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Yes, multiplication.
How do you find the identity element? Under multiplication, the identity element is 1. What about here? Also how to find inverses? eg inverse of 2.
Cayley table for the binary operation multiplication Modulo 10, for the sub set of integers {2,4,6,8}...
HERE
Correct?
Now I understand. In order for the integral to remain unchanged, you must multiply the by the reciprocal 2/5 * 5/2 = 1. Therefore it is unchanged.
So if you were to multiply the numerator by 2, you would have to multiply the whole integral by 1/2.
How much is a 1/2 pound of your brain? I'd like to buy, if possible (you only use 10% after all!).
I'll have a go when I see it.
Another question if I may, why do you multiply the whole integral by 2/5?
You're a teacher? I should have known, there's difference between saying and teaching, it is a skill.
What I meant.. I understood du replaces 10x, since they are both equal, hence, du/√u
.....but why isn't there a dx at the end. Or alternatively, since dx = du/10x, a du/10x on the end? Ahhh...
Just realised...you did it slightly different, see: http://img242.imageshack.us/my.php?image=4atb6.png
You'd still end up with the same answer.
Those pesky du's and dx's.....They never explain what it means, they just tell you it. lol
Kind regards
Thankyou, that was beautifully written. I actually learned alot from that. But what happened to the dx bit? It seemed to disappear. Shouldn't it have become 2/5 ∫ (u)^-1/2 du dx?
Sorry, confused again.
I understood your method completely though.
How do I integrate
(4x)
---------------
sqrt((5x^2)-1)
confused
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