Review question

# If $f(x)$ always satisfies this equation, what is $\int_{-1}^1 f(x)\: dx$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8105

## Suggestion

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.