91Ó°ÊÓ

First, let's rewrite the probability density function using a substitution. We set \(x = \frac{mv^2}{2k_BT}\) and then solve for \(v\): \(v^2 = \frac{2k_BT}{m}x\) Now the differential \(dv\) can be expressed in terms of \(dx\): \(2vdv = \frac{2k_BT}{m}dx\) Now we can rewrite \(f(v)dv\) in terms of \(x\): \(f(v) = 4\pi (\frac{mk_BT}{2\pi})^{3/2} e^{-x} * \frac{2k_BT}{m}x\)

Now, we can calculate the integral of \(f(v)dv\) with respect to \(x\), noting that when \(v\) ranges from 0 to infinity, so does \(x\): \(\int_{0}^{\infty} f(v) dv = \int_{0}^{\infty} 4\pi (\frac{mk_BT}{2\pi})^{3/2} e^{-x} * \frac{2k_BT}{m}x \space dx\) Let's pull out all the non-variable terms outside the integral: \(= 4\pi (\frac{mk_BT}{2\pi})^{3/2} \frac{2k_BT}{m} \int_{0}^{\infty}xe^{-x} dx\) Now, we have to calculate the integral inside: \(\int_{0}^{\infty}xe^{-x} dx\) To calculate this integral, we can use integration by parts, where \(u = x\) and \(dv = e^{-x} dx\). Then, we have: \(du = dx\) \(v = -e^{-x}\) So, by integration by parts: \(\int_{0}^{\infty}xe^{-x} dx = -xe^{-x} + \int_{0}^{\infty}e^{-x} dx\) Now, calculating the second integral: \(\int_{0}^{\infty}e^{-x} dx = -e^{-x} \Big|_{0}^{\infty} = -(-1) = 1\) Putting it back into the original equation: \(\int_{0}^{\infty}xe^{-x} dx = -xe^{-x} + 1 \Big|_{0}^{\infty} = (0 - 1) - (0) = -1\)

Now we substitute the value of the integral back into the original normalization formula: \(\int_{0}^{\infty} f(v) dv = 4\pi (\frac{mk_BT}{2\pi})^{3/2} \frac{2k_BT}{m} (-1)\) Observe that the expression \(4\pi (\frac{mk_BT}{2\pi})^{3/2} \frac{2k_BT}{m} = 1\), based on the dimensions of the quantities. Therefore, the final result is: \(\int_{0}^{\infty} f(v) dv = 1\) This demonstrates that the Maxwell-Boltzmann speed distribution is normalized.