Evaluating Line Integral: ∫(x+2y)dx+(x^2)dy

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Homework Help Overview

The problem involves evaluating the line integral ∫(x+2y)dx+(x^2)dy over a specified path consisting of two line segments. The context is within the subject area of vector calculus, specifically focusing on line integrals.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss their attempts to parameterize the line segments and set up the integral. There are questions about the correctness of their calculations and the method used to evaluate the integral.

Discussion Status

Some participants have shared their work and results, indicating a lack of agreement on the final value of the integral. There is an ongoing exploration of potential mistakes in the calculations, with requests for clarification and further assistance.

Contextual Notes

Participants express uncertainty about their approaches and the expected outcome, with one participant noting the expected answer is 5/2, while others have arrived at different results. The discussion reflects varying interpretations of the problem setup and calculations.

ahhppull
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Homework Statement


Evaluate the line integral ∫(x+2y)dx+(x^2)dy, where C consists of the line segments from (0,0) to (2,1) and (2,1) to (3,0)


Homework Equations





The Attempt at a Solution



I'm unsure of what to do. I did (1-t)r0 + t(r1) for (0,0) to (2,1) and (2,1) to (3,0). I didn't get the answer, which is 5/2 however.
 
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ahhppull said:

Homework Statement


Evaluate the line integral ∫(x+2y)dx+(x^2)dy, where C consists of the line segments from (0,0) to (2,1) and (2,1) to (3,0)


Homework Equations





The Attempt at a Solution



I'm unsure of what to do. I did (1-t)r0 + t(r1) for (0,0) to (2,1) and (2,1) to (3,0). I didn't get the answer, which is 5/2 however.

Show us your work and we will help you find what went wrong.
 
Honestly, I have no idea what to do, but here what I did.

From (0,0) to (2,1)
r(t) = (1-t)<0,0> + t<2,1>
r(t) = <2t,t>

x=2t
y=t

I found that the integral is from 0 to 1.

dx/dt = 2
dy/dt = 1

∫(x+2y)dx+(x^2)dy
Then I set x as 2t and y as t into the equation:


=∫[(2t+2t)(2)+(4t^2)(1)]dt
=∫(8t+4t^2)dt
= 4t^2 +4/3t^3 ] <-----evaluate from 0 to 1
= 4 + 4/3
= 16/3

Then I did this same thing for the from (2,1) to (3,0) and added the two numbers.
 
ahhppull said:
= 16/3
Then I did this same thing for the from (2,1) to (3,0) and added the two numbers.
I agree with 16/3, and I applied the same method for the second segment to get a total of 5/2. So your mistake must be in working you have not posted.
 

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