Approximation in the integral to calculate the age of the universe I'm computing the following integral: $$T=\frac{1}{H_{0}} \int^{1}_{0} \frac{da}{\sqrt{\frac{\Omega_{M}(t_{0})}{a}}\left( \sqrt{1+\frac{\Omega_{R}(t_0)}{\Omega_{M}(t_0) \,a}+\frac{\Omega_{\Lambda}(t_0) \,a^3}{\Omega_{M}(t_0) }}{} \right)}$$
with: $\Omega_{\Lambda}(t_0)= 0.73,\Omega_{M}(t_0)= 0.27,\Omega_{R}(t_0)=8.51 \cdot 10^{-5},H_0=2.26\cdot 10^{-18}s^{-1}$
At this passage I want to do:
$$T=\frac{1}{H_{0}} \int^{1}_{0} \frac{da}{\sqrt{\frac{\Omega_{M}(t_{0})}{a}}\left( \sqrt{1+\frac{\Omega_{R}(t_0)}{\Omega_{M}(t_0) \,a}+\frac{\Omega_{\Lambda}(t_0) \,a^3}{\Omega_{M}(t_0) }}{} \right)} \simeq\frac{1}{H_{0}} \int^{1}_{0} \frac{da}{\sqrt{\frac{\Omega_{M}(t_{0})}{a}}\left( \sqrt{1+\frac{\Omega_{\Lambda}(t_0) \,a^3}{\Omega_{M}(t_0) }}{} \right)} $$
I can't see why this approximation makes sense.
 A: Using the mean value theorem, we have that for any $0 < x < y$, $$\dfrac{1}{\sqrt{1+x}}-\dfrac{1}{\sqrt{1+y}} = \dfrac{y-x}{2(1+\xi_{x,y})^{3/2}}$$ for some $\xi_{x,y}$ between $x$ and $y$. Hence, $$\left|\dfrac{1}{\sqrt{1+x}}-\dfrac{1}{\sqrt{1+y}}\right| \le \dfrac{|y-x|}{2}$$ for all $x,y > 0$.
Using this bound, we have:
\begin{align*}
&\left|\frac{1}{\sqrt{\frac{\Omega_{M}(t_{0})}{a}}\left( \sqrt{1+\frac{\Omega_{\Lambda}(t_0) \,a^3}{\Omega_{M}(t_0) }}{} \right)} - \frac{1}{\sqrt{\frac{\Omega_{M}(t_{0})}{a}}\left( \sqrt{1+\frac{\Omega_{R}(t_0)}{\Omega_{M}(t_0) \,a}+\frac{\Omega_{\Lambda}(t_0) \,a^3}{\Omega_{M}(t_0) }}{} \right)}\right|
\\
&= \frac{1}{\sqrt{\frac{\Omega_{M}(t_{0})}{a}}}\left| \dfrac{1}{\sqrt{1+\frac{\Omega_{\Lambda}(t_0) \,a^3}{\Omega_{M}(t_0)}}} - \dfrac{1}{\sqrt{1+\frac{\Omega_{R}(t_0)}{\Omega_{M}(t_0) \,a}+\frac{\Omega_{\Lambda}(t_0) \,a^3}{\Omega_{M}(t_0)}}}\right|
\\
&\le \frac{1}{\sqrt{\frac{\Omega_{M}(t_{0})}{a}}} \cdot \frac{\Omega_{R}(t_0)}{2\Omega_{M}(t_0) \,a}
\\
&= \dfrac{\Omega_R(t_0)}{\Omega_M(t_0)^{3/2}} \cdot \dfrac{1}{2\sqrt{a}}
\end{align*}
Hence, the two integrals differ by less than $$\dfrac{1}{H_0}\displaystyle\int_{0}^{1}\dfrac{\Omega_R(t_0)}{\Omega_M(t_0)^{3/2}} \cdot \dfrac{1}{2\sqrt{a}}\,da = \dfrac{\Omega_R(t_0)}{H_0\Omega_M(t_0)^{3/2}} \approx \dfrac{6.07 \cdot 10^{-4}}{H_0}$$
The second integral is at least $$\dfrac{1}{H_0}\int_{0}^{1}\dfrac{da}{\sqrt{\tfrac{\Omega_M(t_0)}{a}} \sqrt{1+\tfrac{\Omega_{\Lambda}(t_0) \cdot 1^3}{\Omega_M(t_0)}}} = \dfrac{1}{H_0}\int_{0}^{1}\dfrac{\sqrt{a}\,da}{\sqrt{\Omega_M(t_0)+\Omega_{\Lambda}(t_0)}} = \dfrac{2/3}{H_0}$$
So the second integral is within $0.1\%$ of the first integral.
