What is the integral of e^{-cx^2} ?

The integral of e^{-cx^2} is (sqrt(pi))/(2sqrt(c))\erf(sqrt(c)x)+C

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integral of e^{-cx^2}

\int e^{- c x^{2}} d x

\frac{π}{2 c} erf (c x) + C

Solution steps

\int e^{- c x^{2}} d x

= \int e^{- (c x)^{2}} d x

Apply u-substitution

: \int \frac{e ^{- u^{2}}}{c} d u

= \int \frac{e ^{- u^{2}}}{c} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{c} \cdot \int e^{- u^{2}} d u

Use the nonelementary integral:

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

= \frac{1}{c} \cdot \frac{π}{2} erf (u)

Substitute back

u = c x

= \frac{1}{c} \cdot \frac{π}{2} erf (c x)

Simplify

\frac{1}{c} \cdot \frac{π}{2} erf (c x) : \frac{π}{2 c} erf (c x)

= \frac{π}{2 c} erf (c x)

Add a constant to the solution

= \frac{π}{2 c} erf (c x) + C

\int \frac{x ^{2}}{( x ^{2} - 8 ^{2} ) ^{\frac{3}{2}}} d x

\frac{\partial}{\partial x} (1 + x y^{2} - 2 z^{2})

\int \frac{e ^{x}}{e ^{x}} d x

\frac{1}{y ^{2}} - \frac{9}{y ^{4}} (y + 3 y^{3})

n = 1 \sum \infty \frac{10 n}{n + 1}

What is the integral of e^{-cx^2} ?
The integral of e^{-cx^2} is (sqrt(pi))/(2sqrt(c))\erf(sqrt(c)x)+C