# Thread: Line Integrals Across Line Segment

1. ## Line Integrals Across Line Segment

Hello all,

I have a problem where I need to parametrize a line segment in order to evaluate a line integral, but I don't know how to do that. The problem is as follows:

"Evaluate $\int_C$ $ye^x ds$, where $C$ is the line segment joining $(1,2)$ to $(4,7)$."

The book has a small example about the parametrization of line segments, but I can't follow it. Can anyone give a blow-by-blow description of how to use the equation: $r(t) = (1-t)r_0 + tr_1$ where $0?

I understand that $r_0$ and $r_1$ are just the two points, initial and final, but how do you plug them into the previous formula to get $x = [something]$ and $y = [something else]$?

Austin Martin

2. Originally Posted by auslmar
Hello all,

I have a problem where I need to parametrize a line segment in order to evaluate a line integral, but I don't know how to do that. The problem is as follows:

"Evaluate $\int_C$ $ye^x ds$, where $C$ is the line segment joining $(1,2)$ to $(4,7)$."

The book has a small example about the parametrization of line segments, but I can't follow it. Can anyone give a blow-by-blow description of how to use the equation: $r(t) = (1-t)r_0 + tr_1$ where $0?

I understand that $r_0$ and $r_1$ are just the two points, initial and final, but how do you plug them into the previous formula to get $x = [something]$ and $y = [something else]$?

Austin Martin
Hi

This is how I would have performed the job

One equation of the line segment joining $(1,2)$ to $(4,7)$ is 5x-3y+1=0 or y=5x/3 + 1/3

You can deduce that from one point of the line (x,y) to a very close one (x+dx,y+dy) you have the relationship dy=5dx/3

Now $ds = \sqrt{dx^2+dy^2}=\sqrt{dx^2+\frac{25}{9}dx^2}=\fra c{\sqrt{34}}{3}dx$

Therefore integral becomes

$\int_C ye^x ds = \int_{1}^{4} (\frac{5}{3}x+\frac{1}{3})e^x\frac{\sqrt{34}}{3}dx$

.. but this method does not match with the parametrization you are looking for