1. **State the problem:** We need to find the area of the third trapezium under the curve $y=\frac{4}{\ln(x)}$ between the points $x=\frac{19}{5}$ and $x=\frac{23}{5}$. The trapezium is formed by the vertical lines at $x=\frac{19}{5}$ and $x=\frac{23}{5}$ and the curve values at these points. 2. **Identify the coordinates:** From the table, the $y$-values at these $x$-values are: - At $x=\frac{19}{5}=3.8$, $y=2.996$ - At $x=\frac{23}{5}=4.6$, $y=2.621$ 3. **Calculate the width of the trapezium:** $$\text{width} = x_2 - x_1 = 4.6 - 3.8 = 0.8$$ 4. **Use the trapezium area formula:** $$\text{Area} = \frac{(y_1 + y_2)}{2} \times \text{width}$$ Substitute the values: $$\text{Area} = \frac{(2.996 + 2.621)}{2} \times 0.8 = \frac{5.617}{2} \times 0.8 = 2.8085 \times 0.8$$ 5. **Calculate the area:** $$\text{Area} = 2.2468$$ 6. **Round to 3 decimal places:** $$\boxed{2.247}$$ Thus, the area of the third trapezium is approximately 2.247.