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

- anonymous

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

I just need to know what to do with the absolute value sign.

- anonymous

@SmoothMath @Zarkon @UnkleRhaukus

- anonymous

@Hero

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- Hero

You graph it. Remember
y = |x^2 -4| = x^2 - 4 or 4 - x^2

- anonymous

So it's really 3 curves I am dealing with right?

- Hero

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.

- anonymous

Okay now I am confused.

- anonymous

I get that i consider the or part but then how do I find the area?

- Hero

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

- anonymous

Ohh okay! So I find the area between the curves for both cases and add them up right?

- Hero

In theory

- anonymous

Hmm let me do it...

- Hero

Don't forget to find the enclosing points first.

- anonymous

Yeah that's easy.

- anonymous

yeah I know that @UnkleRhaukus but I jest need to know how to deal with the absolute value sign.

- UnkleRhaukus

|dw:1355700012700:dw|

- UnkleRhaukus

|dw:1355700073022:dw|

- anonymous

Yep I got that... but you have to do this with a calculator don't you?

- UnkleRhaukus

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

- anonymous

I JUST noticed the symmetry :P .

- UnkleRhaukus

|dw:1355700278738:dw|

- anonymous

So I would go:
|dw:1355700438412:dw|

- anonymous

Right? To deal with:|dw:1355700518056:dw|

- UnkleRhaukus

that looks right to me, (you draw your fours funny) but yes thats is the right integration

- anonymous

My teachers said my 4's looked like 9 >.< .

- UnkleRhaukus

looks more like a five to me

- anonymous

|dw:1355700680945:dw|

- UnkleRhaukus

so thats just where the first curve
y=|x^2-4| is really y= 4-x^2

- UnkleRhaukus

(like Hero was trying to saying )

- anonymous

Ohh! So it would be:
|dw:1355700839817:dw|

- anonymous

Yeah I know hero said that :) .

- UnkleRhaukus

yeah thats right ,

- anonymous

Thanks a lot both of you :) @UnkleRhaukus @Hero .

- anonymous

Could you give hero a medal too :) .

- UnkleRhaukus

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

- anonymous

Yep I know :) .

- Hero

No @UnkleRhaukus. I wasn't trying to say it. I did say it.

Looking for something else?

Not the answer you are looking for? Search for more explanations.