Average IQ of Mensa

The $x$ corresponding to the mean value of a density $f$ is not (in general) equal to the mean $x$.

Instead, you want to compute the mean of $x$, weighted by $f$, i.e., \begin{equation} \frac{\int_{130}^\infty xf(x)dx}{\int_{130}^\infty f(x)dx}. \end{equation}

This can also be interpreted probabilistically: we are looking for the expectation $E[X \mid X\geq 130]$, which is obtained by the integral \begin{equation} E[X \mid X\geq 130] = \int_{130}^{\infty} x p(x) dx, \end{equation} where $p$ is the conditional density of the probability distribution, obtained by \begin{equation} p(x) = \frac{f(x)}{\int_{130}^\infty f(t) dt},~x\geq 130 \end{equation}

Note that this answer is focused on the mathematical content, i.e., not questioning the assumptions. For various reasons I suspect this method does not capture the average IQ of Mensa (e.g., probability of joining Mensa might vary as a function of IQ even above the threshold).

If the distribution is continuous uniform, then your idea is correct. But since IQ test scores typically follow what is known as a normal (or Gaussian) distribution, then you should use the left censored and shifted variable method. The expected value using this method can be obtained by $$ \text{E}[(X-d)_+]=\int_d^\infty (x-d)f(x)\ dx. $$ In your case $d=130$ and $X\sim\mathcal{N}(100,15)$. Actually, the minimum accepted IQ score to be a Mensan on the Stanford-Binet Intelligence Scales is $132$.

ADDENDUM :

If you consider this case using the conditional distribution, then for a given value of $d$ with $\Pr[X>d]>0$, in your case $\Pr[X>d]=0.02$ because a Mensa member is a person who scores at or above the 98th percentile on certain standardized IQ or other approved intelligence tests, then for $Y=X-d$ given that $X>d$, its expected value is $$ \text{E}[Y|X>d]=\frac{\int_d^\infty (x-d)f(x)\ dx}{1-\Pr[X\le d]}. $$

Your problem is similar to the problem: Mensa (The High IQ Society) that I posted on Brilliant.org.