# Homework Help: If integral f =0, prove that f(x)=0 for all x in [a,b]

1. Feb 3, 2009

### Unassuming

1. The problem statement, all variables and given/known data

Suppose f is continuous on [a,b] with f(x) >= 0 for all x in [a,b].

If $$\int_a^b \! f = 0$$ , prove that f(x)=0 for all x in [a,b].

2. Relevant equations

3. The attempt at a solution

Assume $$\int_a^b \! f = 0$$.

Then the lower and upper sums equal 0 also. I looked at the definitions of these and came up with,

$$\sum_{i=1}^n ( inf\{ f(x) : x_{i-1} \leq x \leq x_i \} )( \bigtriangleup x_i )=0$$.

This looks obvious to me but that usually means I'm missing something because I don't know where to use that f is continuous.

2. Feb 3, 2009

### Hootenanny

Staff Emeritus
HINT: It may be better to use the Fundamental Theorem of Calculus.

3. Feb 3, 2009

### Unassuming

I don't "know" that yet. I am supposed to prove this without that.

4. Feb 3, 2009

### Dick

Suppose there is an x0 in [a,b] such that f(x0)=c>0. Can you show there is a lower sum that is greater than 0?

5. Feb 3, 2009

### Unassuming

Should I define a partition? If f(x_0)=c and c > 0 then the function must go up to atleast c, and since it is continuous it does not "jump" there.

I was thinking of making a small partition of a,x_0, and b. I then know what the sup or inf values of the two rectangles (speaking of the hieghts) will be greater than or equal to x_0 for the upper sum.

Would this do it? I figure that the upper sum and the lower are equal since the integral in the problem was defined as existing?

6. Feb 3, 2009

### Dick

No. Knowing a particular upper sum is greater than zero doesn't get
you anywhere all you know is that the integral is less or equal to
any upper sum.

But, yes. Make a partition. Use the epsilon-delta definition of continuity to show there is an interval around x0 such that f>c/2 on that interval. The use the endpoints of that interval to define the partition.