How to solve for $\int_{-1}^{1}f(x)\,dx$ in the given experssion We are given function $f(x)$ defined for each real number $x$, such that it satisfies:
$$6 + f(x) = 2f(-x) +3x^2(\int_{-1}^{1}f(t) \, dt)$$
We need to solve this for $\int_{-1}^{1}f(x)\,dx$
I'm not very experienced in integrating and doing calculus, so I tried to look in the solutions, but they aren't very helpful. They say that we should try to integrate the whole given expression and we will end up with $12 + A = 2A + 2A$ where $A = \int_{-1}^1 f(x) \, dx$
 A: Another way to do the question is first by noticing that $$\int_{-1}^1 f(t) dt = \int_{-1}^1 f(x) dx$$ So, one of the terms is what you have to find. 
Let this be $A$. Then, we get $6+f(x) = 2f(-x) + 3Ax^2$
Now, we can see that $x^2$ is even. So, if we replace $x$ with $-x$ in the above equation, that term will not change. Doing so, we get $6+f(-x) = 2f(x) + 3Ax^2$
Now, subtracting the above $2$ equations and solving, we get $f(x) = f(-x)$. Substituting this in the above equation, we get $$f(x) = 6 - 3Ax^2$$
Integrating, we get - $$A = 12 - 2A$$
or $A = 4$.
Side note: We also proved in the process, that $f(x) = 6 - 12x^2$ is the only function that solves your given equation.
A: And what they say is correct. If $A=\int_{-1}^1f(x)\,\mathrm dx$, then$$\int_{-1}^16+f(x)\,\mathrm dx=12+A\tag1$$and$$\int_{-1}^12f(-x)+3x^2A\,\mathrm dx=4A,\tag2$$since $\int_{-1}^1x^2\,\mathrm dx=2$ and $\int_{-1}^1f(-x)\,\mathrm dx=A$. Since $(1)=(2)$, $A=4$.
A: $$6 + f(x) = 2f(-x) +3x^2(\int_{-1}^{1}f(t) \, dt)=  2f(-x) +3x^2A\implies $$
$$f(x)-2f(-x)=3Ax^2-6$$
Change $x$ to $-x$ and you get $$ f(-x)-2f(x)=3Ax^2-6$$
Therefore $$f(x)-2f(-x) = f(-x)-2f(x)$$ which imlies $$f(x)=f(-x)$$
Plugging in $$f(x)-2f(-x)=3Ax^2-6$$ we get $$f(x)=6-3Ax^2$$
Thus $$A=\int_{-1}^1 f(t)dt = \int_{-1}^16-3At^2 dt =12-2A $$
Thus $3A=12$ which gives us $A=4$
