Lookback Time & Age of the Universe Calculations I try to calculate the age of the universe with the FLRW model:
$$ H(a) = H_0 \sqrt{\Omega_{\mathrm{R},0} \left(\frac{a_0}{a}\right)^4 + \Omega_{\mathrm{M},0} \left(\frac{a_0}{a}\right)^3 + (1-\Omega_{\mathrm{T},0}) \left(\frac{a_0}{a}\right)^2 + \Omega_{\Lambda,0}}. $$
I set $\Omega_{\mathrm{M},0} = 0.317$ (matter density) and $\Omega_{\Lambda,0} = 0.683$ (dark energy), as delivered by Planck 2013; $\Omega_{\mathrm{T},0} = 1.02$ (space curvature), according to this site; and $\Omega_{\mathrm{R},0} = 4.8\times10^{-5}$ (radiation density), according to this document.
For the time $t(a)$ I take the scale factor $a$ and divide it through the integrated recessional velocity
$$ t(a) = \frac{a}{\int_0^a{H(a')a'\ \mathrm{d}a'}/(a-0)} $$
and finally simplify to
$$ t(a) = \frac{a^2}{\int_0^a{H(a')a'\ \mathrm{d}a'}}. $$
But the problem is, I then get about $8\times10^9$ years for the age of the universe, but it should be around $12\times10^9$ years (which I get when I set $\Omega_{\mathrm{R},0}$ to zero):
$\Omega_{\mathrm{R},0} = 4.8\times10^{-5}$:

$\Omega_{\mathrm{R},0} = 0 \to 0.00001$:

Do I have to use some other models than FLRW/ΛCDM, or is one of my parameters outdated?
 A: The total energy density is by definition
$$ \Omega_{T,0} = \Omega_{R,0} + \Omega_{M,0} + \Omega_{\Lambda,0},$$
so with the values you cite ($\Omega_{R,0}=4.8\times 10^{-5}$, $\Omega_{M,0}=0.317$, $\Omega_{\Lambda,0}=0.683$), we get $\Omega_{T,0} = 1$, or in a more common notation $\Omega_{K,0}=1-\Omega_{T,0}=0$, i.e. a space with zero curvature.
It is also common to define the present-day value of the scale-factor as $a_0=1$, so that
$$
H(a) = H_0\sqrt{\Omega_{R,0}a^{-4} + \Omega_{M,0}a^{-3} + \Omega_{K,0}a^{-2} + \Omega_{\Lambda,0}}.
$$
The age of the universe can then be derived as follows: from
$$
\frac{\text{d}a}{\text{d}t} = \dot{a},
$$
we get
$$
\begin{align}
\text{d}t &= \frac{\text{d}a}{\dot{a}} = \frac{\text{d}a}{aH(a)} = \frac{a\,\text{d}a}{a^2H(a)}\\
&= \frac{1}{H_0}\frac{a\,\text{d}a}{a^2\sqrt{\Omega_{R,0}a^{-4} + \Omega_{M,0}a^{-3} + \Omega_{K,0}a^{-2} + \Omega_{\Lambda,0}}}\\
&= \frac{1}{H_0}\frac{a\,\text{d}a}{\sqrt{\Omega_{R,0} + \Omega_{M,0}a + \Omega_{K,0}a^2 + \Omega_{\Lambda,0}a^4}}.
\end{align}
$$
Integrating yields the difference between the time that a signal is emitted and the time it is observed:
$$
t_{\text{ob}} - t_{\text{em}} = \frac{1}{H_0}\int_{a_{\text{em}}}^{a_{\text{ob}}}
\frac{a\,\text{d}a}{\sqrt{\Omega_{R,0} + \Omega_{M,0}a + \Omega_{K,0}a^2 + \Omega_{\Lambda,0}a^4}},
$$
and the age of the universe is
$$t_0 = \frac{1}{H_0}\int_0^1
\frac{a\,\text{d}a}{\sqrt{\Omega_{R,0} + \Omega_{M,0}a + \Omega_{K,0}a^2 + \Omega_{\Lambda,0}a^4}}.$$
This should give you the correct age.
