What is the integral of 1/(4pit)e^{-(r^2)/(4t)}r ?

The integral of 1/(4pit)e^{-(r^2)/(4t)}r is -1/(2pi)e^{-(r^2)/(4t)}+C

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integral of 1/(4pit)e^{-(r^2)/(4t)}r

\int \frac{1}{4 π t} e^{- \frac{r ^{2}}{4 t}} r d r

- \frac{1}{2 π} e^{- \frac{r ^{2}}{4 t}} + C

Solution steps

\int \frac{1}{4 π t} e^{- \frac{r ^{2}}{4 t}} r d r

Take the constant out:

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

= \frac{1}{4 t π} \cdot \int e^{- \frac{r ^{2}}{4 t}} r d r

Apply u-substitution

: \int \frac{e ^{- \frac{u}{4 t}}}{2} d u

= \frac{1}{4 t π} \cdot \int \frac{e ^{- \frac{u}{4 t}}}{2} d u

Take the constant out:

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

= \frac{1}{4 t π} \cdot \frac{1}{2} \cdot \int e^{- \frac{u}{4 t}} d u

Apply u-substitution

: \int - 4 t e^{v} d v

= \frac{1}{4 t π} \cdot \frac{1}{2} \cdot \int - 4 t e^{v} d v

Take the constant out:

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

= \frac{1}{4 t π} \cdot \frac{1}{2} (- 4 t \cdot \int e^{v} d v)

Use the common integral:

\int e^{v} d v = e^{v}

= \frac{1}{4 t π} \cdot \frac{1}{2} (- 4 t e^{v})

Substitute back

= \frac{1}{4 t π} \cdot \frac{1}{2} (- 4 t e^{- \frac{r ^{2}}{4 t}})

Simplify

\frac{1}{4 t π} \cdot \frac{1}{2} (- 4 t e^{- \frac{r ^{2}}{4 t}}) : - \frac{1}{2 π} e^{- \frac{r ^{2}}{4 t}}

= - \frac{1}{2 π} e^{- \frac{r ^{2}}{4 t}}

Add a constant to the solution

= - \frac{1}{2 π} e^{- \frac{r ^{2}}{4 t}} + C

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What is the integral of 1/(4pit)e^{-(r^2)/(4t)}r ?
The integral of 1/(4pit)e^{-(r^2)/(4t)}r is -1/(2pi)e^{-(r^2)/(4t)}+C