How do I evaluate a double integral over a specific region?

Click For Summary

Homework Help Overview

The discussion revolves around evaluating a double integral over a specified region, defined by the inequalities y ≤ x ≤ 1 and 0 ≤ y ≤ 1. The original poster attempts to integrate the function x^2 e^y with respect to x and expresses uncertainty about completing the integration process.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the method of integration, with some suggesting to integrate with respect to y first, while others express concerns about the resulting expressions containing variables instead of numerical answers. There is also a focus on correctly interpreting the region of integration.

Discussion Status

The discussion is ongoing, with participants providing guidance on how to approach the integration and clarifying the setup of the region R. There is a mix of interpretations regarding the order of integration and its implications on the resulting expressions.

Contextual Notes

Participants note the importance of sketching the region of integration and rewriting it in a suitable form for integration. There are indications of confusion regarding the limits of integration and the implications of the chosen order of integration.

kieranl
Messages
22
Reaction score
0

Homework Statement



Sketch the region of integration and then evaluate the double integral:


Homework Equations



[tex]\int\int[/tex]x2exydA over the region R= {(x,y), y<=x<=1, 0<=y<=1}

The Attempt at a Solution



I have managed to do half of the problem and integrate it respect to x but then have no idea how to finish the problem.

I mangaged to get upto [tex]\int[/tex]ey(2/y2-2/y3-1/y)+ey2(y-2/y+2/y3) dy

can someone please tell me how to finish it or if i have done something wrong so far??

cheers
 
Last edited:
Physics news on Phys.org
Welcome to PF!

Hi kieranl! Welcome to PF! :smile:

(have a ≤ :wink:)
kieranl said:
[tex]\int\int[/tex]x2exydA over the region R= {(x,y), y<=x<=1, 0<=y<=1}

I mangaged to get upto [tex]\int[/tex]ey(2/y2-2/y3-1/y)+ey2(y-2/y+2/y3) dy

uhh? :confused:

start again, and integrate over dy first! :smile:
 
but if i integrated by dy first wouldn't i end up with an answer containing y's and not numbers??
 
kieranl said:
but if i integrated by dy first wouldn't i end up with an answer containing y's and not numbers??

If you integrate with respect to dy, then you eliminate y, and only have x's.
 
but i have to integrate over y<=x<=1 for dx so the final answer would then contain y's?
 
Draw a picture of the region R. Then use the picture to write R in the form

R={(x,y):a<x<b, f(x)<y<g(x)}.
 
Hi kieranl! :smile:

(have an integral: ∫ and a ≤ :wink:)
kieranl said:
but i have to integrate over y<=x<=1 for dx so the final answer would then contain y's?

No, you have to integrate over 0 ≤ y ≤ x ≤ 1 …

whether you do it over x first or over y first is up to you. :wink:
 

Similar threads

  • · Replies 19 ·
Replies
19
Views
3K
Replies
10
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
2
Views
2K
Replies
3
Views
2K
Replies
4
Views
2K
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 17 ·
Replies
17
Views
4K