2. In $\triangle A B C$ with an area of 1, $P$ is the midpoint of side $B C$, and point $Q$ is on side $A C$ such that $A Q=2 Q C$. Connecting $A P$ and $B Q$ intersects at point $R$. Then the area of $\triangle A B R$ is $\qquad$
2. $\frac{2}{5}$.

As shown in Figure 4, draw $P M / / A C$, intersecting $B Q$ at point $M$. Given $P M=\frac{1}{2} C Q=\frac{1}{4} A Q$, we have
$$
\begin{array}{l}
\frac{P R}{R A}=\frac{P M}{A Q}=\frac{1}{4} \\
\frac{A R}{A P}=\frac{4}{5}
\end{array}
$$

Therefore, $S_{\triangle A B R}=\frac{4}{5} S_{\triangle A B P}$
$$
=\frac{4}{5} \times \frac{1}{2} S_{\triangle A B C}=\frac{2}{5} .
$$