1. **State the problem:** We are asked to estimate the integral $$\int_{0.5}^{2.5} \sqrt{\frac{x}{1+x}} \, dx$$ using the trapezium rule with given values of $y = \sqrt{\frac{x}{1+x}}$ at $x = 0.5, 1, 1.5, 2, 2.5$. 2. **Formula for trapezium rule:** The trapezium rule for $n$ intervals with equally spaced points is: $$\int_a^b f(x) \, dx \approx \frac{h}{2} \left(y_0 + 2y_1 + 2y_2 + \cdots + 2y_{n-1} + y_n\right)$$ where $h = \frac{b-a}{n}$ is the width of each subinterval. 3. **Apply the trapezium rule:** Here, $a=0.5$, $b=2.5$, and there are 4 intervals between 5 points, so: $$h = \frac{2.5 - 0.5}{4} = \frac{2}{4} = 0.5$$ The $y$ values are: $$y_0 = 0.5774, y_1 = 0.7071, y_2 = 0.7746, y_3 = 0.8165, y_4 = 0.8452$$ Using the formula: $$\int_{0.5}^{2.5} \sqrt{\frac{x}{1+x}} \, dx \approx \frac{0.5}{2} \left(0.5774 + 2(0.7071 + 0.7746 + 0.8165) + 0.8452\right)$$ Calculate the sum inside the parentheses: $$0.5774 + 2(0.7071 + 0.7746 + 0.8165) + 0.8452 = 0.5774 + 2(2.2982) + 0.8452 = 0.5774 + 4.5964 + 0.8452 = 6.019$$ Therefore: $$\int_{0.5}^{2.5} \sqrt{\frac{x}{1+x}} \, dx \approx \frac{0.5}{2} \times 6.019 = 0.25 \times 6.019 = 1.505$$ Rounded to 3 significant figures: $$\boxed{1.51}$$ 4. **Part (b):** Estimate $$\int_{0.5}^{2.5} \sqrt{\frac{9x}{1+x}} \, dx$$ using the answer from part (a). Note that: $$\sqrt{\frac{9x}{1+x}} = 3 \sqrt{\frac{x}{1+x}} = 3y$$ So the integral is: $$\int_{0.5}^{2.5} 3y \, dx = 3 \int_{0.5}^{2.5} y \, dx$$ Using the estimate from part (a): $$3 \times 1.505 = 4.515$$ Rounded to 3 significant figures: $$\boxed{4.52}$$ 5. **Part (c):** Given the exact value is 4.535 (to 4 significant figures), our estimate 4.52 is very close, differing by 0.015, which indicates the trapezium rule estimate is quite accurate for this integral.