- #1

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## Homework Statement

Reverse the order of integration:

[tex]\int_{0}^{1}\int_{y^2}^{2-y} f(x,y) \ dx\ dy[/tex]

**This was on a test I took today. I was having trouble with it and was wondering how I should have gone about it?**

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- Thread starter Fusi
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- #1

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Reverse the order of integration:

[tex]\int_{0}^{1}\int_{y^2}^{2-y} f(x,y) \ dx\ dy[/tex]

- #2

Mark44

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## Homework Statement

Reverse the order of integration:

[tex]\int_{0}^{1}\int_{y^2}^{2-y} f(x,y) \ dx\ dy[/tex]

This was on a test I took today. I was having trouble with it and was wondering how I should have gone about it?

Sketch a graph of the region over which integration is taking place. In the integral above, the inner integral runs from x = y

- #3

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So x=2-y and x=[itex]y^{2}[/itex] so y=2-x y=[itex]\sqrt{x}[/itex]

intersect at: [itex]2-x=\sqrt{x}[/itex] [itex]x=1[/itex] [itex]y=1[/itex]

y at 0 is (0,2)

I got confused because at y=1,x=1 each side can be explained differently so my first attempt at a solutions was:

[tex]\int_{0}^{1}\int_{0}^{\sqrt{x}} f(x,y) \ dy\ dx + \int_{1}^{2}\int_{2-x}^{1} f(x,y) \ dy\ dx[/tex]

I used some random logic in my head to get that and we've never seen anything like that in class before so I crossed it out and just wrote:

[tex]\int_{0}^{2}\int_{2-x}^{\sqrt{x}} f(x,y) \ dy\ dx[/tex]

intersect at: [itex]2-x=\sqrt{x}[/itex] [itex]x=1[/itex] [itex]y=1[/itex]

y at 0 is (0,2)

I got confused because at y=1,x=1 each side can be explained differently so my first attempt at a solutions was:

[tex]\int_{0}^{1}\int_{0}^{\sqrt{x}} f(x,y) \ dy\ dx + \int_{1}^{2}\int_{2-x}^{1} f(x,y) \ dy\ dx[/tex]

I used some random logic in my head to get that and we've never seen anything like that in class before so I crossed it out and just wrote:

[tex]\int_{0}^{2}\int_{2-x}^{\sqrt{x}} f(x,y) \ dy\ dx[/tex]

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- #4

Mark44

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This isn't correct, but it's closer than what you have below. Here's what you should get:So x=2-y and x=[itex]y^{2}[/itex] so y=2-x y=[itex]\sqrt{x}[/itex]

intersect at: [itex]2-x=\sqrt{x}[/itex] [itex]x=1[/itex] [itex]y=1[/itex]

y at 0 is (0,2)

I got confused because at y=1,x=1 each side can be explained differently so my first attempt at a solutions was:

[tex]\int_{0}^{1}\int_{0}^{\sqrt{x}} f(x,y) \ dy\ dx + \int_{1}^{2}\int_{2-x}^{1} f(x,y) \ dy\ dx[/tex]

[tex]\int_{0}^{1}\int_{0}^{\sqrt{x}} f(x,y) \ dy\ dx + \int_{1}^{2}\int_{0}^{2 - x} f(x,y) \ dy\ dx[/tex]

I used some random logic in my head to get that and we've never seen anything like that in class before so I crossed it out and just wrote:

[tex]\int_{0}^{2}\int_{2-x}^{\sqrt{x}} f(x,y) \ dy\ dx[/tex]

- #5

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Yea I made a mistake on my last post. I actually did write my second bound as [itex]{0}\leq{y}\leq{2-x}[/itex]

We've never seen a problem like that before so I got so nervous I just crossed the entire answer out. The answer I decided to write down in the last couple seconds was actually:

[tex]\int_{0}^{2}\int_{\sqrt{x}}^{2-x} f(x,y) \ dy\ dx[/tex]

I kind of just tried to push the two integrals from my past answer together so I wouldn't have two iterated integrals.

We've never seen a problem like that before so I got so nervous I just crossed the entire answer out. The answer I decided to write down in the last couple seconds was actually:

[tex]\int_{0}^{2}\int_{\sqrt{x}}^{2-x} f(x,y) \ dy\ dx[/tex]

I kind of just tried to push the two integrals from my past answer together so I wouldn't have two iterated integrals.

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- #6

Mark44

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You understand why a single integral won't work, right?

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Yea, I have a bad habit of writing something that I know can't work when I get nervous.

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