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

## Solution

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.