Average kinetic energy in 1 dimension according to Maxwell-Boltzmann Distribution

The format of the 3 dimensional MB distribution is $A cdot e^{-frac{E}{k_BT}} cdot g(E)$ in which $A$ can be derived using normalization (integration up to $infty$ must be 1) and $g(E)$ being the degeneracy according to $g(E)=frac{Vpi cdot 2^{2.5}m^{1.5}}{h^3}$

The 3 dimensional average kinetic energy $bar E$ of a particles system can then be calculated by multiplying this MB distribution with $E$ and integrating it over infinity, which yields:
$$bar E = int_0^{infty} frac{2}{sqrt pi} cdot (frac{1}{k_BT})^{frac{3}{2}} cdot e^{-frac{E}{k_BT}} cdot sqrt{E} cdot E cdot dE = frac{3}{2}k_BT$$

The format for the 1 dimensional MB distribution (e.g. the x-coordinate) is $Acdot e^{-frac{E_x}{k_BT}}$ where $A$ is derived by normalizing the integration to 1, which gives $A= frac{1}{k_BT}$ When calculating the 1 dimensional average energy $bar E_x$, this MB distribution should also be multiplied by the energy $E_x$ and integrated up to $infty$ which gives:
$$bar E_x=int^{infty}_0 frac{1}{k_BT}cdot e^{-frac{E_x}{k_BT}}cdot E_xcdot dE = k_BT$$
But this should be $frac{1}{2}k_BT$ instead. The peculiar thing is that when writing $E_x$ in terms of $frac{1}{2}mv_x^2$ within the formula $Acdot e^{-frac{E_x}{k_BT}}$, normalizing $A$ to that, multiplying the formula with $frac{1}{2}mv_x^2$ and integrating it up to $infty$, then one would indeed get $frac{1}{2}k_BT$.
$$int^{infty}_0frac{sqrt{2m}}{sqrt{pi k_BT}}cdot e^{-frac{mv_x^2}{2k_BT}}cdot frac{1}{2}mv_x^2 cdot dv=frac{1}{2}k_BT$$
But it wasn’t necessary for the 3 dimensional MB distribution to write the format down in terms of $v$ to get the correct average kinetic energy.

Why does the 1 dimensional MB distribution in terms of $E_x$ give an incorrect average energy and how would one realise that this is the wrong way to do it?