Evaluating an Integral on a Line Segment

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SUMMARY

The discussion focuses on evaluating the integral ∫(y-x)dx + xydy along the line segment from (3,4) to (2,1). The user derived the slope of the line segment as y = 3x - 5 and parametrized it with x = t, resulting in y = 3t - 5, dx = dt, and dy = 3. The integral was transformed into ∫(-9t + 13t - 15)dt, but the user struggled with determining the correct limits of integration. The correct value of the integral is -39/2, and the user was advised to find the appropriate values of t corresponding to the endpoints of the line segment.

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Mdhiggenz
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Homework Statement



∫(y-x)dx+xydy

c: the line segment from (3,4)to(2,1)

First thing I did was get the formula for the slop of the line segment, getting the relation

y=3x-5

I then parametrizided with x=t which gave y=3t-5

dx=dt and dy=3

plugging in I get the following integral

∫-9t+13t-15dt

I don't know how to find the limits of integration though. I tried using (4,1) and just inserting a negative in front of the integral in order to reverse it to the correct order, but still did not get the correct value.

Which is -39/2

Thanks again



Homework Equations





The Attempt at a Solution

 
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Mdhiggenz said:

Homework Statement



∫(y-x)dx+xydy

c: the line segment from (3,4)to(2,1)

First thing I did was get the formula for the slop of the line segment, getting the relation

y=3x-5

I then parametrizided with x=t which gave y=3t-5

dx=dt and dy=3

plugging in I get the following integral

∫-9t+13t-15dt

I don't know how to find the limits of integration though. I tried using (4,1) and just inserting a negative in front of the integral in order to reverse it to the correct order, but still did not get the correct value.

Which is -39/2

Thanks again



Homework Equations





The Attempt at a Solution


Your initial point is ##(x(t),y(t)) = (3,4)##, so for which value of t does x=3, y=4 in your parametrization? Similarly, your final point is (2,1), so for what value of t is x=2, y=1? If the the time for the final point is less than the time for the initial point, you can do as you suggested and flip the limits by introducing an overall negative sign to the integral. I'm not sure if by "(4,1)" that is supposed to be your limits of integration, but if so, those are not the correct limits.
 

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