Double Integrals Homework: Solve (a) & (b)

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In summary: There are a few ways to do this, but I think the easiest is to break up the integral into two parts, as you tried to do before: one part for which x <= 0 and the other for x >= 0. Make sure to change the limits of integration appropriately for each part of the integral.In summary, The problem asks to evaluate the integral of f(x^4)dx from a to b, where a and b are the limits of integration and f(x) = 1 + x. A substitution of u = x^4 is used, resulting in the new integral of 1 + u * (du/4x^3). The method breaks down because both limits of integration in the new integral are
  • #1
roam
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Homework Statement



[PLAIN]http://img413.imageshack.us/img413/6839/49887281.gif [Broken]


The Attempt at a Solution



(a)

du=4x3dx

[tex]\int^b_{a} f(x^4)dx= \int^{1^4}_{(-1)^4} 1+u \frac{du}{4x^3}[/tex]

So, why does the method break down? Is it because a=b=1 (it should've been a < b)?

(b) What does it mean for the Jacobian [tex]\frac{du}{dx}[/tex] to "vanish"? And how do we find this point? :confused:
 
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  • #2
roam said:

Homework Statement



[PLAIN]http://img413.imageshack.us/img413/6839/49887281.gif [Broken]


The Attempt at a Solution



(a)

du=4x3dx

[tex]\int^b_{a} f(x^4)dx= \int^{1^4}_{(-1)^4} 1+u \frac{du}{4x^3}[/tex]

So, why does the method break down? Is it because a=b=1 (it should've been a < b)?
Both limits of integration in the new integral are 1, so the value of the definite integral is 0.
roam said:
(b) What does it mean for the Jacobian [tex]\frac{du}{dx}[/tex] to "vanish"? And how do we find this point? :confused:
An expression "vanishes" if its value becomes zero. For what value of x does your Jacobian become zero?
 
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  • #3
Mark44 said:
Both limits of integration in the new integral are 1, so the value of the definite integral is 0.

An expression "vanishes" if its value becomes zero. For what value of x does your Jacobian become zero?

So, the first step is to find the Jacobian? Can I get some clues on how to form the Jacobian matrix for this problem? I know that the u=x4, but I can't work out what the four entries in J are supposed to be. :confused: Any help is appreciated.
 
  • #4
There aren't four entries, since u is a function of only one variable, x. The problem even tells you what it is when it asks in part b, at what point does the "Jacobian" du/dx vanish. "Jacobian" is in quotes here because it's normally applied to functions of more than one variable.
 
  • #5
Mark44 said:
There aren't four entries, since u is a function of only one variable, x. The problem even tells you what it is when it asks in part b, at what point does the "Jacobian" du/dx vanish. "Jacobian" is in quotes here because it's normally applied to functions of more than one variable.

Oh, I see. Since u=x4

[tex]\frac{du}{dx}=4x^3[/tex]

It is only zero when x=0. So is this the point we were required to find (the "Jacobian" is zero here)?

And for part (c) in order to exclude this point, do I need to treat the integral as a "type I region" or a "type II" region? I sketched the function and it is symmetric about the y axis...
 
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  • #6
I don't remember what "Type I" and "Type II" regions are. When you make the substitution u = x^4, what is the new integral you get? It should not have x or dx in it.
 
  • #7
Mark44 said:
I don't remember what "Type I" and "Type II" regions are. When you make the substitution u = x^4, what is the new integral you get? It should not have x or dx in it.

So, the question says I need to separate it into two integrals (using the substitution u=x4) to avoid the point found in part (b) as an interior point. So is the following correct?

[tex]\frac{du}{dx}=4x^3[/tex], so [tex]dx=\frac{du}{4x^3}[/tex]

[tex]\int^0_{-1} 1+u \frac{du}{4x^3} + \int^1_0 1+u \frac{du}{4x^3}[/tex]
 
  • #8
No. As I said before, expressions involving x should not appear in the new integrals.
 
  • #9
Mark44 said:
No. As I said before, expressions involving x should not appear in the new integrals.

I see, so is the following correct:

Since 4x3=4u3/4

[tex]\int^0_{-1} (1+u) \frac{du}{4u^{3/4}} + \int^1_0 (1+u) \frac{du}{4u^{3/4}}[/tex]

since the graph is even about the y=axis we can write it as

[tex]2 \int^1_{0} (1+u) \frac{du}{4u^{\frac{3}{4}}}[/tex]

[tex]=2 \int^1_{0} 1+u 4u^{-\frac{4}{3}}[/tex]

[tex]=\frac{2}{4} \int^1_{0} 1+u^{-\frac{1}{3}}[/tex]

[tex]=\frac{1}{2} \int^1_{0} u+\frac{3}{2}u^{\frac{2}{3}}[/tex]

[tex]=\frac{1}{2} \frac{5}{2} = \frac{5}{4} =1.25[/tex]

Is this correct now?
 
  • #10
No, not correct. Evaluate the first integral you started with, which is a very easy one. The transformed integral should have the same value.

You have made several mistakes in your work.
[tex](1+u) \frac{du}{4u^{\frac{3}{4}}} \neq 1+u 4u^{-\frac{4}{3}}[/tex]

(1 + u)/(4u^(3/4)) is not symmetric about the y-axis. It's not even defined for u <= 0 . Your substitution was u = x^4, so u must be >= 0. If you solve for x, you need one formula for x >= 0 and another for x <= 0.

Finally, part b asked you where du/dx was equal to zero, which happens to be in the interval [-1, 1]. You will have to account for the fact that both integrands are undefined at some point in the interval of integration.
 

1. What is a double integral?

A double integral is a type of mathematical operation that involves integrating a function of two variables over a specific region in a 2D plane. It is essentially the extension of a single integral to two dimensions.

2. How do I solve a double integral?

To solve a double integral, you need to first identify the limits of integration for both variables and set up the integral in the correct form. Then, you can use various techniques such as iterated integrals, change of variables, or integration by parts to solve the integral.

3. What is the purpose of double integrals?

Double integrals are used to calculate the area under a surface in a 3D space or to find the volume of a solid. They are also used in various fields of science and engineering to model and solve real-world problems involving functions of two variables.

4. Can I use a calculator to solve double integrals?

Yes, most scientific calculators have a built-in function for solving double integrals. However, it is important to understand the concept and steps involved in solving a double integral before relying on a calculator.

5. How can I check if my double integral solution is correct?

There are a few ways to check the correctness of your double integral solution. You can use the Fundamental Theorem of Calculus, verify your answer graphically, or use online integral calculators to compare your solution. It is also helpful to double-check the limits of integration and the integrand to ensure they are correct.

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