+33615 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!
+33615 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.
+118609 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? 
+33615 Melody, does this help
average x distance = area so average = area/distance.
+118609 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.
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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! 
+33615 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!
+118609 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.
+33615 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!
