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# find the average slope

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find the average slope of 2x^2 -2x +5.

Aug 4, 2014

#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! Aug 5, 2014 ### 8+0 Answers #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. 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? Aug 5, 2014 #3 +5 Melody, does this help average x distance = area so average = area/distance. 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! 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! 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. 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
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Yes okay, thanks Alan. Aug 5, 2014