Answer
$6\pi$
Work Step by Step
To find the the line integral, we simply compute $\displaystyle \int_{a}^bf(\vec{r}(t))||\vec{r}'(t)||dt$
where $\quad\displaystyle f(x,y) = x - y + 3 \quad$ , $\quad \vec{r}(t) = \langle \cos(t), \sin(t) \rangle\quad$ , $\quad a = \displaystyle 0\quad$ , and $\quad b = 2\pi$
$\vec{r}'(t) = \langle -\sin(t), \cos(t) \rangle$
$||\vec{r}'(t)|| = \sqrt{(-\sin(t))^2 + (\cos(t))^2} = \sqrt{\sin^2(t)+\cos^2(t)} = \sqrt{1} = 1$
$f(\vec{r}(t)) = \displaystyle \cos(t) - \sin(t) + 3$
$\displaystyle \int_{2\pi}^0(\cos(t) - \sin(t) + 3)(1)dt$
$[\sin(t)+\cos(t)+3t]_{x=0}^{x=2\pi}$
$(\sin(2\pi) + \cos(2\pi)+3(2\pi))-(\sin(0) + \cos(0)+3(0))$
$(0+1+6\pi) - (0+1+0)$
$6\pi$
