Just a Cal 2 Problem

$\int_{-a}^a x^{2n}f(x)\,dx \; = \; \int_{-a}^a x^{2n}f(-x)\,dx \; = \; \int_{-a}^a x^{2n}\big[k - f(x)\big]\,dx \; = \; \tfrac12k\int_{-a}^a x^{2n}\,dx \; = \; \tfrac{1}{2n+1}ka^{2n+1}$ with $a=n=k=1$ , the answer is $\boxed{\tfrac13}$ .

Here's a quick and easy solution that takes advantage of the wording of the problem and can be used for timed tests where the reasoning of the answer does not need to be given.

It can be safely assumed that the value of the integral we need to find is constant for any function $f(x)$ such that $f(x)+f(-x)=1$ .

Obviously, $f(x)=\frac{1}{2}$ satisfies the condition, and we have

$\int_{-1}^{1}\ \frac{1}{2}x^2\ dx=\left [\frac{1}{6}x^3\right ]_{-1}^{1}=\frac{1}{3}$

Answer to the bonus question is k(a^(2n+1))/(2n+1)

Let's try a substitution approach. Let $y = -x$ so that the original functional equation transforms into:

$f(x) + f(y) = 1 + x + y$ (i).

For $x = y = 0$ , we obtain the initial condition: $2f(0) = 1 \Rightarrow f(0) = \frac{1}{2}.$ Differentiating (i) with respect to x and y respectively yields:

$f'(x) = 1$ and $f'(y) = 1$ ;

which equates to $f'(x) = f'(y) = A$ , and integrating with respect to x gives $f(x) = Ax + B$ (where $A,B \in \mathbb{R})$ (ii). At $f(0) = \frac{1}{2}$ , we obtain $B = \frac{1}{2}$ , or $f(x) = Ax + \frac{1}{2}.$ (iii). Substituting (iii) into (i) now gives:

$(Ax + \frac{1}{2}) + (Ay + \frac{1}{2}) = 1 + x + y \Rightarrow A(x+y) = x+y \Rightarrow A = 1$ ;

thus, $\boxed{f(x) = x + \frac{1}{2} }.$ The required integral above now computes to:

$\int_{-1}^{1} (x + \frac{1}{2})x^2 dx = \frac{x^4}{4} + \frac{x^3}{6} |_{-1}^{1} = \boxed{\frac{1}{3}}.$