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

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SUMMARY

To prove that if the integral of a continuous function f over the interval [a,b] is zero, then f(x) must equal zero for all x in [a,b], one must utilize the properties of continuity and the definition of the integral. Given that f is non-negative and continuous, if there exists any point x0 in [a,b] where f(x0) = c > 0, then there exists a partition that results in a lower sum greater than zero, contradicting the assumption that the integral is zero. Therefore, f(x) must equal zero throughout the interval.

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Homework Statement



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].

Homework Equations





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.
 
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HINT: It may be better to use the Fundamental Theorem of Calculus.
 
Hootenanny said:
HINT: It may be better to use the Fundamental Theorem of Calculus.

I don't "know" that yet. I am supposed to prove this without that.
 
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?
 
Dick said:
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?

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?
 
Unassuming said:
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?

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.
 

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