From a statistical standpoint, the mean energy of a system is given by: $$\langle E \rangle =E \cdot P(x)=\frac{\int_{-\infty}^{+\infty} E {e^{- \beta E}}}{\int_{-\infty}^{+\infty} {e^{- \beta E}}}
$$
Where $\beta =\frac{1}{k_B T}$ and $P(x)$ is the probabily of the system being at a particular energy, $E$.
Now, if your energy dependance is quadric in some variable, this is if $E=ax^2$, where $a$ is just some constant, the mean energy becomes (thanks to some pretty cool math and Gaussian Integrals):
$$\langle E \rangle = \frac{\int_{-\infty}^{+\infty} {ax^2 e^{- \beta ax^2}}}{\int_{-\infty}^{+\infty} {e^{- \beta ax^2}}} = \frac{1}{2 \beta}= \frac{1}{2}k_BT$$
You should note that this result is independent of $x$ and $a$. Actually, it is easy to show that if instead of one variable, the energy depends on $n$ quadratic variables, often called the modes of the system, each mode contributes the same amount of energy ($\frac{1}{2}k_BT$) to the system (all you have to do is repeat the calculations for $E=\sum_{i=0}^n {a_i}x_i^2$) - this is known as the equipartition theorem. Hence, the mean energy of a system with $n$ quadratic modes becomes:
$$\langle E \rangle = \frac{1}{2}nk_BT$$
In an ideal monoatomic gas, the internal energy of the system, $U$, is just its kinetic energy (there is no energy due to vibration, rotation and intermolecular interactions), and therefore: $$E=KE=\frac{1}{2}mv^2 = \frac{1}{2}mv_x^2 + \frac{1}{2}mv_y^2 + \frac{1}{2}mv_z^2 $$
The energy is the sum of $3$ quadratic modes, so the internal energy of the system is simply (substituting $n$ by 3 in our expression for $U$): $$U=\frac{3}{2}k_BT = \frac{3}{2}nRT $$
PS: This might have been a bit more calculus then you'd asked for, but it is the reason this expression exists. I always find it helpful to know where things come from, so I hope this helps you as well
