Question about an integral trick This is an excerpt from brilliant.org on integral tricks:

A similar method to the above is to reverse the interval of
  integration: to "integrate backwards." For a function $f$ and real
  numbers $a < b$,
$\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx.$
Instead of the function being centered at $0$, the function is now
  centered at $(\tfrac{a+b}{2})$. Then,
$\int_a^b f(x) \, dx = \frac{1}{2} \int_a^b f(x) + f(a+b-x) \, dx.$

I understand that the trick is to integrate backwards, from $b$ towards $a$, but I don't understand what is meant by that the function is centered at $(\tfrac{a+b}{2})$. What is meant by that and why is it significant? Also, how is the last integral derived?
 A: If we let $x=a+b-y$ then when $x=a$, $y=a+b-a=b$ and when $x=b$, $y=a+b-b=a$ and $dx=-dy$, so
$$\int_a^bf(x)dx=\int_b^af(a+b-y)(-dy)=\int_a^bf(a+b-y)dy=\int_a^bf(a+b-x)dx$$
Where running the integral backwards changes its sign and we renamed the variable of integration back to $x$. Then
$$\begin{align}\int_a^bf(x)dx&=\frac12\int_a^bf(x)dx+\frac12\int_a^bf(x)dx\\
&=\frac12\int_a^bf(x)dx+\frac12\int_a^bf(a+b-x)dx\\
&=\frac12\int_a^b\left(f(x)+f(a+b-x)\right)dx\end{align}$$
The function is centered at $\frac{b+a}2$ means it is even about that point: let $g(x)=\frac12\left(f(x)+f(a+b-x)\right)$. Then
$$\begin{align}g\left(\frac{b+a}2-x\right)&=\frac12\left(f\left(\frac{b+a}2-x\right)+f\left(b+a-\frac{b+a}2+x\right)\right)\\
&=\frac12\left(f\left(b+a-\frac{b+a}2-x\right)+f\left(\frac{b+a}2+x\right)\right)\\
&=\frac12\left(f\left(b+a-\left(\frac{b+a}2+x\right)\right)+f\left(\frac{b+a}2+x\right)\right)\\
&=g\left(\frac{b+a}2+x\right)\end{align}$$
Usually when we talk about even functions the point of reflection is implicitly $0$ so $g(-x)=g(x)$.  
Here's an integral from the CRC Handbook of Chemistry and Physics and the Putnam exam:
$$\begin{align}\int_0^{\frac{\pi}2}\frac{d\theta}{1+\tan^{\sqrt2}\theta}&=\frac12\int_0^{\frac{\pi}2}\left(\frac{1}{1+\tan^{\sqrt2}\theta}+\frac{1}{1+\tan^{\sqrt2}\left(\frac{\pi}2-\theta\right)}\right)d\theta\\
&=\frac12\int_0^{\frac{\pi}2}\left(\frac{1}{1+\tan^{\sqrt2}\theta}+\frac{1}{1+\cot^{\sqrt2}\left(\theta\right)}\right)d\theta\\
&=\frac12\int_0^{\frac{\pi}2}\left(\frac{1}{1+\tan^{\sqrt2}\theta}+\frac{\tan^{\sqrt2}\theta}{\tan^{\sqrt2}\theta+1}\right)d\theta\\
&=\frac12\int_0^{\frac{\pi}2}1d\theta=\frac{\pi}4\end{align}$$
Cool how folding about $\theta=\frac{\pi}4$ knocked down what seemed like a very daunting integral.
A: From here
$$\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx\\\iff \int_a^b f(x) \, dx +\int_a^b f(x) \, dx = \int_a^b f(x) \, dx+\int_a^b f(a+b-x) \, dx\\\iff 2\int_a^b f(x) \, dx = \int_a^b f(x)+f(a+b-x) \, dx\\\iff \int_a^b f(x) \, dx = \frac{1}{2} \int_a^b f(x) + f(a+b-x) \, dx$$
