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

1. Feb 7, 2009

### irresistible

1. The problem statement, all variables and given/known data
f:[0,1] $$\rightarrow$$ R is a continous function such that
$$\int$$f(t)dt (from 0 to x) = $$\int$$ f(t)dt( from x to 1) for all x$$\in$$[0,1] .
Describe f.

2. Relevant equations
integral represents area

3. 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?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 7, 2009

### Staff: Mentor

One possibility is that f(t) = 0 for all t in [0, 1]. If there are other functions, none come to mind.

3. Feb 7, 2009

### slider142

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.

4. Feb 7, 2009

### irresistible

I think so,Thank you