Showing something involving integrals is an inner product I have this problem:

Let $C([0,1])$ be the real vector space of continuous functions on the interval [0,1]. 
Show that
$<. , .>: C([0,1]) \times C([0,1]) \rightarrow \mathbb{R}$
given by
$<f,g> :=  \int_{0}^{1}f(x)g(x)dx$
is an inner product.

So, to do this, I have to show that several properties hold true. I'm having a lot of trouble showing that $<f,f> \ge 0$ for all $f$. This is what I have so far:
$<f,f> = \tfrac{f(x)^3}{3} \Big|_0^1 $
Then I thought about what happens when you actually evaluate this. I broke it up into cases depending on if $f$ is strictly increasing or decreasing along $[0,1]$ and if $f(1) = f(0)$. Is this right?  I'm not sure if it is because I'm having trouble with what happens if $f$ is strictly decreasing.
I'm also having my doubts that this actually is an inner product space. When you calculate a definite integral, you are calculating the area under a curve. If a function consists mostly of negative values, then its 'area' will be negative. So isn't it true that the inner product here can be negative? Even if we take the inner product of $f$ and itself? 

Are you sure about what is $<f,f>$? Try an example, let say $f(x)=\cos x$. Observe that $$<f,f>=\int_0^1 [f(x)]^2\, dx$$ and $[f(x)]^2\geqslant 0$ for all $x$. Now you can use the fact that "the integral is the area under the curve" to argue that $<f,f>$ is always nonnegative.
