Integrate along the Curve

Let $C_1$ be the line segment from $(1,0,1)$ to $(2,3,1)$ . Since

$\boldsymbol{r}(t)=(1-t)<1,0,1>+t<2,3,1>=<1+t, 3t, 1>,$

$C_1$ can be redefined as $C_1$ : $x=1+t$ , $y=3t$ , $z=1$ , and $0 \leq t \leq 1$ .

Similarly, let $C_2$ be the line segment from $(2,3,1)$ to $(2,5,2)$ . Since

$\boldsymbol{r}(t)=(1-t)<2,3,1>+t<2,5,2>=<2, 3+2t, 1+t>,$

$C_2$ can be redefined as $C_2$ : $x=2$ , $y=3+2t$ , $z=1+t$ , and $0 \leq t \leq 1$ .

Thus, the given integral value can be calculated as follows:

$\int_{C_1} \left[(x+yz)\ dx + 2x\ dy + xyz\ dz\right]$

$\begin{aligned} &= \int_0^1 \left[\left( 1+t+(3t) (1) \right)\ dt + 2(1+t) 3\ dt + (1+t) (3t) (1) \cdot 0\ dt\right] \\ &= \int_0^1 (10t+7)\ dt = \left[ 5t^2+7t \right]_0^1 = 12, \\ \end{aligned}$

$\int_{C_2} \left[(x+yz)\ dx + 2x\ dy + xyz\ dz\right]$ $\begin{aligned} &= \int_0^1 \left[(2+(3+2t)(1+t)) \cdot 0\ dt + 2(2) \cdot 2\ dt + (2)(3+2t)(1+t)\ dt\right] \\ &= \int_0^1 (4t^2+10t+14)\ dt = \left[ \frac{4}{3}t^3 + 5t^2 + 14t \right]_0^1 = \frac{61}{3}. \\ \end{aligned}$

Therefore, summing the two integrals, we have

$\int_C ((x+yz)\ dx + 2x\ dy + xyz\ dz) = 12+\frac{61}{3} = \frac{97}{3}.$

This implies $a=97$ and $b=3$ , hence $a+b= 100.$