I have two functions and I want to show that they are orthogonal over a given interval. Do I just multiply them together and show that the integral over the interval is zero?
If your definition of inner product is <f, g>= $\displaystyle \int_a^b f(x)g(x)dx$. But there are other ways of defining the inner product in function spaces, usually involving a "weighting function", so the inner product is $\displaystyle <f(x),g(x)>= \int_a^b f(x)g(x)\mu(x)dx$. For example, the Leguerre polynomials are orthogonal using the inner product defined by $\displaystyle <f, g>= \int_0^\infty f(x)g(x)e^{-x}dx$. In that case the "$\displaystyle e^-{x}$ is necessary because the interval of definition is infinite. Bessel functions are orthogonal using the inner product defined by $\displaystyle <f, g>= \int_0^1 x f(x)g(x)dx$. Here, the "x" factor is necessary because the Bessel functions may have a pole of order 1 at 0.
In general, two vectors in an innerproduct space are orthgonal if and only if their inner product is 0. Check to see what inner product you should be using.
actually it turn out quite nice:
$\displaystyle h(x) = f(x) \cdot g(x) = \frac{{x\left( {3x^2 - 1} \right)\left( {5x^2 - 3} \right)}}
{4}$
now you can easily see that h(x) is an odd function since h(x) = -h(-x)
and we all know that the integral of an odd function over a symmetrical interval is 0.