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

 Aug 4, 2014

Best Answer 

 #7
avatar+30086 
+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
 #1
avatar+30086 
+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
avatar+109520 
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
avatar+30086 
+5

Melody, does this help

average

 average x distance = area  so average = area/distance.

 Aug 5, 2014
 #4
avatar+109520 
+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
avatar+30086 
+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
avatar+109520 
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
avatar+30086 
+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
 #8
avatar+109520 
0

Yes okay, thanks Alan.  

 Aug 5, 2014

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