Figure 3 shows a sketch of the curve with equation $y = (x - 1) \ln{x}$, $x > 0$ - Edexcel - A-Level Maths Pure - Question 1 - 2006 - Paper 6

Using the trapezium rule for points $y$ at $x = 1, 2$, and $3$:

$I \approx \frac{1}{2} \times (1 - 1) + (2 - 1) \times 0 + (3 - 2) \times 2 \ln 3$

Calculating:

1. Calculate the segment areas:
   • Between $1$ and $2$: area is $\frac{1}{2} \times (0 + \ln 2)$
   • Between $2$ and $3$: area is $\frac{1}{2} \times (\ln 2 + 2 \ln 3)$

Combine and simplify to find the approximate value.

Using the trapezium rule for points $y$ at $x = 1, 1.5, 2, 2.5$, and $3$:

1. Calculate segment areas:
   • Between $1$ and $1.5$: $\text{Area} = \frac{1}{2} \times (0 + 0.20273) \times 0.5$
   • Between $1.5$ and $2$: $\text{Area} = \frac{1}{2} \times (0.20273 + \ln 2) \times 0.5$
   • Between $2$ and $2.5$: $\text{Area} = \frac{1}{2} \times (\ln 2 + 1.25276) \times 0.5$
   • Between $2.5$ and $3$: $\text{Area} = \frac{1}{2} \times (1.25276 + 2 \ln 3) \times 0.5$

Combine areas for total approximate value to 4 significant figures.

An increase in the number of values allows for smaller trapezoidal sections, providing a closer approximation to the actual curve shape. As noted in Figure 3, the curve's concavity results in segments above and below the curve over larger sections. More subdivisions mean that the greater number of trapezoids capture variations in the curve closer to its actual area, thus enhancing accuracy.

To solve:

Using integration by parts: Let $u = \ln{x}$ and $dv = (x - 1)dx$.

Thus, $I = \left[ x \ln{x} - \int x \cdot (1) dx \right]_{1}^{3}$

Solving:

1. Evaluate limits at $3$ and $1$,
2. Combine results:
   • It yields the exact value calculated as: $\frac{1}{2} \ln 3$ This matches the integral's definition.