b) Evaluate \( \int_{C} y d x+z d y+x d z \) where \( C \) is the line segment from \( (2,0,0) \) to \( (3,4,5) \). \( [13 \) marks \( ] \)

To evaluate the integral \( \int_{C} y \, dx + z \, dy + x \, dz \) along the line segment \( C \) from \( (2,0,0) \) to \( (3,4,5) \), we can first set up a parameterization for the line segment. Let \( \mathbf{r}(t) = (x(t), y(t), z(t)) \) be the parameterization of the path where \( t \) ranges from 0 to 1. We can define: \[ x(t) = 2 + t(3-2) = 2 + t \quad (from \, 2 \, to \, 3) \] \[ y(t) = 0 + t(4-0) = 4t \quad (from \, 0 \, to \, 4) \] \[ z(t) = 0 + t(5-0) = 5t \quad (from \, 0 \, to \, 5) \] Thus, the parameterization becomes: \[ \mathbf{r}(t) = (2+t, 4t, 5t) \quad \text{for } t \in [0, 1] \] Next, we calculate \( dx, dy, \text{ and } dz \): \[ dx = dt, \quad dy = 4 dt, \quad dz = 5 dt \] Now substitute \( x(t) \), \( y(t) \), \( z(t) \), \( dx \), \( dy \), and \( dz \) into the integral: \[ \int_{C} (y \, dx + z \, dy + x \, dz) = \int_{0}^{1} \left( 4t \cdot dt + 5t \cdot 4dt + (2 + t) \cdot 5dt \right) \] This simplifies to: \[ \int_{0}^{1} \left( 4t + 20t + 10 + 5t \right) dt = \int_{0}^{1} (29t + 10) dt \] Now we can integrate: \[ \int_{0}^{1} (29t + 10) dt = \left[ \frac{29}{2} t^2 + 10t \right]_{0}^{1} = \left( \frac{29}{2} \cdot 1^2 + 10 \cdot 1 \right) - \left( 0 \right) \] \[ = \frac{29}{2} + 10 = \frac{29}{2} + \frac{20}{2} = \frac{49}{2} \] Thus, the final value of the integral is: \[ \int_{C} y \, dx + z \, dy + x \, dz = \frac{49}{2} \]