Area Between Curves

  • Thread starter Rapier
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  • #1
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Homework Statement



Find the area bounded by the curves y=x^2 and y= 2 - x^2 for 0 ≤ x ≤ 2.


Homework Equations



∫top - ∫bottom


The Attempt at a Solution



∫(2-x^2)dx - ∫x^2dx

What I'm confused about is that the two equations only cross on [-1,1] so within the interval of the problem I only have an enclosed area on [0,1]. But the problem asks for the area on [0,2]. How do I reconcile the differing intervals?
 

Answers and Replies

  • #2
LCKurtz
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If you draw the vertical line x = 2 it gives a right boundary just like x = 0 gives the left boundary. Your curves cross so you have to do it in two parts.
 
  • #3
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OH! I think I see that now.

So I'll have:

[∫(0→1)(2-x^2)dx - ∫(0→1)x^2dx] + [∫(1→2)(2-x^2)dx - ∫(1→2)x^2dx]

Basically the area between the curves on [0,1] plus the bits hanging off on [1,2].

A = 4/3 un^2

I knew there was something I was missing and it's been a couple of weeks since we did that.

Thanks for the helps!
 
  • #4
LCKurtz
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OH! I think I see that now.

So I'll have:

[∫(0→1)(2-x^2)dx - ∫(0→1)x^2dx] + [∫(1→2)(2-x^2)dx - ∫(1→2)x^2dx]

Basically the area between the curves on [0,1] plus the bits hanging off on [1,2].

A = 4/3 un^2

I knew there was something I was missing and it's been a couple of weeks since we did that.

Thanks for the helps!

Your integrand is always y-upper - y-lower. Check that on the interval [1,2].
 
  • #5
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Oh! Yep. I forgot that my lines crossed.

One step at a time.... :)

A = 4 un^2

Thanks again.
 

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