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

  • Thread starter Thread starter Unassuming
  • Start date Start date
  • Tags Tags
    Integral
Click For Summary

Homework Help Overview

The problem involves proving that a continuous function \( f \) on the interval \([a,b]\) must be zero everywhere if its integral over that interval is zero, given that \( f(x) \geq 0 \) for all \( x \) in \([a,b]\).

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the implications of the integral being zero and consider the definitions of lower and upper sums. Some question how to incorporate the continuity of \( f \) into their reasoning.
  • There is a suggestion to consider a point where \( f(x_0) > 0 \) and explore the consequences for the lower sum, questioning whether a partition should be defined.
  • One participant proposes using the epsilon-delta definition of continuity to establish a relationship between \( f \) and the integral.

Discussion Status

The discussion is ongoing with various approaches being considered. Some participants are exploring the implications of continuity and the properties of integrals, while others are questioning assumptions about the function's behavior at specific points.

Contextual Notes

Participants are working under the constraint of proving the statement without relying on the Fundamental Theorem of Calculus. There is an emphasis on the continuity of \( f \) and its implications for the integral and sums.

Unassuming
Messages
165
Reaction score
0

Homework Statement



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

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

Homework Equations





The Attempt at a Solution



Assume [tex]\int_a^b \! f = 0[/tex].

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

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

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.
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 105 ·
4
Replies
105
Views
11K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
26
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
4
Views
1K
  • · Replies 26 ·
Replies
26
Views
3K