Integration over four lines! Tricky one?? 
I don't know how to seal the deal on this one. If you just read the comments on the answer by @RecklessReckoner you will see where I am still stuck.




I would appreciate if someone could overlook my method here please :). I was asked specifically to utilize $\oint_C \mathrm{F\cdot T \;ds}$
Consider the vector field:

$F(x,y) = xy \;\boldsymbol{i} + x^2 \;\boldsymbol{j} $

Let $C$ be the rectangle with vertices $(0,0),(3,0),(3,1),(0,1)$, let $T$ denote the unit tangent vector to $C$ directed anticlockwise around $C$.
Calculating:

$\oint_C \mathrm{F\cdot T \;ds}{}$

Parameters:  
$$\begin{align}
\\r_1(t) =  \left(3t,0\right), 0\leq t\leq 1
\\r_2(t) =  \left(3,t\right), 0\leq t\leq 1
\\r_3(t) =  \left(3-3t, 1\right), 0\leq t\leq 1
\\r_4(t) =  \left(0, 1 -t\right), 0\leq t\leq 1
\end{align}$$
Which using $T(t) = \cfrac{r'(t)}{||r'(t)||}$, we get:
$$\begin{align}
\\T_1(t) =  \left(1,0\right)
\\T_2(t) =  \left(0,1\right)(*),
\\T_3(t) =  \left(-1,0\right)(*),
\\T_4(t) =  \left(0, -1\right)
\end{align}$$
$\oint_C \mathrm{F\cdot T \;ds} = \int_0^1 xy \mathrm{ds} + \int_0^1 x^2\; \mathrm{ds} + \int_0^1 -xy \;\mathrm{ds} + \int_0^1 -x^2\; \mathrm{ds}$
$ \;= xy + x^2 - xy - x^2 = 0 $
 A: Your unit tangent vectors are fine; it's the dot products that are not evaluated properly. Your line integral should look like
$$\oint_C \ \mathbf{F} \cdot \mathbf{T} \ \ ds \ \  = \ \int_0^1 \  (3t) \ \cdot \ 0 \ \  (3 \ dt) \  + \  \int_0^1 \ 3^2 \ \ dt \ + \ \int_0^1  (3t) \ \cdot \ 1 \ \ (-3 \ dt) \ + \  \int_0^1 0^2 \ \ (-dt) \ \ . $$
On each leg, you would have:
$$ \vec{F} \ \cdot \ \vec{T_1} \ \vert_{y=0} \ = \ xy \ \vert_{y=0} \ = \ 0 \ \ ; $$
$$ \vec{F} \ \cdot \ \vec{T_2} \ \vert_{x=3} \ = \ x^2 \ \vert_{x=3} \ = \ 9 \ \ ; $$
$$ \vec{F} \ \cdot \ \vec{T_3} \ \vert_{y=1} \ = \ -xy \ \vert_{y=1} \ = \ -x \ \ ; $$
$$ \vec{F} \ \cdot \ \vec{T_4} \ \vert_{x=0} \ = \ -x^2 \ \vert_{x=0} \ = \ 0 \ \ . $$
In the last two line integral terms above, I've attached the negative orientation to the differentials.
[After you've had Green's Theorem, you can form a double integral over the rectangle, thus
$$ \iint_A \ \left( \ \frac{d}{dx} [x^2] \ - \ \frac{d}{dy}[xy] \ \right) \ \ dx \ dy \ \ = \ \ \iint_A \ ( \ 2x \ - \ x \ ) \ \ dx \ dy  $$
$$ = \ \ \int_0^1 \int_0^3 \ x \ \ dx \ dy  \ \ . $$
Using either method, the integral value is $ \ \frac{9}{2} \  $ . ]
A: Your computations are correct up until the end. The notation
$$\int_0^1 xy\,ds$$
for instance does not mean to anti-differentiate with respect to $s$ to get $xys$ and then plug in $1$ and $0$, it denotes the line integral of the function $xy$ over the bottom of the rectangle. You need to use the general formula
$$\int_C f(x,y)\,ds = \int_a^b f(x(t),y(t))\lVert r'(t)\rVert\,dt$$
to compute these.
Edit: Or rather, perhaps you're just forgetting to plug in for $x$ and $y$ before you evaluate the integrals.
A: $\oint_C F \cdot T \; \mathrm{ds} = \int_a^bf(x(t),y(t)) ||r'(t)|| dt$
Where we have $r(t) = x(t)\boldsymbol{i} + y(t)\boldsymbol{j},\; a\leq t\leq b$
From this I am sure you can appreciate:
$$\begin{align}\\x_1(t) = 3t, y_1(t)=0\\x_2(t)=3,y_2(t)=t\\x_3(t)=3-3t,y_3(t)=1\\x_4(t)=0,y_4(t)=1-t\end{align}$$
So we can now see that:
$$\begin{align}1) F \cdot T = (3t,0)\cdot(1,0)=3t\end{align}\\2)F \cdot T = (3,t)\cdot(0,1)=t\\3)F \cdot T = (3-3t,1)\cdot(-1,0)=3t-3\\4)F \cdot T = (0,1-t)\cdot(0,-1)=t-1$$
Which gives us $\int_0^1 3t * 3 dt + \int_0^1 t * 1 dt + \int_0^1 3t-3 * -3 dt + \int_0^1 t-1 * -1 dt$
= $\frac92+\frac12-\frac92+6-\frac12+1=7$
