Does absolute makes a difference in area

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Homework Help Overview

The problem involves finding the area between the curves defined by the equations y = 2 − x² and y = |x|. The original poster expresses confusion regarding the role of the absolute value in the integration process and seeks clarification on determining the intersection points of the curves.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss splitting the problem into cases based on the definition of |x|, questioning how to handle the absolute value in the integral. There are inquiries about finding intersection points for both x < 0 and x >= 0, and the implications of symmetry in the graphs.

Discussion Status

Some participants have provided guidance on how to approach the problem by suggesting the need to split the integration into two parts based on the cases for |x|. There is an ongoing exploration of the intersection points, particularly for the case where x < 0, and the discussion reflects a mix of understanding and uncertainty.

Contextual Notes

Participants note that the original poster is grappling with the concept of absolute values in the context of integration and the need to find valid intersection points for both cases of x. There is mention of symmetry in the graphs, which influences the approach to calculating the area.

jwxie
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Homework Statement



y = 2 − x^2, y = |x|
sketch and find the area of the region determined
by the intersections of the curves

Homework Equations



regular f - g of integral

The Attempt at a Solution



The absolute value puzzles me. How would it work then?

I mean we know how each equation looks like.
So I would do
integral of (2-x^2) - |x| from a to b

Now how would I get the intersection point and evaluate this integral with |x|?

Thanks
 
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Split into the cases x<0 and x>=0. What's a simpler expression for |x| in each case? What's the intersection point in each case?
 
2-x-x^2 = 0
this gives me -2, 1

and how do i integrate it?
treat it as a regular x but keep || when i do the calculation?
 
The point here is that for x>=0 |x|=x. For x<0 |x|=(-x). You have to split the integration into two parts. And you also have to find the intersection separately for each part. x=(-2) is NOT an intersection.
 
hi thanks for the help.
i still can't figure out the x <0
but after examining the graph, i see they are symmetry, so instead, i take 2 * integral of the positive, will produce the answer

yet, i still want to know how to find the intersection point for x <0
 
jwxie said:
2-x-x^2 = 0
this gives me -2, 1

and how do i integrate it?
treat it as a regular x but keep || when i do the calculation?

Your values of x above are the x-coordinates at the points of intersection of y = 2 - x^2 and y = x. The only one that is valid is x = 1, because |x| = x only if x >= 0.

The other point of intersection is found in almost the same way that you found the first one. The only difference is that you are looking for the point of intersection of y = 2 - x^2 and y = -x. The latter equation is due to the fact that |x| = -x for x <=0.

So, solve 2 - x^2 = - x for x. The domain this time is x <= 0, so discard any positive x values. You have already commented on the symmetry of the two graphs, so your result should not be surprising.
 
Hi, thanks for your help. Now I get it. ^^
 

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