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I was given the task to find the line integral $\int _C (x+y)ds$ where $C$ is the line segment from $(0,1,1)$ to $(3, 2, 2)$.

I parameterised $C$ as $3t\vec{i}+(1+t)\vec{j}+(1+t)\vec{k}$, which means that $\vec{r'}(t)=3\vec{i}+\vec{j}+\vec{k}$ and that $||\vec{r'}(t)||=\sqrt11$.

This gave me the integral $\sqrt11 \int(4t+1)dt$, but then I wasn't sure what the bounds of $t$ are supposed to be. In class, we've been using $t \space∈\space [0,1]$, and if I use that, I get $3\sqrt11$.

The only thing I'm uncertain of here is why the bounds of $t$ go from $0$ to $1$. And if they don't, how can I determine what the bounds are? (Also, if I made any other mistakes in my calculation, please correct me!) Thank you very much!

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    $\begingroup$ It depends on your parameterization of the curve. Does $t = 0$ correspond to the initial point and $t = 1$ correspond to the terminal point? If yes, cool. If no, use other bounds. $\endgroup$ – user296602 Dec 4 '18 at 1:26
  • $\begingroup$ @T.Bongers when you say 'correspond,' do you mean if I substitute t=0 and t=1 back into the parameterisation, I get the initial and terminal points respectively? Thank you! $\endgroup$ – bletillafoliosa Dec 4 '18 at 1:28
  • $\begingroup$ Yes. ${}{}{}{}{}$ $\endgroup$ – user296602 Dec 4 '18 at 1:28
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In short this is dependent on the choice of parametrization. As a simple example, let's set up the line integral of $x+y$ along the line segment $C$ connecting $(0,0)$ and $(1,1)$ in $\mathbb{R}^2$. We can parametrize $C$ as $C(t)=(t,t)$ for $0\le t\le 1$ or we can parametrize $C$ "more quickly" using $C(t)=(2t,2t)$ for $0\le t\le \frac{1}{2}$. Using the first parametrization: $$ \int_C (x+y)ds=\int_0^1x(t)\frac{dx}{dt}\cdot dt+\int_0^1 y(t)\frac{dy}{dt}\cdot dt=\int_0^1t\cdot dt+\int_0^1t\cdot dt=\int_0^12t\cdot dt=t^2\bigg|_0^1=1.$$ Using the second parametrization: $$ \int_C (x+y)ds=\int_0^{\frac{1}{2}}x(t)\frac{dx}{dt}\cdot dt+\int_0^\frac{1}{2} y(t)\frac{dy}{dt}\cdot dt=\int_0^{\frac{1}{2}}4t\cdot dt+\int_0^{\frac{1}{2}}4t\cdot dt=\int_0^{\frac{1}{2}}8t\cdot dt=4t^2\bigg|_0^{\frac{1}{2}}=1.$$ So, the integral does not care about the parametrization. More particularly, we can see that the $\frac{dx}{dt}$ and $\frac{dy}{dt}$ terms "correct" the error introduced by reparametrization.

This is actually a special case of the Change of Variables formula, which in this case tells us that given a reparametrization $\varphi:[a,b]\to [c,d]$ then $$ \int_c^df(x)dx=\int_a^b f(\varphi(t))\cdot \frac{d\varphi}{dt}dt.$$

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