Given the probability density function $f(x)$, and the $[X] = \frac{2}{\sqrt{\pi\lambda}} $, how best should I go about deducing the even moments of this distribution?
$f(x) = 4\sqrt{\frac{\lambda^3}\pi}$ $x^2$ $ (-\lambda x^2)$
I am trying to use integration by parts on $[X^2] = \int{{(\frac{2}{\sqrt{\pi\lambda}}})^2 4\sqrt{\frac{\lambda^3}\pi}x^2 (-\lambda x^2)dx}$
but am ultimately going wrong somewhere.
Any guidance would be very much appreciated.
First of all you need for the probability density function
$f_X(x) \geq 0$ and for it to integrate to $1$.
$$\int_{-\infty}^{\infty}f_X(x)\:\text{dx} = \int_{x \in D}{}f_X(x)\:\text{dx} = 1$$
Then for the moments
$$E[X^n] = \int_{-\infty}^{\infty}x^n f_X(x)\:\text{dx} = \int_{x \in D}{}x^nf_X(x)\:\text{dx} $$
and since you used the moment-generating function tag here is how you calculate your moments using that method instead of using the defintion above
$$ M_X(t) = E(e^{tX}) = \begin{cases} \sum_i e^{tx_i}p_X(x_i), & \text{(discrete case)} \\ \\ \int_{-\infty}^{\infty} e^{tx}f_X(x)\:\text{dx}, & \text{(continuous case)} \end{cases} $$
and the $n$th moment of $X$ is given by
$$E[X^n] = M_X^{(n)}(0) \:\:\:\:\:\:n = 1, 2...$$
$$M_X^{(n)}(0) = \frac{d^n}{dt^n} M_X(t) |_{t=0}$$