91Ó°ÊÓ

For a probability density function \(f(x)\), the total probability must equal 1. Mathematically, this is represented as: \[\int_{-\infty}^{\infty} f(x) dx = 1\] Since the function \(f(x) = 0\) for \(x < 0\), we only need to integrate the function over the interval \([0, \infty]\): \[\int_{0}^{\infty} ax^2e^{-bx^2} dx = 1\]

Now, we will integrate the function \(ax^2e^{-bx^2}\) with respect to \(x\) over the interval \([0, \infty]\): \[\int_{0}^{\infty} ax^2e^{-bx^2} dx\] To evaluate this integral, we can use substitution. Let \(u = bx^2\), then \(\frac{du}{dx} = 2bx\). Therefore, \(dx = \frac{du}{2b}\). Now substitute \(u\) and \(dx\): \[\int_{0}^{\infty} a\left(\frac{u}{b}\right)e^{-u} \frac{du}{2b}\] Now, we can factor out the constant terms \(\frac{a}{2kb}\): \[\frac{a}{2b}\int_{0}^{\infty} ue^{-u} du\] This is the integration by parts problem, where we have \(u\) and \(dv\): \[\begin{cases}v = e^{-u} \\ du = u\, du\end{cases}\] Now we compute the integration by parts: \[\frac{a}{2b} \left[-ue^{-u}\Big|_0^{\infty} + \int_0^{\infty} e^{-u} du \right]\] After solving the integration by parts: \[\frac{a}{2b} \left[-\lim_{u\to \infty}(ue^{-u}) - (0 \times 1) + 1\right] = \frac{a}{2b} \cdot 1\] Hence, we obtain: \[\frac{a}{2b} = 1\]

Now we need to solve for \(a\) in terms of \(b\). From the previous step: \[\frac{a}{2b} = 1\] Multiply both sides by \(2b\): \[a = 2b\] So, the value of \(a\) in terms of \(b\) is: \[a = 2b\] Therefore, our probability density function \(f(x)\) is: \[f(x) = \begin{cases} 2bx^{2} e^{-b x^{2}} & x \geq 0 \\\ 0 & x<0\end{cases}\]