1. **State the problem:** We want to approximate the area of the region $R$ bounded by the curve $y=\log_{10}x$, the lines $x=2$ and $x=14$, and the $x$-axis using the trapezium rule with four strips of equal width. 2. **Formula for trapezium rule:** The trapezium rule for $n$ strips of equal width $h$ approximates the integral $\int_a^b f(x)\,dx$ as $$\frac{h}{2}\left[f(x_0) + 2f(x_1) + 2f(x_2) + \cdots + 2f(x_{n-1}) + f(x_n)\right]$$ where $x_0=a$, $x_n=b$, and $x_i = a + ih$. 3. **Calculate width $h$ and $x$ values:** $$h = \frac{14 - 2}{4} = 3$$ The $x$ values are $2, 5, 8, 11, 14$. 4. **Evaluate $f(x) = \log_{10} x$ at these points:** $$f(2) = \log_{10} 2 \approx 0.3010$$ $$f(5) = \log_{10} 5 \approx 0.6990$$ $$f(8) = \log_{10} 8 \approx 0.9031$$ $$f(11) = \log_{10} 11 \approx 1.0414$$ $$f(14) = \log_{10} 14 \approx 1.1461$$ 5. **Apply trapezium rule:** $$\text{Area} \approx \frac{3}{2} \left[0.3010 + 2(0.6990 + 0.9031 + 1.0414) + 1.1461\right]$$ Calculate the sum inside brackets: $$0.3010 + 2(0.6990 + 0.9031 + 1.0414) + 1.1461 = 0.3010 + 2(2.6435) + 1.1461 = 0.3010 + 5.2870 + 1.1461 = 6.7341$$ So, $$\text{Area} \approx \frac{3}{2} \times 6.7341 = 1.5 \times 6.7341 = 10.10115 \approx 10.10$$ 6. **Answer to (a):** The area of $R$ is approximately $10.10$ using the trapezium rule with four strips. 7. **Answer to (b):** To obtain a more accurate estimate, increase the number of strips (i.e., use more trapeziums with smaller width $h$). This reduces the error by better approximating the curve. 8. **Answer to (c)(i):** Estimate $$\int_2^{14} \log_{10} \sqrt{x} \, dx$$ Note that $$\log_{10} \sqrt{x} = \log_{10} x^{1/2} = \frac{1}{2} \log_{10} x = \frac{1}{2} f(x)$$ So, $$\int_2^{14} \log_{10} \sqrt{x} \, dx = \int_2^{14} \frac{1}{2} f(x) \, dx = \frac{1}{2} \int_2^{14} f(x) \, dx$$ Using the trapezium rule result from (a): $$\approx \frac{1}{2} \times 10.10 = 5.05$$ 9. **Answer to (c)(ii):** Estimate $$\int_2^{14} \log_{10} (100 x^3) \, dx$$ Use log properties: $$\log_{10} (100 x^3) = \log_{10} 100 + \log_{10} x^3 = 2 + 3 \log_{10} x = 2 + 3 f(x)$$ So, $$\int_2^{14} \log_{10} (100 x^3) \, dx = \int_2^{14} (2 + 3 f(x)) \, dx = \int_2^{14} 2 \, dx + 3 \int_2^{14} f(x) \, dx$$ Calculate: $$\int_2^{14} 2 \, dx = 2 \times (14 - 2) = 24$$ Using trapezium rule result: $$3 \times 10.10 = 30.30$$ Sum: $$24 + 30.30 = 54.30$$ **Final answers:** (a) Area $\approx 10.10$ (b) Use more strips for better accuracy. (c)(i) $\approx 5.05$ (c)(ii) $\approx 54.30$