What is the integral of e^{-10r^2} ?

The integral of e^{-10r^2} is (sqrt(pi))/(2sqrt(10))\erf(sqrt(10)r)+C

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integral of e^{-10r^2}

\int e^{- 10 r^{2}} d r

\frac{π}{2 10} erf (10 r) + C

Solution steps

\int e^{- 10 r^{2}} d r

= \int e^{- (10 r)^{2}} d r

Apply u-substitution

: \int e^{- u^{2}} \frac{1}{10} d u

= \int e^{- u^{2}} \frac{1}{10} d u

Take the constant out:

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

= \frac{1}{10} \cdot \int e^{- u^{2}} d u

Use the nonelementary integral:

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

= \frac{1}{10} \cdot \frac{π}{2} erf (u)

Substitute back

u = 10 r

= \frac{1}{10} \cdot \frac{π}{2} erf (10 r)

Simplify

\frac{1}{10} \cdot \frac{π}{2} erf (10 r) : \frac{π}{2 10} erf (10 r)

= \frac{π}{2 10} erf (10 r)

Add a constant to the solution

= \frac{π}{2 10} erf (10 r) + C

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\int \frac{7 x ^{4} - 6 x ^{2} - x + 5}{x + 2} d x

x \to 3 - lim (\frac{1 - 2 x}{x - 3})

\frac{d}{d x} (sin^{2} (2 b x))

What is the integral of e^{-10r^2} ?
The integral of e^{-10r^2} is (sqrt(pi))/(2sqrt(10))\erf(sqrt(10)r)+C