2. Given the ellipse $\frac{x^{2}}{2}+y^{2}=1$ with two foci $F_{1}$ and $F_{2}$, a chord $A B$ is drawn through the right focus $F_{2}$ with an inclination angle of $45^{\circ}$. Then the area of $\triangle A B F_{1}$ is $\qquad$.
2. $\frac{4}{3}$ Detailed explanation: $F_{2}(1,0), B(0,-1), F_{1}(-1,0)$, so $\angle A B F_{1}=90^{\circ}$, and since $\left|F_{1} A\right|+\left|F_{2} A\right|=2 a=2 \sqrt{2}$, hence $\left|F_{1} A\right|=2 \sqrt{2}-\left|F_{2} A\right|$. Also, $\left|F_{1} A\right|^{2}=\left|F_{1} B\right|^{2}+|A B|^{2}$, so $\left(2 \sqrt{2}-\left|F_{2} A\right|\right)^{2}=\left|F_{1} B\right|^{2}+\left(\left|F_{2} B\right|+\right.$ $\left.\left|F_{2} A\right|\right)^{2}=\sqrt{2}^{2}+\left(\sqrt{2}+\left|F_{2} A\right|\right)^{2}$, hence $\left|F_{2} A\right|=\frac{\sqrt{2}}{3},|A B|=\frac{4 \sqrt{2}}{3}$. Then $S_{\triangle A B F_{1}}=\frac{1}{2}|A B| \cdot\left|B F_{1}\right|=\frac{4}{3}$.