Question in an exercise of line integral

[Anto]

Hello, Can someone help me with this exercise, please ?

∫_g (x − y)dx + (x + y)dy, where g is the line connecting (1,0) with (0,2).

I've parametrized the function as r = (t,2-2t) where t E [0,1]. Then ||r(t)|| = √5.
So, when I replace it in the integral i get ∫¹ ₀ (t-2+2t) + (t+2-2t) = √5 and the result should be 7/2.
What Am I doing wrong ?

Thanks.

 

[pka]

Hello, Can someone help me with this exercise, please ?

∫_g (x − y)dx + (x + y)dy, where g is the line connecting (1,0) with (0,2).

I've parametrized the function as r = (t,2-2t) where t E [0,1]. Then ||r(t)|| = √5.
So, when I replace it in the integral i get ∫¹ ₀ (t-2+2t) + (t+2-2t) = √5 and the result should be 7/2.
What Am I doing wrong ?

First, where are the \(\displaystyle dt's~?\)
I would use \(\displaystyle (-t+1,2t)\) so that \(\displaystyle dx=-dt~\&~dy=2dt\)
\(\displaystyle \int_0^1 {\left[ {( - t + 1) - 2t} \right]\left( { - dt} \right) + \left[ {( - t + 1) + 2t} \right]\left( {2dt} \right)} \)