# The average f(x) value, from a mechanical point of view?

• RandomMystery
Like I said, it's been bugging me for a while and I thought maybe someone could help. In summary, the average f(x) value, from a mechanical point of view, is x2/3. Algebra I and II method?There is no clear answer, but the average point value of the function x2 is x2/3 from a mechanical point of view. For example, if you tilt your head to the right to see the graphs in standard orientation or even better, alt-ctrl-left, you can see that the average is k.

#### RandomMystery

The average f(x) value, from a mechanical point of view? Algebra I and II method?

I don't quite understand why the average point value of the function x2 is x2/3 from a mechanical point of view -> no integrals, I also don't prefer derivative approach.

For example:
* tilt your head to the right to see the graphs in standard orientation or even better alt ctrl left​

k=y
J=J​
the average is k because everything is k

=====y====
==lllllllllllllllllllllllll
==lllllllllllllllllllllllll
x=llllllllllllllllllllllllll
==lllllllllllllllllllllllll
==lllllllllllllllllllllllll

kx=y
the average is kx/2

=====y====
==llll
==llllllll
x=llllllllllll
==llllllllllllllll
==llllllllllllllllllll

kx2 = y
the average is kx2/3

=====y======
==ll
==llll
x=llllllll
==lllllllll

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This post above was fine a two hours ago. Something cut off the rest of the post.

y = kx2

llIII
llllII
llllllll
lllllllliiiiiiii
lllllllliiiiiiiillllllllllllllll

If you can see it, the I's and i cancel out to form the average (also creating a rectangle that is 1/3 of the box)
If I fill the rest of the space to make a box:

llIIIBBBBBBBBBBBBB
llllIIBBBBBBBBBBBBB
llllllllBBBBBBBBBBBB
lllllllliiiiiiiiBBBBBBBB
lllllllliiiiiiiillllllllllllllll

The area under the kx2 is 1/3 of the box, which I can't figure out why from a mechanical, graphical, point of view. My old post showed a modification of a physics equation which may help explain, but I don't quite understand that either.

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The average of a function f(x) within the interval [a,b] is given by

$$\frac{1}{b-a} \int_a ^b f(x) dx$$

so if f(x) = x2 then ∫f(x) dx = x2/3, so the average would be

$$\frac{1}{b-a} \left[ \frac{x^2}{3} \right]_a ^b$$

I want to know it without doing the anti-derivative. There has to a valid reason to why a third of the x cubed function is it's average. I don't know. It's been bugging me since physics. The closest I think I've gotten to doing it without the derivative is:
RandomMystery said:
I have the same question regarding jerking, and then the change of jerking based on time
I derived this equation from the anti-derivative of the constant jerking, distance equation-

$$\underline{(Final Velocity+ Final* Velocity without jerking + Initial Velcoity)}$$
3

(Final Velocity+ Final Velocity without jerking + Initial Velcoity)/3 =

(Jerking t^2)/6+ (Initial acceleration t)/2+ Initial Velocity =

Average Velocity with Constant Jerking =

This equation can also be used to find the average point between any two points in any parabola.
I don't quite understand why this works. I hope that by rearranging the equation I could understand it. This is the best most helpful form I came up with.

I also want to reform equations with a change in Jerking and so on... or the average point of higher polynomials for better understanding.

My friend saw something very similar in statistics- yet he couldn't explain it to me.

Any help would be greatly appreciated!
Maybe I just need to study the derivative more.

*edit, I mean Final Velocity without jerking not Average velocity without jerking

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You can think of it as summing the values of f(x)dx from a to b.

So the sum of all the values of f(x)dx is

$$\int_a ^b f(x) dx$$

and then you divide by the number of intervals which work out as b-a.

Well that's how I think of it.

