= Solution
Assume the usual nondegenerate <interval> $a<b$. The integral is bilinear and symmetric because the functions and weight are real. It is nonnegative on $(f,f)$. If $f$ is nonzero at a point, <continuity> makes $|f|$ bounded below by a positive number on a nontrivial relative <interval> nearby, and the positive continuous weight is also bounded below there. Hence $\int_a^bf^2w>0$. Thus the formula is positive definite and defines an <inner product>.
The functions are orthogonal precisely when
$$
\boxed{\langle f,g\rangle=\int_a^bf(x)g(x)w(x)\,dx=0.}
$$
A singleton <interval> would make this integral zero for every function, so the condition $a<b$ is necessary.
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