Find the area of the plane region that is enclosed by the curves:

$$y=\left| x^2-4 \right|$$ and$$y=(\frac{ x^2 }{ 2 }) +4$$

Question

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 · Answer

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

anonymous · Answer

@SmoothMath @Zarkon @UnkleRhaukus

anonymous · Answer

@Hero

hero · Answer

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

anonymous · Answer

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

hero · Answer

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 · Answer

Okay now I am confused.

anonymous · Answer

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

hero · Answer

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 · Answer

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

hero · Answer

In theory

anonymous · Answer

Hmm let me do it...

hero · Answer

Don't forget to find the enclosing points first.

anonymous · Answer

Yeah that's easy.

anonymous · Answer

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

unklerhaukus · Answer

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