Integral calculus involving Fubini

[Yami]

Integral calculus involving Fubini's Theorem

Homework Statement

$$f(x,y) = x + y, if: x^2 ≤ y ≤ 2x^2$$
$$f(x,y) = 0, otherwise$$
Evaluate $$\iint_\textrm{I}f$$ where I = [0,1] x [0,1]

Homework 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
$$\int_\textrm{D}f = \int_\textrm{K}\int_{h(x)}^{g(x)}f(x,y)dydx$$

The Attempt at a Solution

This was my answer
$$\iint_\textrm{I}f = \int_{0}^{1}\int_{x}^{2x^2}(x + y)dydx$$
$$= \int_{0}^{1}\left[xy + \frac{1}{2}y^2\right]_{x}^{2x^2}dx$$ 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?

 

[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.

 

[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?

 

[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?

 

[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?

 

[Dick]

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?

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

 

[Yami]

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

 

[Dick]

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

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

 

[Yami]

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

Thank you for your help.

 

[Dick]

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

Thank you for your help.

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.

 

[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?

 

[Dick]

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?

The second one looks good.