We are asked to evaluate the line integral:

$\int_{\Gamma} x \, ds$

where $\Gamma$ is the line segment joining the points $(0, 1)$ and $(1, 0)$ on the $xy$-plane.

Step 1: Parametrize the Line Segment

The line segment $\Gamma$ can be parametrized as: $\mathbf{r}(t) = (1 - t)(0, 1) + t(1, 0) = (t, 1 - t), \quad t \in [0, 1]$ So, $x(t) = t$ and $y(t) = 1 - t$.

Step 2: Compute $ds$

The differential arc length element $ds$ is given by: $ds = \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt$ First, compute $\frac{dx}{dt}$ and $\frac{dy}{dt}$: $\frac{dx}{dt} = 1, \quad \frac{dy}{dt} = -1$ Now compute $ds$: $ds = \sqrt{(1)^2 + (-1)^2} \, dt = \sqrt{2} \, dt$

Step 3: Set Up the Integral

Now, substitute $x = t$ and $ds = \sqrt{2} \, dt$ into the integral: $\int_{\Gamma} x \, ds = \int_0^1 t \cdot \sqrt{2} \, dt$

Step 4: Evaluate the Integral

$\int_0^1 t \cdot \sqrt{2} \, dt = \sqrt{2} \int_0^1 t \, dt = \sqrt{2} \left[ \frac{t^2}{2} \right]_0^1 = \sqrt{2} \cdot \frac{1}{2} = \frac{\sqrt{2}}{2}$

Final Answer:

The value of the line integral is $\frac{\sqrt{2}}{2}$, which corresponds to option B.

Do you want more details, or do you have any other questions?

Here are 5 related questions:

1. How would the integral change if the function was $y \cdot ds$ instead of $x \cdot ds$?
2. What happens if we change the curve $\Gamma$ to a circular arc instead of a straight line?
3. How do you compute the line integral of a vector field over a curve?
4. How does this process generalize to higher dimensions (e.g., surface integrals)?
5. What role does the parametrization of the curve play in evaluating line integrals?

Tip: Always double-check your parametrization and the limits of integration, as they are crucial for computing line integrals correctly.