4. Given $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$ are two points on the graph of the inverse proportion function $y=\frac{1}{x}$ in the first quadrant of the Cartesian coordinate system $x O y$, satisfying $y_{1}+y_{2}=\frac{7}{2}, x_{2}-x_{1}=\frac{5}{3}$. Then $S_{\triangle A O B}=$ ( ).
(A) $2 \frac{10}{11}$
(B) $2 \frac{11}{12}$
(C) $2 \frac{12}{13}$
(D) $2 \frac{13}{14}$
4. B.

As shown in Figure 2, add auxiliary lines, and let $O B$ intersect $A C$ at point $P$.
$$
\begin{array}{l}
\text { Given } x_{1} y_{1}=x_{2} y_{2}=1 \\
\Rightarrow S_{\triangle A O C}=S_{\triangle B O D} \\
\Rightarrow S_{\triangle A O B}=S_{\text {quadrilateral } A O D B}-S_{\triangle B O D} \\
\quad=S_{\text {quadrilateral } A O D B}-S_{\triangle A O C}=S_{\text {trapezoid } A C D B} \\
\quad=\frac{1}{2}\left(y_{1}+y_{2}\right)\left(x_{2}-x_{1}\right) \\
\quad=\frac{1}{2} \times \frac{7}{2} \times \frac{5}{3}=\frac{35}{12}=2 \frac{11}{12} .
\end{array}
$$