Here is the Oxford MAT 2014 question 1. J). The problem is to find $\int_{-1}^1 \! f(x) \, \mathrm{d}x $ given that f(x) satisfies
$ 6+f(x)=2f(-x) +3x^2 \int_{-1}^1 \! f(t) \, \mathrm{d}t. $
for all real x. The examiners solve the problem by integrating the above identity from 1 to -1, giving
$ 12+ \int_{-1}^1 \! f(x) \, \mathrm{d}x = 2 \int_{-1}^1 \! f(x) \, \mathrm{d}x +[x^3]^1_{-1} \int_{-1}^1 \! f(x) \, \mathrm{d}x $.
Noting that $\int_{-1}^1 \! f(x) \, \mathrm{d}x=\int_{-1}^1 \! f(-x) \,\mathrm{d}x$ as reflecting the function in the y-axis does not change the area between the function and the x-axis. They then continue by substituting $A=\int_{-1}^1 \! f(x) \, \mathrm{d}x$ and solving the resulting linear equation for A.
Now I do not understand what happens in line 2 with the original equation. More specifically what happens to $\int_{-1}^1 \! f(t) \, \mathrm{d}t$? Thanks for any help! I would also appreciate if you could direct me to similar questions, thanks!
Bear in mind that $\int_{-1}^{1}f(t)dt$ is a constant because the integral is given numberical bounds.
This means that $\int_{-1}^{1}f(t)dt$ is analogous to $\int_{-1}^{1}f(x)dx$
Thus, when integrating the whole thing, you treat that part as a constant, which remains the same after integration.