Evaluate $\int_{C}(x+y) d s, C: x=3 t, y=-3 t+3$, for $-2 \leq t \leq-1$
$-31 \times 0^{6}$

Step 1 ：Given the line integral \(\int_{C}(x+y) d s\), where C is the curve defined by the parametric equations \(x = 3t\) and \(y = -3t + 3\) for \(-2 \leq t \leq -1\).

Step 2 ：Substitute the parametric equations into the integrand to get \(f(t) = 3t + (-3t + 3) = 3\).

Step 3 ：Calculate the differential of the arc length ds as \(ds = \sqrt{(dx/dt)^2 + (dy/dt)^2} dt = \sqrt{(3)^2 + (-3)^2} dt = 3\sqrt{2} dt\).

Step 4 ：Substitute f(t) and ds into the line integral to get \(\int_{C} f ds = \int_{-2}^{-1} 3 \cdot 3\sqrt{2} dt = 9\sqrt{2}\).

Step 5 ：Finally, evaluate the integral to get the final answer.

Step 6 ：The value of the line integral \(\int_{C}(x+y) d s, C: x=3 t, y=-3 t+3\), for \(-2 \leq t \leq-1\) is \(\boxed{9\sqrt{2}}\).