# Homework Help: Integral calculus involving Fubini

1. May 2, 2012

### Yami

Integral calculus involving Fubini's Theorem

1. The problem statement, all variables and given/known data
$$f(x,y) = x + y, if: x^2 ≤ y ≤ 2x^2$$
$$f(x,y) = 0, otherwise$$
Evaluate $$\iint_\textrm{I}f$$ where I = [0,1] x [0,1]

2. Relevant equations

For a Jordan domain K in ℝ^n, let h: K → ℝ and g: K → ℝ be continuous bounded functions with the property that
$$h(x) ≤ g(x)$$ for all points x in K.
Define
D = {(x,y) in ℝ^(n+1): x in K, h(x) ≤ y ≤ g(x)}.
Suppose that the function f: D → ℝ is continuous and bounded. Then
$$\int_\textrm{D}f = \int_\textrm{K}\int_{h(x)}^{g(x)}f(x,y)dydx$$
3. The attempt at a solution
$$\iint_\textrm{I}f = \int_{0}^{1}\int_{x}^{2x^2}(x + y)dydx$$
$$= \int_{0}^{1}\left[xy + \frac{1}{2}y^2\right]_{x}^{2x^2}dx$$ etc
until I eventually came to 1/5 as my answer.

However the grader wrote that this $$\int_{0}^{1}\int_{x}^{2x^2}(x + y)dydx$$ is the wrong integral. Can anyone help me figure out why?

Last edited: May 2, 2012
2. May 2, 2012

### Dick

Sure. If x=1 and y=2x^2, the upper limit of your integral, gives you the point (x,y)=(1,2). That's not in I. Did you draw a graph of the integration region and your boundary curves? I'd suggest you try that.

3. May 2, 2012

### Yami

I did

Okay, I would guess that it's
$$\iint_\textrm{I}f = \int_{0}^{\frac{1}{\sqrt{2}}}\int_{x^2}^{2x^2}(x + y)dydx$$

Am I warm?

Last edited: May 3, 2012
4. May 2, 2012

### Dick

Getting better. From x=0 to x=1/sqrt(2) your upper bound for y is 2*x^2. From x=1/sqrt(2) to x=1 it's just 1. You should probably think about splitting the integral up into parts if you are doing it that way. And shouldn't the lower bound be x^2?

Last edited: May 2, 2012
5. May 3, 2012

### Yami

I just realized I've been writing h(x) = x instead of x^2 as the lower bound. That's not the main reason it's wrong though, yes?

6. May 3, 2012

### Dick

It's one. The other is that the upper bound is wrong between x=1/sqrt(2) and x=1.

7. May 3, 2012

### Yami

Okay my next guess is this:
$$\iint_\textrm{I}f = \int_{0}^{\frac{1}{\sqrt{2}}}\int_{x^2}^{2x^2}(x + y)dydx + \int_{\frac{1}{\sqrt{2}}}^{1}\int_{x^2}^{1}(x + y)dydx$$

8. May 3, 2012

### Dick

That looks ok to me. Does it work? It's getting kind of late here, so I might be a little fuzzy.

9. May 3, 2012

### Yami

I've got this:
$$\frac{45\sqrt{2} - 32}{80\sqrt{2}}$$
which is about 0.5. It feels right I guess.

10. May 3, 2012

### Dick

I get something different. Might want to check that. You could also integrate dx first. Then you wouldn't need to split the integral up and it should give you the same answer.

11. May 3, 2012

### Yami

Okay I went over it again and found a mistake so the new answer is:
$$\frac{21\sqrt{2} - 18}{40\sqrt{2}}$$

I also tried your suggestion of integrating dx first and set this up:

$$\int_{0}^{1}\int_{\sqrt{0.5y}}^{\sqrt{y}}(x+y) dx dy$$
which led to a different answer:
$$\frac{21\sqrt{2} - 16}{40\sqrt{2}}$$

I guess I made a mistake somewhere that I can't spot.
Either of those match what you got?

12. May 3, 2012

### Dick

The second one looks good.