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

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The discussion centers on a continuous function f defined on the interval [0,1] with the property that the integral from 0 to x equals the integral from x to 1 for all x in [0,1]. This condition implies that the area under the curve from 0 to x is balanced by the area from x to 1, suggesting that f must either be the zero function or cross the x-axis at least once within the interval. The geometric interpretation highlights that if f remains entirely above or below the x-axis, the areas would not balance, leading to contradictions. An algebraic approach using the fundamental theorem of calculus confirms that the function must exhibit similar behavior. Overall, f must either be identically zero or oscillate to ensure the areas remain equal.
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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?
 
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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
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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