If f:[0,1] -> R is a continuous function, describe f.

Click For Summary

Homework Help Overview

The problem involves a continuous function f defined on the interval [0,1] with a specific property relating the integrals of f over two subintervals. Participants are tasked with describing the function based on the given integral condition.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the implications of the integral condition, noting that the area under the curve from 0 to x equals the area from x to 1. Some suggest that f could be the zero function, while others explore the necessity for f to cross the x-axis based on the signed nature of the integral.

Discussion Status

The discussion is ongoing, with participants exploring various interpretations of the integral condition. Some have provided geometric and algebraic insights, suggesting that the function must exhibit specific behaviors, such as crossing the x-axis or being identically zero.

Contextual Notes

Participants are considering the implications of continuity and the properties of integrals, including the concept of signed area, which may influence their understanding of the function's behavior over the interval.

irresistible
Messages
15
Reaction score
0

Homework Statement


f:[0,1] \rightarrow R is a continuous function such that
\intf(t)dt (from 0 to x) = \int f(t)dt( from x to 1) for all x\in[0,1] .
Describe f.




Homework Equations


integral represents area



The Attempt at a Solution



what ever the function is, I know that the area under the graph from 0 to x is equal to the area under the graph from x to 1.
But how else can I describe f?
 
Physics news on Phys.org
One possibility is that f(t) = 0 for all t in [0, 1]. If there are other functions, none come to mind.
 
Note that the integral is a signed area, so areas under the x-axis are negatively signed. If you draw a picture of a function that is entirely on one side of the axis (positive or negative) and start integrating from 0 to small x, it implies the remaining curve must take a nose-dive in order to account for the small amount of area between 0 and x. On the other hand, if we place x close to 1, it implies the small region between x and 1 must be higher than the previous region in order to account for the larger area, which doesn't agree with the previous scenario (or the function is the 0 function).
This implies that the function must cross the x-axis between 0 and 1 (or be the 0 function). Can you continue from there?
Another less geometric and more algebraic approach is to use the fundamental theorem of calculus to note that there exists some function F on [0,1] such that the first integral is F(x) - F(0) and the second integral is F(1) - F(x), which leads to the same conclusion.
 
I think so,Thank you
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
8
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
5
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K