Integration with absolute value

In summary, the problem is in the context of a probability problem; however, my question in regards to a computation regarding a particular integral. All that is needed to know is that the probability density function is 1 in the range 0 < y < 1 , y-1 < x < 1 - y, and 0 otherwise.
  • #1
Gridvvk
56
1

Homework Statement


The problem is in the context of a probability problem; however, my question in regards to a computation regarding a particular integral. All that is needed to know is that the probability density function is 1 in the range 0 < y < 1 , y-1 < x < 1 - y, and 0 otherwise.

I need to find E(XY) and E(|XY|).

Homework Equations


[tex] E(XY) = \int_0^{1} \int_{y-1}^{1-y}\ XY \ dx dy[/tex]
[tex] E(|XY|) = \int_0^{1} \ \int_{y-1}^{1-y} \ |XY| \ dx dy[/tex]

The Attempt at a Solution



E(XY) = 0. Simple calculation.

For E(|XY|) however, I am getting 1/24, but WolframAlpha computes it out to be 1/12.

I am doing:
[tex] E(|XY|) = \int_0^{1}\ \int_{y-1}^{1-y}\ |XY| \, dx dy[/tex] = [tex]

\int_0^{1} \int_0^{1-y} \ XY \, dx dy[/tex]

I am not sure why this is incorrect. Making the lower bound of x 0 worked for me when I was computing E(|X|), but is giving me half the answer for E(|XY|).
 
Last edited:
Physics news on Phys.org
  • #2
Fixed your LaTeX, which was garbling your integrals.

For definite integrals, the LaTeX markup should look like this:

\int_{lower_bound}^{upper_bound} <integrand> dx dy

If lower_bound or upper_bound is a single character, you don't need the braces, but for expressions such as y - 1, you do need braces, not parentheses as you used.

Also, you had stuff like \1 - y, which was causing problems.
 
  • Like
Likes 1 person
  • #3
Mark44 said:
Fixed your LaTeX, which was garbling your integrals.

For definite integrals, the LaTeX markup should look like this:

\int_{lower_bound}^{upper_bound} <integrand> dx dy

If lower_bound or upper_bound is a single character, you don't need the braces, but for expressions such as y - 1, you do need braces, not parentheses as you used.

Also, you had stuff like \1 - y, which was causing problems.

Thanks! I was surprised it magically fixed itself. I'll keep that in mind for future posts.
 
  • #4
I don't think this is right:
$$ \int_0^{1} \int_0^{1-y} \ XY \, dx dy$$
You are changing the shape of the region of integration, removing half of it. The region over which integration should be taking place is the triangle whose vertices are (-1, 0), (0, 1), and (1, 0). The region you are using in the integral above is also a triangle, with vertices (0, 0), (0, 1), and (1, 0), that has half the area of the larger triangle. I believe that's why your answer (1/24) is half of what Wolfram shows (1/12).
 
  • #5
Mark44 said:
I don't think this is right:
$$ \int_0^{1} \int_0^{1-y} \ XY \, dx dy$$
You are changing the shape of the region of integration, removing half of it. The region over which integration should be taking place is the triangle whose vertices are (-1, 0), (0, 1), and (1, 0). The region you are using in the integral above is also a triangle, with vertices (0, 0), (0, 1), and (1, 0), that has half the area of the larger triangle. I believe that's why your answer (1/24) is half of what Wolfram shows (1/12).

That does make sense geometrically, I rechecked and it turns out.

$$ 2 * \int_0^{1} \int_0^{1-y} \ X \, dx dy = \int_0^{1} \int_{y-1}^{1-y} \ |X| \, dx dy $$

So you would need a factor of 2 for |X| as well. Thanks for the help!
 

What is integration with absolute value?

Integration with absolute value is a mathematical process that involves finding the area under a curve when the function being integrated contains an absolute value term.

What is the purpose of integrating with absolute value?

The purpose of integrating with absolute value is to find the total change in a quantity over a given interval, even when the quantity may have negative values.

How is integration with absolute value different from regular integration?

Integration with absolute value is different from regular integration because it involves breaking the function into two parts, one for the positive values and one for the negative values, and then finding the area under each part separately.

What are some real-life applications of integration with absolute value?

Integration with absolute value is commonly used in physics and engineering to calculate displacement, velocity, and acceleration of objects. It is also used in economics to calculate total profit or loss.

What are some techniques for solving integrals with absolute value?

Some techniques for solving integrals with absolute value include using the properties of absolute value, breaking the function into two parts, and using substitution or integration by parts when necessary.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
482
  • Calculus and Beyond Homework Help
Replies
6
Views
541
  • Calculus and Beyond Homework Help
Replies
10
Views
415
  • Calculus and Beyond Homework Help
Replies
4
Views
836
  • Calculus and Beyond Homework Help
Replies
2
Views
529
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
440
  • Calculus and Beyond Homework Help
Replies
27
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
670
Back
Top