Surprising result?

A

apple2357

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Integrating this function for different values of k between 0 and pi/2 appears to give the same result. I can't see why this might be the case?
I put it on Geogebra to test out..

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Etc...
 
It looks to me like a matter of symmetry; the curve under which you are integrating is symmetrical about the point (pi/4, 1/2).

The area in each case is half of a rectangle with base pi/2 and height 1, because the curve bisects that rectangle. And that happens because f(pi/2 - x) = 1 - f(x).
 
Good spot! It does look like it is.

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Just thinking about your transformations explanation... ( I imagine integrating these functions wouldnt be so easy!)
 
Yes, certain definite integrals are much easier than their indefinite integrals ...

I saw the answer purely visually; the last graph was most obvious, then I saw the symmetry in the others, and then I realized how to see that in the function itself. It's all about cofunctions.
 
I can see your transformations explanation works, i can't quite see why f(pi/2-x) is the same as f(x) but reflected in the way it is ( as below)
If it was f(x-pi/2) i would understand that as a simple translation but is f(pi/2-x) is a combination of transformations? Is this a translation and a reflection?


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How did you come up with f(pi/2-x) = 1-f(x) as the explanation for the graphical observation?
 
I'd call it a double reflection, first around x=pi/4 and then around y=1/2. Equivalently, and the way I described it initially, it is actually a rotation around the point (pi/4, 1/2). In the same way, reflecting in both y=0 and x=0 amounts to rotating by 180 degrees about the origin.

Replacing x with pi/2 - x reflects in x=pi/4, and replacing y with 1-y (that is, subtracting the result from 1) reflects in y=1/2. These are worth pondering until you see why!
 
Dr.Peterson said:
I'd call it a double reflection, first around x=pi/4 and then around y=1/2. Equivalently, and the way I described it initially, it is actually a rotation around the point (pi/4, 1/2). In the same way, reflecting in both y=0 and x=0 amounts to rotating by 180 degrees about the origin.

Replacing x with pi/2 - x reflects in x=pi/4, and replacing y with 1-y (that is, subtracting the result from 1) reflects in y=1/2. These are worth pondering until you see why!
Click to expand...

So If we reflect y=f(x) in x=a, we get the curve y=f(−x+2a) ?

Is this because if you want to move every point to a position the same distance the other side of x=a we end up at x+2(a-x). Which gives us 2a -x
Is that how you would explain it?
 
Yes, that is one good way to explain this fact.

And, of course, the same fact is why the other reflection is 1 - y.
 
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