To evaluate the line integral of the function $f(x, y, z) = xy + x + z$ along the curve $\vec{r}(t) = 2t \, \mathbf{i} + t \, \mathbf{j} + (2 - 2t) \, \mathbf{k}$, for $0 \leq t \leq 1$, we can follow these steps:

1. Identify $f(x, y, z)$ and parameterize the path.

   The vector function of the path is given by: $\vec{r}(t) = 2t \, \mathbf{i} + t \, \mathbf{j} + (2 - 2t) \, \mathbf{k}.$

   From this, we can see that: $x = 2t, \quad y = t, \quad z = 2 - 2t.$

2. Compute $ds$:

   To find $ds$, we calculate $\frac{d\vec{r}}{dt}$ and then its magnitude $\left| \frac{d\vec{r}}{dt} \right|$.

   $\frac{d\vec{r}}{dt} = \frac{d}{dt}(2t) \, \mathbf{i} + \frac{d}{dt}(t) \, \mathbf{j} + \frac{d}{dt}(2 - 2t) \, \mathbf{k} = 2 \, \mathbf{i} + 1 \, \mathbf{j} - 2 \, \mathbf{k}.$

   Now, we find the magnitude: $\left| \frac{d\vec{r}}{dt} \right| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3.$

   Therefore, $ds = 3 \, dt$.

3. Substitute $x$, $y$, $z$, and $ds$ into the integral:

   Substitute $x = 2t$, $y = t$, $z = 2 - 2t$, and $ds = 3 \, dt$ into the integrand $f(x, y, z) \, ds = (xy + x + z) \, ds$:

   $\int_C (xy + x + z) \, ds = \int_0^1 \left((2t)(t) + (2t) + (2 - 2t)\right) \cdot 3 \, dt.$

4. Simplify the integrand:

   Simplify inside the parentheses: $(2t)(t) + (2t) + (2 - 2t) = 2t^2 + 2t + 2 - 2t = 2t^2 + 2.$

   Substitute this back into the integral: $\int_0^1 \left(2t^2 + 2\right) \cdot 3 \, dt = 3 \int_0^1 (2t^2 + 2) \, dt = 3 \left( \int_0^1 2t^2 \, dt + \int_0^1 2 \, dt \right).$

5. Evaluate each integral separately:

   • For the first part: $\int_0^1 2t^2 \, dt = 2 \cdot \frac{t^3}{3} \Big|_0^1 = \frac{2}{3}.$

   • For the second part: $\int_0^1 2 \, dt = 2t \Big|_0^1 = 2.$

6. Combine the results:

   $3 \left( \frac{2}{3} + 2 \right) = 3 \cdot \frac{8}{3} = 8.$

Therefore, the value of the integral is: $\int_C (xy + x + z) \, ds = 8.$

Would you like further explanation on any of these steps, or perhaps related questions?

Related Questions:

1. How would this approach change if the limits of $t$ were different?
2. What if we had a different parameterization for $\vec{r}(t)$?
3. How would we approach this problem if the function $f(x, y, z)$ were different?
4. Can we solve this integral using a different coordinate system, like cylindrical or spherical?
5. What would the solution be if $ds$ were replaced with $dx$, $dy$, or $dz$?

Tip:

When evaluating line integrals, carefully verify each substitution step to avoid errors, especially when parameterizing $ds$.