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
\[\text{(a)}\quad 4, \quad\text{(b)}\quad 6, \quad\text{(c)}\quad 11, \quad\text{(d)}\quad \dfrac{17}{2}, \quad\text{(e)}\quad 23.\]
The value of the definite integral in the given equation is a constant. We could call it \(A\).
The equation holds for all values of \(x\). What happens if we substitute \(-x\) for \(x\)?
To find \(\int_{-1}^1 f(x)\: dx\), we might try integrating both sides of the equation.
