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

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

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integral from 0 to x of (e^{-t^2})

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

\frac{π}{2} erf (x)

Solution steps

\int_{0}^{x} 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}^{x}

Compute the boundaries:

\frac{π}{2} erf (x)

= \frac{π}{2} erf (x)

x 1 + y^{2} d x = y 1 + x^{2} d y

\frac{d}{d x} (\frac{f ( x )}{g ( x ) - 2})

\frac{d y}{d x}, 3 x y + 1 = 0

\int_{0}^{33} 3 (x - 1)^{- \frac{1}{5}} d x

x y \frac{d y}{d x} = 2 y^{2} - x^{2}

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