+0  
 
0
643
8
avatar

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

Guest Aug 4, 2014

Best Answer 

 #7
avatar+27057 
+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
avatar+27057 
+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
avatar+93691 
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
 #3
avatar+27057 
+5

Melody, does this help

average

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

Alan  Aug 5, 2014
 #4
avatar+93691 
+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
avatar+27057 
+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
avatar+93691 
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
avatar+27057 
+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+93691 
0

Yes okay, thanks Alan.  

Melody  Aug 5, 2014

15 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.