Why must orthogonal functions satisfy f(a)=f(b)=0?

[dirk_mec1]

Homework Statement

http://img55.imageshack.us/img55/8494/67023925dy7.png

The Attempt at a Solution

All functions orthogonal to 1 result in the fact that: $$\int_a^b f(t)\ \mbox{d}t =0$$

Now the extra condition is that f must be continous. (because of the intersection).

But where does the fact that f(a)=f(b)=0 comes from? And why look at the deratives?

 

[Dick]

Remember way back in first year calc, when you learned that to do that integral you find an antiderivative F(x) and evaluate F(b)-F(a). This is that same problem in disguise.

 

[dirk_mec1]

Well I thought of this: $$\int_a^b \int_a^t f(s)\ \mbox{d}s \mbox{d}t =0$$

 

[Dick]

Well I thought of this: $$\int_a^b \int_a^t f(s)\ \mbox{d}s \mbox{d}t =0$$

Fine. What are you going to do with it? Why don't you just define $$F(x)=\int_a^x f(s)\ \mbox{d}s$$
What are some of the properties of F(x)?