Why can I simply say that multiplying the Boltzmann distribution function with $v^2$ and integrating from 0 to $\infty$ leads to the mean square speed?
$\langle v^2\rangle = \int\limits_{0}^{\infty} v^2 \cdot 4\pi \left( \frac{M}{2\pi RT}\right)^{\frac{3}{2}}v^2\exp\left(-Mv^2/RT\right)\;\mathrm{d}v=\frac{3RT}{M}$
My Physical Chemistry textbook says only that it is calculated that way and doesn't go any further.
Thanks for helping
The Boltzmann distribution states that the probability of having a particle with velocity $\vec v$ is proportional to $e^{-\frac{E}{kT}}d^3v$, where $E=\frac12 mv^2$ and $k$ is Boltzmann's constant. So $$p(\vec v)=Ce^{-\frac{mv^2}{2kT}}$$ You get the integration constant $C$ from $\int p(\vec v)d^3v=1$, where you integrate over the entire space. Since the energy is direction independent, you can integrate the angular part, and you get $$\int_{angles}d^3v=4\pi v^2dv$$Then $$C4\pi\int_0^\infty v^2e^{-\frac{mv^2}{2kT}}dv=1$$ This yields $$C=\left(\frac{m}{2\pi kT}\right)^{3/2}$$ You can use $R=N_Ak$ and $M=N_A m$to get $$p(\vec v)=\left(\frac{M}{2\pi RT}\right)^{3/2}e^{-\frac{Mv^2}{2RT}}$$ To get the average value of $v^2$, we need to do $$\langle v^2\rangle=\int v^2p(\vec v)d^3v=4\pi\int_0^\infty v^2v^2p(v)dv$$ One of the $v^2$ comes from the quantity you want to average, the other from transformation of the integral from $3D$ to $1D$. Plugging in the probability you now get the formula in your question. Note that there is a factor of $1/2$ in the exponential missing from your formula.