Figure 4 shows a sketch of part of the curve C with equation $$y = \frac{x^2 \ln x}{3} - 2x + 5, \; x > 0$$ The finite region S, shown shaded in Figure 4, is bounded by the curve C, the line with equation $x = 1$, the x-axis and the line with equation $x = 3$ - Edexcel - A-Level Maths Pure - Question 1 - 2017 - Paper 1

To achieve a more accurate estimate using the trapezium rule, one can:

• Increase the number of strips: This reduces the width of each strip, making the approximation closer to the actual area under the curve.
• Decrease the width of the strips: By taking smaller intervals, we capture more fluctuations in the function's value.
• Use more trapezia: This essentially means applying the trapezium method repeatedly over smaller ranges for better accuracy.

To find the exact area of S, we calculate:

$\text{Area} = \int_1^3 \left( \frac{x^2 \ln x}{3} - 2x + 5 \right) dx$

We can split this integral: $= \int_1^3 \frac{x^2 \ln x}{3} dx - \int_1^3 2x dx + \int_1^3 5 dx$

Now, we integrate each part:

1. Integration by parts for $\int \frac{x^2 \ln x}{3} dx$: Let $u = \ln x$ and $dv = \frac{x^2}{3} dx$. Then, $du = \frac{1}{x} dx, \quad v = \frac{x^3}{9}$ This gives: $= \left[ \frac{x^3}{9} \ln x \right]_1^3 - \int_1^3 \frac{x^3}{9} \cdot \frac{1}{x} dx$

2. The second integral: $\int_1^3 2x dx = [x^2]_1^3 = 9 - 1 = 8$

3. The third integral: $\int_1^3 5 dx = [5x]_1^3 = 15 - 5 = 10$

Putting all the calculations together will yield the area in the desired form, plus some simplification, resulting in: $\text{Area} = \frac{28}{27} \; \text{where} \; a=28, b=27, c=27$.