The line integral $$\int\limits_{{{\text{P}}_1}}^{{{\text{P}}_2}} {\left( {{\text{ydx}} + {\text{xdy}}} \right)} $$ for P1(x1, y1) to P2(x2, y2) along the semicircle P1, P2 shown in the figure is A. x2y2 – x1y1 B. $$\left( {{\text{y}}_2^2 – {\text{y}}_1^2} \right) + \left( {{\text{x}}_2^2 – {\text{x}}_1^2} \right)$$ C. (x2 – x1) (y2 – y1) D. (y2 – y1)2 + (x2 – x1)2

The correct answer is $\boxed{\left( {{\text{y}}_2^2 – {\text{y}}_1^2} \right) + \left( {{\text{x}}_2^2 – {\text{x}}_1^2} \right)}$.

The line integral $\int\limits_{{{\text{P}}_1}}^{{{\text{P}}_2}} {\left( {{\text{ydx}} + {\text{xdy}}} \right)} $ is the integral of the vector field $\left( {{\text{y}}, {\text{x}}} \right)$ along the curve $C$ from $P_1$ to $P_2$. In this case, the vector field is $\left( {{\text{y}}, {\text{x}}} \right)$ and the curve $C$ is the semicircle $P_1, P_2$ shown in the figure.

To evaluate the line integral, we can use the formula $$\int\limits_{{{\text{P}}_1}}^{{{\text{P}}_2}} {\left( {{\text{ydx}} + {\text{xdy}}} \right)} = \left( {{\text{y}}_2^2 – {\text{y}}_1^2} \right) + \left( {{\text{x}}_2^2 – {\text{x}}_1^2} \right)$$

Substituting the values of $P_1$ and $P_2$, we get $$\int\limits_{{{\text{P}}_1}}^{{{\text{P}}_2}} {\left( {{\text{ydx}} + {\text{xdy}}} \right)} = \left( {{\text{y}}_2^2 – {\text{y}}_1^2} \right) + \left( {{\text{x}}_2^2 – {\text{x}}_1^2} \right)$$

Therefore, the line integral $\int\limits_{{{\text{P}}_1}}^{{{\text{P}}_2}} {\left( {{\text{ydx}} + {\text{xdy}}} \right)} $ is $\boxed{\left( {{\text{y}}_2^2 – {\text{y}}_1^2} \right) + \left( {{\text{x}}_2^2 – {\text{x}}_1^2} \right)}$.

The other options are incorrect because they do not take into account the fact that the curve $C$ is a semicircle.