Okay, I'm going to do three integrals of the following equation:

Final Jerk = Initial Jerk

Final Acceleration = Jerk(t) + Initial Acceleration

Final Velocity = Jerk(t2)/2 + Initial Acceleration(t) + Initial Velocity

Final Displacement = Jerk(t3)/6 + Initial Acceleration(t2)/2 + Initial Velocity(t) + Initial Displacement

When I had done this, I did not know of the integral. I just didn't understand why it was 1/6 of jerk.

I found the instantaneous velocity by factoring out the (t):

(Jerk(t2)/6 + Initial Acceleration(t)/2 + Initial Velocity)t = (Instantaneous Velocity) t

So, I then rearranged some things and got:

[Jerk(t2)/2 + Initial Acceleration(t)/2 + Initial Velocity] + Initial Acceleration(t)/2 + Initial Velocity + Initial Velocity
3

Using the original Jerk = Jerk equations we can sub in and get:

Final Velocity + Final Velocity as if there was no Jerk + Initial Velocity​
3​

At this point some friend from statistics told me that this made sense from statistics (the Final velocity without jerking part), but he didn't or couldn't explain to me why.

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rock.freak667 said:
You can think of it as summing the values of f(x)dx from a to b.

So the sum of all the values of f(x)dx is

$$\int_a ^b f(x) dx$$

and then you divide by the number of intervals which work out as b-a.

Well that's how I think of it.

Yeah, I think that I'm trying to approach it from a statistical method by using averages. Unfortunately I don't know about line of regression or have taken statistics to explain it. I don't even think they explain why the formulas work in statistics, just plug and chug. But your right. I'm just trying to figure out, if you can find the average point of a parabola, just by knowing the mechanics of algebra I and II, not Calculus.

RandomMystery said:
Yeah, I think that I'm trying to approach it from a statistical method by using averages. Unfortunately I don't know about line of regression or have taken statistics to explain it. I don't even think they explain why the formulas work in statistics, just plug and chug. But your right. I'm just trying to figure out, if you can find the average point of a parabola, just by knowing the mechanics of algebra I and II, not Calculus.

If you wanted to do it from a statistical perspective, you would need to find the probability distribution.

I don't know an algorithmic way to do this, but an intuitive way is to draw horizontal lines starting from x = a to x = b and count the number of points that each horizontal line goes through, and based on that create a probability density function.

From that you can take the expectation of your so called "random" variable E[X] and you will get your statistical average.

If your pdf obeys Kolmogorov's axioms, this should work.

Here is the integral of the distance equation (with jerking)
http://latex.codecogs.com/gif.latex?\frac{ax^4}{24}&space;&plus;&space;\frac{bx^3}{6}&space;&plus;&space;\frac{cx^2}{2}&space;&plus;d=\left&space;[&space;\frac{\frac{ax^3}{6}&space;&plus;&space;\frac{4bx^2}{6}&space;&plus;&space;\frac{4cx}{2}&space;&plus;4d}{4}\right&space;]x

I want to put it in terms of the average of four function, but I don't know what are the ??

http://latex.codecogs.com/gif.latex?=\frac{&space;\langle&space;\frac{\left&space;[&space;\left&space;(&space;\frac{ax&plus;b&space;&plus;b}{2}\right&space;)x&space;&plus;&space;c&space;\right&space;

I think this may be one of the ?

http://latex.codecogs.com/gif.latex?=\frac{&space;\langle&space;\frac{\left&space;[&space;\left&space;(&space;\frac{ax&plus;b&space;&plus;b}{2}\right&space;)x&space;&plus;&space;c&space;\right&space;

This may have to do with Simpson's Rule, which I don't understand why it works.
If you can put the ? in a form that makes sense.

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http://latex.codecogs.com/gif.latex?=\frac{&space;\langle&space;\frac{\left&space;[&space;\left&space;(&space;\frac{ax&plus;b&space;&plus;b}{2}\right&space;)x&space;&plus;&space;c&space;\right&space;

I got this equation, but it's wrong. If your wondering about how I got this form, I just rearranged the average rate of change of the derivative of this equation. s:b...?

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