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

[roam]

Homework Statement

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

The Attempt at a Solution

(a)

du=4x³dx

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

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 \frac{du}{dx} to "vanish"? And how do we find this point?

 

[Mark44]

Homework Statement

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

The Attempt at a Solution

(a)

du=4x³dx

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

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.

(b) What does it mean for the Jacobian \frac{du}{dx} to "vanish"? And how do we find this point?

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

 

[roam]

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. Any help is appreciated.

 

[Mark44]

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.

 

[roam]

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

\frac{du}{dx}=4x^3

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...

 

[Mark44]

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.

 

[roam]

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?

\frac{du}{dx}=4x^3, so dx=\frac{du}{4x^3}

\int^0_{-1} 1+u \frac{du}{4x^3} + \int^1_0 1+u \frac{du}{4x^3}

 

[Mark44]

No. As I said before, expressions involving x should not appear in the new integrals.

 

[roam]

No. As I said before, expressions involving x should not appear in the new integrals.

I see, so is the following correct:

Since 4x³=4u^3/4

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

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

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

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

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

=\frac{1}{2} \int^1_{0} u+\frac{3}{2}u^{\frac{2}{3}}

=\frac{1}{2} \frac{5}{2} = \frac{5}{4} =1.25

Is this correct now?

 

[Mark44]

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.
(1+u) \frac{du}{4u^{\frac{3}{4}}} \neq 1+u 4u^{-\frac{4}{3}}

(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.