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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I just need to know what to do with the absolute value sign.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@SmoothMath @Zarkon @UnkleRhaukus

Hero
 4 years ago
Best ResponseYou've already chosen the best response.1You graph it. Remember y = x^2 4 = x^2  4 or 4  x^2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So it's really 3 curves I am dealing with right?

Hero
 4 years ago
Best ResponseYou've already chosen the best response.1You 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay now I am confused.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I get that i consider the or part but then how do I find the area?

Hero
 4 years ago
Best ResponseYou've already chosen the best response.1On 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

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

Hero
 4 years ago
Best ResponseYou've already chosen the best response.1Don't forget to find the enclosing points first.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah I know that @UnkleRhaukus but I jest need to know how to deal with the absolute value sign.

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1355700012700:dw

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1355700073022:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yep I got that... but you have to do this with a calculator don't you?

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.1i 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I JUST noticed the symmetry :P .

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1355700278738:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So I would go: dw:1355700438412:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Right? To deal with:dw:1355700518056:dw

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0My teachers said my 4's looked like 9 >.< .

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.1looks more like a five to me

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1355700680945:dw

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

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.1(like Hero was trying to saying )

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ohh! So it would be: dw:1355700839817:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah I know hero said that :) .

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.1yeah thats right ,

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks a lot both of you :) @UnkleRhaukus @Hero .

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Could you give hero a medal too :) .

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.1so take those integrals, add them together, then dont forget to Double your answer because we were only looking at the RHS of the graph

Hero
 4 years ago
Best ResponseYou've already chosen the best response.1No @UnkleRhaukus. I wasn't trying to say it. I did say it.
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