Proof on a property of Riemann integrals

[kidmode01]

Homework Statement

Let f be continuous on [a,b] and suppose that $$f(x) \geq 0$$ for all x Є [a,b]
Prove that if there exists a point c Є [a,b] such that $$f(c) > 0$$, then

$$\int_{a}^{b} f > 0$$

Homework Equations

The Attempt at a Solution

Using my books notation,

Suppose P = {x0,x1,...,xn} is a partition of [a,b]. For each i = 1,...,n we let:
$$M_i(f) = sup\{f(x) : x \epsilon [x_{i-1},x_i]\}$$
$$m_i(f) = inf\{f(x) : x \epsilon [x_{i-1},x_i]\}$$

The upper sum of f with respect to P: U(f,P)
The lower sum of f with respect to P: L(f,P)

The upper integral of f on [a,b]: U(f) = inf{ U(f,P) : P is a partition of [a,b] }
The lower integral of f on [a,b]: L(f) = sup{ L(f,P) : P is a partition of [a,b] }

Suppose there exists a c Є [a,b] such that f(c) > 0 then
there exists a $$m_i(f) > 0$$
thus L(f,P) > 0

but then L(f) > 0, but since f is Riemann integrable:

$$0 < L(f) = U(f) = \int_{a}^{b} f$$

I know I'm kinda leaving out some details but is the outline alright? I have a feeling it's wrong since it was really short...

 

[rochfor1]

Looks pretty decent to me.

 

[kidmode01]

I should also include:

$$\Delta x_i = x_i - x_{i-1}$$

$$U(f,P) = \sum_{i=1}^n M_i \Delta x_i$$
$$L(f,P) = \sum_{i=1}^n m_i \Delta x_i$$