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

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

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

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

\frac{π}{2 2}

Decimal Notation

0.62665 \dots

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

= \frac{π}{2 2} - 0

Simplify

= \frac{π}{2 2}

\int sec^{2} (4 x t a) n^{3} 4 x d x

5 e^{2 t}

\frac{d}{d x} (\frac{1}{2 1 + x})

\frac{\partial}{\partial x} (x^{2} - 2 x y + 2 y^{2} + x - 2 y)

\frac{d}{d x} (cos (π \cdot x))

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