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I just need to know what to do with the absolute value sign.
@SmoothMath @Zarkon @UnkleRhaukus
You graph it. Remember y = |x^2 -4| = x^2 - 4 or 4 - x^2
So it's really 3 curves I am dealing with right?
You have to pay attention to the OR part. You will have two separate graphs. However, my theory is that the areas of both will be the same.
Okay now I am confused.
I get that i consider the or part but then how do I find the area?
On one graph, you will find the area enclosing y = x^2 - 4 and y = x^2/2 + 4 On the other graph you will find the are enclosing y = 4 - x^2 and y = x^2/2 + 4
Ohh okay! So I find the area between the curves for both cases and add them up right?
Hmm let me do it...
Don't forget to find the enclosing points first.
Yeah that's easy.
yeah I know that @UnkleRhaukus but I jest need to know how to deal with the absolute value sign.
Yep I got that... but you have to do this with a calculator don't you?
i would break the graph in half first , because it is symmetric about, the y axis then i would calculate the area when x is from 0 to 2, and add this to the area from x=2 to x=4
I JUST noticed the symmetry :P .
So I would go: |dw:1355700438412:dw|
Right? To deal with:|dw:1355700518056:dw|
that looks right to me, (you draw your fours funny) but yes thats is the right integration
My teachers said my 4's looked like 9 >.< .
looks more like a five to me
so thats just where the first curve y=|x^2-4| is really y= 4-x^2
(like Hero was trying to saying )
Ohh! So it would be: |dw:1355700839817:dw|
Yeah I know hero said that :) .
yeah thats right ,
Thanks a lot both of you :) @UnkleRhaukus @Hero .
Could you give hero a medal too :) .
so take those integrals, add them together, then dont forget to Double your answer because we were only looking at the RHS of the graph
Yep I know :) .
No @UnkleRhaukus. I wasn't trying to say it. I did say it.