{π}{2} erf (y)

Solution steps

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

Use the nonelementary integral:

\int e^{- t^{2}} d t = \frac{π}{2} erf (t)

= [\frac{π}{2} erf (t)]_{0}^{y}

Compute the boundaries:

\frac{π}{2} erf (y)

= \frac{π}{2} erf (y)

\int_{0}^{1} sin (t^{2}) d t

\int_{0}^{3} (9 - x^{2})^{2} d x

\int_{0}^{l n (8)} e^{x} d x

\int_{- 3}^{0} \frac{1}{x ^{2}} d x

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

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