How to find parametric equation between two points in line integral?

Question

[In this example how can we find parametric equations of x and y.]

[1] [question]: https://i.stack.imgur.com/lTOnW.png

[1] [Solution]: https://i.stack.imgur.com/l8ao7.jpg

Answer 1

Okay, you have that your segment goes from $(0,0)$ to $(1,2)$. If you let $t$ be a parameter that expresses the value of $1$, then $2$ will be $2t$.

So your $x=t$ and your $y=2t$ but they are bounded from $0$ to $1$ since your ending point is $(1,2)$. So, $ t \in [0,1]$.

Then you get $dy = d(2t) = 2$ and your integral becomes :

$\int_C 3xydy = \int_0^1 3(t)(2t)(2)dt = \int_0^1 12t^2dt = [4t^3]_0^1 = 4 $

The second solution is when you go backwards, from $(1,2)$ to $(0,0)$. So your starting point is $(1,2)$ which means $1$ for $x$ and $2$ for $y$.

Since you go backwards to $(0,0)$, it's your staring point minus something that is still in the boundaries. So you get :

$x= 1-t$ and $y=2-2t$ with $ t \in [0,1]$. Thus $dy = d(2-2t) = -2dt$

Then solve the integral : $\int_0^1 3(1-t)(2-2t)(-2)dt$.