Area Between Curves: Find Area for 0 to 2

[Rapier]

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?

 

[LCKurtz]

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.

 

[Rapier]

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!

 

[LCKurtz]

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].

 

[Rapier]

Oh! Yep. I forgot that my lines crossed.

One step at a time... :)

A = 4 un^2

Thanks again.