Are the bounds of t always [0, 1] for line integrals? 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!
 A: 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 
$$ \int_c^df(x)dx=\int_a^b f(\varphi(t))\cdot \frac{d\varphi}{dt}dt.$$
