For all real values of \(x\), the function \(f(x)\) satisfies

\[6+f(x)=2f(-x)+3x^2\left(\int_{-1}^1 f(t)\: dt\right).\]

It follows that \(\displaystyle\int_{-1}^1 f(x)\: dx\) equals

\[\text{(a)}\quad 4, \quad\text{(b)}\quad 6, \quad\text{(c)}\quad 11, \quad\text{(d)}\quad \dfrac{17}{2}, \quad\text{(e)}\quad 23.\]

Let \(\int_{-1}^1 f(x) dx = A\).

This means that \(\int_{-1}^1 f(t) dt = A\) also; the choice of variable within the integral is unimportant.

Thus for all \(x, 6+f(x)=2f(-x)+3x^2A\), and so replacing \(x\) with \(-x\), we also have \(6+f(-x)=2f(x)+3x^2A\).

Subtracting these two equations gives us \(f(x)-f(-x)= 2f(-x)-2f(x) \implies f(x) = f(-x)\) for all \(x\), and so \(f(x)\) is even.

So returning to our equation, \(6=f(x) +3x^2A\).

Now we can integrate this from \(-1\) to \(1\), giving us \(\bigl[6x\bigr]_{-1}^1=A + \bigl[x^3\bigr]_{-1}^1A\).

So we have \(12=A + 2A\), so \(A = 4\), and the answer is (a).

Strictly speaking we could have found \(A\) without establishing the evenness of \(f(x)\), but it is a good technique to see.