How can a value of $x$ have an expectation value? Surely there must be a distribution of values of $x$ for the expectation value to be calculated. Is this the reason for the normal distribution function? When finding the mean of the normal distribution function do you have to do: $$\int_{-\infty}^\infty xN(x)dx$$ Where $N(x)$ is the normal distribution funtion.
Or should it be: $$\int_{-\infty}^\infty N(x)p(x)dx$$ Where $p(x)$ is some probability density function.
Also, am I right in thinking that the expectation value is essentially the mean?
The first integral is correct since you are computing $$ E(X) = \int_{-\infty}^{\infty} xdP = \int_{-\infty}^{\infty} xf(x)dx $$ where $dP$ is the probability measure, and $f(x)$ is the probability density function (in your case the Normal). The above also evaluates to the mean.
To answer another question..the value of $x$ is better described as $$ E(X) = \int_{x\in A} xf(x)dx $$ where the set of values of $x$ are only considered if they exist in $A$
$$ E(g(X)) = \int_{x\in A} g(x)f(x)dx $$ I am not caring about the subtle/or not subtle cases of existence of the integral..but if we take $$ g(X) = X^2 $$ we can compute $$ E(g(X)) = E(X^2) = \int_{x\in A} x^2f(x)dx $$