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!

• 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. – user296602 Dec 4 '18 at 1:26
• @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! – bletillafoliosa Dec 4 '18 at 1:28
• Yes. ${}{}{}{}{}$ – user296602 Dec 4 '18 at 1:28

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 $$\int_c^df(x)dx=\int_a^b f(\varphi(t))\cdot \frac{d\varphi}{dt}dt.$$