Evaluate the work done integral_C F* dr by the vector field F(x,y) = y−x, x along the straight line segment from (2,0) to (0,2).

The line segment goes from $P_0(2,0)$ to $P_1(0,2)$. We can parameterize it as $r(t) = (1-t)P_0 + tP_1$ for $0 \le t \le 1$.

$$r(t) = (1-t)\langle 2,0 \rangle + t\langle 0,2 \rangle$$ $$r(t) = \langle 2(1-t), 0 \rangle + \langle 0, 2t \rangle$$ $$r(t) = \langle 2-2t, 2t \rangle$$

So, $x(t) = 2-2t$ and $y(t) = 2t$.

First, find the derivative of $r(t)$:

$$r'(t) = \left\langle \frac{d}{dt}(2-2t), \frac{d}{dt}(2t) \right\rangle = \langle -2, 2 \rangle$$

Then, $d\mathbf{r} = r'(t) \, dt = \langle -2, 2 \rangle \, dt$.

The given vector field is $\mathbf{F}(x,y) = \langle y-x, x \rangle$.
Substitute $x=2-2t$ and $y=2t$:

$$\mathbf{F}(x(t), y(t)) = \langle (2t) - (2-2t), (2-2t) \rangle$$ $$\mathbf{F}(x(t), y(t)) = \langle 2t - 2 + 2t, 2-2t \rangle$$ $$\mathbf{F}(x(t), y(t)) = \langle 4t - 2, 2-2t \rangle$$

$$\mathbf{F} \cdot d\mathbf{r} = \langle 4t - 2, 2-2t \rangle \cdot \langle -2, 2 \rangle \, dt$$ $$\mathbf{F} \cdot d\mathbf{r} = ((4t - 2)(-2) + (2-2t)(2)) \, dt$$ $$\mathbf{F} \cdot d\mathbf{r} = (-8t + 4 + 4 - 4t) \, dt$$ $$\mathbf{F} \cdot d\mathbf{r} = (-12t + 8) \, dt$$

The work done is $\int_C \mathbf{F} \cdot d\mathbf{r} = \int_0^1 (-12t + 8) \, dt$.

$$\int_0^1 (-12t + 8) \, dt = \left[ -12\frac{t^2}{2} + 8t \right]_0^1$$ $$= \left[ -6t^2 + 8t \right]_0^1$$

Now, evaluate at the limits of integration:

$$= (-6(1)^2 + 8(1)) - (-6(0)^2 + 8(0))$$ $$= (-6 + 8) - (0)$$ $$= 2$$

The work done is 2.