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Find the area of the plane region that is enclosed by the curves:
\[y=\left x^24 \right\] and\[y=(\frac{ x^2 }{ 2 }) +4\]
 one year ago
 one year ago
Find the area of the plane region that is enclosed by the curves: \[y=\left x^24 \right\] and\[y=(\frac{ x^2 }{ 2 }) +4\]
 one year ago
 one year ago

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Dido525Best ResponseYou've already chosen the best response.0
I just need to know what to do with the absolute value sign.
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
@SmoothMath @Zarkon @UnkleRhaukus
 one year ago

HeroBest ResponseYou've already chosen the best response.1
You graph it. Remember y = x^2 4 = x^2  4 or 4  x^2
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
So it's really 3 curves I am dealing with right?
 one year ago

HeroBest ResponseYou've already chosen the best response.1
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.
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
Okay now I am confused.
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
I get that i consider the or part but then how do I find the area?
 one year ago

HeroBest ResponseYou've already chosen the best response.1
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
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
Ohh okay! So I find the area between the curves for both cases and add them up right?
 one year ago

HeroBest ResponseYou've already chosen the best response.1
Don't forget to find the enclosing points first.
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
yeah I know that @UnkleRhaukus but I jest need to know how to deal with the absolute value sign.
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.1
dw:1355700012700:dw
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.1
dw:1355700073022:dw
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
Yep I got that... but you have to do this with a calculator don't you?
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.1
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
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
I JUST noticed the symmetry :P .
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.1
dw:1355700278738:dw
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
So I would go: dw:1355700438412:dw
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
Right? To deal with:dw:1355700518056:dw
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.1
that looks right to me, (you draw your fours funny) but yes thats is the right integration
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
My teachers said my 4's looked like 9 >.< .
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.1
looks more like a five to me
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.1
so thats just where the first curve y=x^24 is really y= 4x^2
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.1
(like Hero was trying to saying )
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
Ohh! So it would be: dw:1355700839817:dw
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
Yeah I know hero said that :) .
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.1
yeah thats right ,
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
Thanks a lot both of you :) @UnkleRhaukus @Hero .
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
Could you give hero a medal too :) .
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.1
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
 one year ago

HeroBest ResponseYou've already chosen the best response.1
No @UnkleRhaukus. I wasn't trying to say it. I did say it.
 one year ago
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