Given that $d$ follows an equation, $d= (v^2/g)\sin2\theta$, in which v and g are constants and $\theta$ follows a uniform distribution on $[\pi/6, \pi/3]$, try to find the expected value and the variance of $d$.
I had the idea of directly integrating d, but I guess the uniform distribution of theta does have an effect and I should also be considering that, but I'm not really sure how to factor that in. The answer is $(3v^2)/(\pi g)$ for the expected value. It would be great if anyone can point me in the right direction... Thank you so much!
Just use the definition
$$\mathbb{E}[g(X)]=\int_{-\infty}^{+\infty} g(x)f(x)dx$$
$$\mathbb{E}[g(\theta)]=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{v^2}{g}sin(2\theta)\frac{6}{\pi}d\theta=\dots=\frac{3v^2}{g\pi}$$
Use similar arguments for the variance, remembering that
$$\mathbb{V}[X]=\mathbb{E}[X^2]-\mathbb{E}^2[X]$$