How do you find the point of a function where its integral is equal on both sides?

[SFNQuestions]

I think I have my own methodology but I'm curious to see what the formal definition of this is. I'm looking for an x-value of a function where its integral is the same on both sides of the x-value over some interval [a,b]. With some testing it doesn't seem like the average value theorem directly relates to the answer, but it seems like the general answer should be something close to that.

The answer I got is [math]x=F^{-1}(\frac{F(a)+F(b)}{2})[/math] where F(x) is the integral of f(x).

Edited February 22, 2017 by SFNQuestions

[Country Boy]

It sounds like you are looking for "x" such that [math]\int_a^x f(t)dt= \int_x^b f(t)dt[/math].
For example, if [math]f(x)= x^2[/math], a= 0, and b= 1, then [math]\int_0^x t^2 dt= \frac{1}{3}x^3[/math] while [math]\int_x^1 t^2 dt= \frac{1}{3}- \frac{1}{3}x^3[/math]. They will be equal when [math]\frac{1}{3}x^3= \frac{1}{3}- \frac{1}{3}x^2[/math]. Multiplying both sides by [math]\frac{1}{3}[/math] and then adding [math]x^3[/math] to both sides, [math]2x^3= 1[/math], [math]x^3= \frac{1}{2}[/math] so [math]x= \frac{1}{\sqrt[3]{x}}[/math].

[SFNQuestions]

It sounds like you are looking for "x" such that [math]\int_a^x f(t)dt= \int_x^b f(t)dt[/math].

For example, if [math]f(x)= x^2[/math], a= 0, and b= 1, then [math]\int_0^x t^2 dt= \frac{1}{3}x^3[/math] while [math]\int_x^1 t^2 dt= \frac{1}{3}- \frac{1}{3}x^3[/math]. They will be equal when [math]\frac{1}{3}x^3= \frac{1}{3}- \frac{1}{3}x^2[/math]. Multiplying both sides by [math]\frac{1}{3}[/math] and then adding [math]x^3[/math] to both sides, [math]2x^3= 1[/math], [math]x^3= \frac{1}{2}[/math] so [math]x= \frac{1}{\sqrt[3]{x}}[/math].

Right but I'm looking for a general answer for any function.

[Mathematical]

So let me get this straight, because I want to make sure I understand what you are asking, you want to prove that given some function, [latex] f [/latex], continuous over [latex] [ a , b ] [/latex] that there is some [latex] x \in [a,b] [/latex] such that [latex] \int_a^x f(t)dt = \int_x^b f(t)dt [/latex]? If so, this is the same as asking for any area, can I divide it in two (with a straight line), such that both sides have equal area. The answer is clearly yes, in fact the intermediate value theorem necessitates that this is true.

 

Like you wrote, the simplest way I can think of to find x is to do what you did,

 

[latex] \int_a^x f(t)dt = \int_x^b f(t)dt \rightarrow F(x) - F(a) = F(b) - F(x) \rightarrow 2F(x) = F(b) + F(a) \implies x = arcF( \frac{F(b) + F(a)}{2} ) [/latex]

 

If you want to prove that it is true, then I'd use this method:

 

IVT states that given some [latex] c \in [f(a),f(b)] [/latex], assuming the function is generally increasing over our interval, then there is some [latex] u \in [a,b] [/latex] such that [latex] f(u) = c [/latex]

 

Changing the names to match our problem yields

 

If there is some [latex] c \in [F(a),F(b)] [/latex], assuming the function is generally increasing over our interval, then there is some [latex] x \in [a,b] [/latex] such that [latex] F(x) = c [/latex]

 

Since c is exactly halfway between [latex] F(a) [/latex] and [latex] f(b) [/latex], then we've proven that there is some [latex]x[/latex] that satisfies our conditions. Replacing [latex]c[/latex] with [latex]\frac{F(a)+f(b)}{2}[/latex] shows us that the formula that you came up with is exactly the formula that you need to find x, and while you might be able to find other such methods, this is probably the simplest out there.

It sounds like you are looking for "x" such that [math]\int_a^x f(t)dt= \int_x^b f(t)dt[/math].
For example, if [math]f(x)= x^2[/math], a= 0, and b= 1, then [math]\int_0^x t^2 dt= \frac{1}{3}x^3[/math] while [math]\int_x^1 t^2 dt= \frac{1}{3}- \frac{1}{3}x^3[/math]. They will be equal when [math]\frac{1}{3}x^3= \frac{1}{3}- \frac{1}{3}x^2[/math]. Multiplying both sides by [math]\frac{1}{3}[/math] and then adding [math]x^3[/math] to both sides, [math]2x^3= 1[/math], [math]x^3= \frac{1}{2}[/math] so [math]x= \frac{1}{\sqrt[3]{x}}[/math].

 

 

I just want to point out that you meant [math]x= \frac{1}{\sqrt[3]{2}}[/math]

[SFNQuestions]

Is there any practical application for this? The original inspiration for this came from an experimental way to create a "functional average" like for sums exponents representing different characteristics.

[Mathematical]

Is there any practical application for this? The original inspiration for this came from an experimental way to create a "functional average" like for sums exponents representing different characteristics.

A functional average? As in the average of a function over an interval? Can you explain more clearly?

[SFNQuestions]

A functional average? As in the average of a function over an interval? Can you explain more clearly?

No, I can't explain it because as far as I know it doesn't exist.