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

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

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integral of e^{-2bx^2}

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

\frac{π}{2 2 b} erf (\frac{2 b x}{2}) + C

Solution steps

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

Apply Integral Substitution

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

Take the constant out:

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

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

Apply exponent rule:

\frac{a ^{n}}{b ^{n}} = (\frac{a}{b})^{n}

\frac{b u ^{2}}{2} = (\frac{b u}{2})^{2}

= \frac{1}{2} \cdot \int e^{- (\frac{b u}{2})^{2}} d u

Apply u-substitution

: \int e^{- v^{2}} \frac{2}{b} d v

= \frac{1}{2} \cdot \int e^{- v^{2}} \frac{2}{b} d v

Take the constant out:

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

= \frac{1}{2} \frac{2}{b} \cdot \int e^{- v^{2}} d v

Use the nonelementary integral:

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

= \frac{1}{2} \frac{2}{b} \frac{π}{2} erf (v)

Substitute back

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

Simplify

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

= \frac{π}{2 2 b} erf (\frac{2 b x}{2})

Add a constant to the solution

= \frac{π}{2 2 b} erf (\frac{2 b x}{2}) + C

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