Why does a line integral evaluated using two different parameterizations of a curve give two different answers?

Question

For example, I'm asked to find the line integral of $C_1$ above for a vector field $\vec F=x\vec i + y\vec j$

To do so I first try and find a parameterization of $C_1$. I parameterize $C_1$ as $\vec r(t)=t\vec i+(-2t+2)\vec j$.

But apparently this is already incorrect, as the correct parameterization was $\vec r(t) = (1-t)\vec i + 2t\vec j $.

The latter parameterization seems to be the same as mine, but it results in a different final answer. The question tells us to use the formula $$\int_{C_1}\vec F\ \cdot d\vec r= \int_a^b\vec F(\vec r(t)) \cdot \vec r'(t)dt$$ and using that formula with my parameterization results in a different answer than using it with the textbook's parameterization.

Why do two different but equivalent parameterizations of the curve result in different values for the line integral? And how do I know which parameterization to use/which the question is asking for?

Answer 1

Because your parameterisation describes a path which begins at $2\vec j$ and ends at $\vec i$, whereas it should begin at $\vec i$ and end at $2\vec j$ (as the other one does).

Answer 2

Your parameterization differs in that, for all intents and purposes, it's "backwards."

Plug in $t=0$, the "start" point. Where your parameterization starts incorrectly at $2 \vec j$ in that parameterization, the suggested one correctly starts at $\vec i$.

Likewise, plug in $t=1$, the "end" point. You obtain $\vec i$ and the suggested one obtains $2 \vec j$.

You can tell yours are wrong owing to the direction $C_1$ has its arrow pointing in. Both answers travel along the same path, but in different directions!

In general, if you want to parameterize a line, starting at some vector $\vec v$ and ending at some vector $\vec w$, the formula would be

$$\vec r(t) = (1-t) \vec v + t \vec w$$

where $t \in [0,1]$.