I'll start by saying I have never studied probability before, so forgive me if this question is very basic.
Anyway, let $T$ be the set of all real triplets $(a,b,c)$ so that $1 \leq a \leq 2$, $1 \leq b \leq a^3$, and $c = a+b^2$, and let $(x_a,x_b,x_c)$ be a triplet chosen uniformly randomly from $T$.
I have a few questions. My first: what is the expected value of $x_a$? Intuitively, I think "the larger the value of $a$, the 'more' triplets in $T$ there are having the value $a$, so it cannot be $1/2$." That is, suppose $a=1.2$, then $b$ can assume values between $1$ and $1.2^2$, an interval larger than the range of values it could assume if $a$ were less then $1.2$. In this sense, if I want to determine the expected value of $x_a$, I need to account for the varying quantity $a^3-1$. But I am not sure how.
My second question: How may I determine the expected value of $x_c$?
Edit: To clarify, when I say that a triplet is chosen uniformly randomly, I mean that the joint distribution of $x_a$ and $x_b$ has constant density.
I don’t quite see why you introduced the separate triples $a,b,c$ and $x_a,x_b,x_c$; I’ll use $a,b,c$ to keep the notation simple.
The expected value of $a$ is
\begin{eqnarray} \frac{\int_1^2\mathrm da\int_1^{a^3}\mathrm db\,a}{\int_1^2\mathrm da\int_1^{a^3}\mathrm db} &=& \frac{\int_1^2\mathrm da\left(a^3-1\right)a}{\int_1^2\mathrm da\left(a^3-1\right)} \\ &=& \frac{\frac15(32-1)-\frac12(4-1)}{\frac14(16-1)-(2-1)} \\ &=& \frac{94}{55}\;. \end{eqnarray}
The expected value of $b^2$ is
\begin{eqnarray} \frac{\int_1^2\mathrm da\int_1^{a^3}\mathrm db\,b^2}{\int_1^2\mathrm da\int_1^{a^3}\mathrm db} &=& \frac{\frac13\int_1^2\mathrm da\left(a^9-1\right)}{\int_1^2\mathrm da\left(a^3-1\right)} \\ &=& \frac13\cdot\frac{\frac1{10}(1024-1)-(2-1)}{\frac14(16-1)-(2-1)} \\ &=& \frac{2026}{165}\;. \end{eqnarray}
By the linearity of expectation, the expected value of $c=a+b^2$ is the sum,
$$ \frac{94}{55}+\frac{2026}{165}=\frac{2308}{165}\;. $$