What is the integral of e^{-9x^2} ?

The integral of e^{-9x^2} is (sqrt(pi))/6 \erf(sqrt(9)x)+C

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integral of e^{-9x^2}

\int e^{- 9 x^{2}} d x

\frac{π}{6} erf (9 x) + C

Solution steps

\int e^{- 9 x^{2}} d x

= \int e^{- (9 x)^{2}} d x

Apply u-substitution

: \int e^{- u^{2}} \frac{1}{3} d u

= \int e^{- u^{2}} \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int e^{- u^{2}} d u

Use the nonelementary integral:

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

= \frac{1}{3} \cdot \frac{π}{2} erf (u)

Substitute back

u = 9 x

= \frac{1}{3} \cdot \frac{π}{2} erf (9 x)

Simplify

\frac{1}{3} \cdot \frac{π}{2} erf (9 x) : \frac{π}{6} erf (9 x)

= \frac{π}{6} erf (9 x)

Add a constant to the solution

= \frac{π}{6} erf (9 x) + C

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What is the integral of e^{-9x^2} ?
The integral of e^{-9x^2} is (sqrt(pi))/6 \erf(sqrt(9)x)+C