#7**+5 **

What I'm really trying to do in the denominator is to find the x-axis distance between two limits. If I integrate a constant (k, say) with respect to x, I just get kx, which, if my limits are hi and lo gives me k(hi - lo). If I set k to 1 then I just get hi - lo, the distance between the two limits. (In the case of $$$\int dx$$$ the 1 is implicit).

Perhaps better to ignore some of the calculus and just think in terms of *average x distance = area*. The area can be found using calculus, and the distance using common sense!

Alan
Aug 5, 2014

#1**+5 **

The slope at any point x is given by s(x) = 4x - 2. The average of this depends on the interval over which you want to calculate it. If you want it from x = lo to x = hi, say then the average is given by $$\frac{\int_{lo}^{hi}{s(x)dx}}{hi-lo} = 2(hi+lo-1)$$

If lo = -hi then the average is just -2.

Alan
Aug 4, 2014

#2**0 **

This is interesting Alan,

You use calculus in a way that I have not seen it used before (or I don't remember anyway)

I have just enough of knowlege to attempt to understand

2x^2 -2x +5

tangent function : 4x-2

Average = sum of all tangents(numerator makes sense) over number of tangents (**okay I am having trouble with the denominator)**

Maybe I just need to think it through. Can you help me get my mind around it?

Melody
Aug 5, 2014

#4**+3 **

Thanks Alan,

I can see that a denominator of 5 works for your example but I still don't comprehend where

$$\int1 dx$$ comes from. Why is it 1? What are you finding the sum of?

I obvioulsy have a major gap in my understanding.

------------------------------------------------------------------

Actually I am just thinking about the units as rates The average of anything can be thought of as whatever per unit or item. That makes sense. So thinking of it like this I absolutely get the difference on the bottom.

For example (for the benefit of others) The average of 6m,3m and 2m = 9m/3items = 3m/item

So $$\int1 dx$$ is the sum of units on the x axis which is 5.

So for the given question the average is ** sum of the gradients/unit of x**

Does that make sense, sort of anyway?

Maybe I have got it!

Melody
Aug 5, 2014

#5**+5 **

Melody,

$$$\int1dx$ is really just $\int dx$\\\\

Had I done this in LaTeX I would probably just have written $\int dx$, but I did it using Mathcad, which requires an argument to be specified.$$

$$$\int_{lo}^{hi}dx$ is just the distance along the x-axis from x=lo to x=hi.$$

I think you've got it!

Alan
Aug 5, 2014

#6**0 **

Yes but did what I said make sense?

Actually if you'd presented the integral without anything there I would have been even more confused. I didn't know you could have an integral with nothing there.

Melody
Aug 5, 2014

#7**+5 **

Best Answer

What I'm really trying to do in the denominator is to find the x-axis distance between two limits. If I integrate a constant (k, say) with respect to x, I just get kx, which, if my limits are hi and lo gives me k(hi - lo). If I set k to 1 then I just get hi - lo, the distance between the two limits. (In the case of $$$\int dx$$$ the 1 is implicit).

Perhaps better to ignore some of the calculus and just think in terms of *average x distance = area*. The area can be found using calculus, and the distance using common sense!

Alan
Aug 5, 2014