Figure 3 shows a sketch of the curve with equation $y = (x - 1) ext{ln}x$, $x > 0$ - Edexcel - A-Level Maths Pure - Question 7 - 2006 - Paper 6

Using the trapezium rule, we first calculate the area under the curve as:

$I \approx \frac{1}{2} \times (x_2 - x_1)(y_1 + y_2) + \frac{1}{2} \times (x_3 - x_2)(y_2 + y_3)$

Substituting the values:

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

Calculating this yields:

1.792 (to 4 significant figures).

Here we use the trapezium rule again but with more subdivisions:

$I \approx \frac{1}{2} \times 0.5 \times (0 + 0.5 \text{ln}(1.5)) + \frac{1}{2} \times 0.5 \times (0.5 \text{ln}(1.5) + \text{ln}(2)) + \frac{1}{2} \times 0.5 \times (\text{ln}(2) + 1.5 \text{ln}(2.5)) + \frac{1}{2} \times 0.5 \times (1.5 \text{ln}(2.5) + 2 \text{ln}(3))$

Calculating this sums to approximately 1.684 (to 4 significant figures).

As seen in Figure 3, the curve is not linear. By increasing the number of intervals (values), we can better capture the shape of the curve. Each segment of the trapezium rule approximates a straight line; more divisions reduce the error by providing a closer fit to the curve, leading to a more accurate overall estimate.

To find the integral, we use integration by parts. Let:

$u = \text{ln}x \quad \Rightarrow \quad du = \frac{1}{x} dx, \quad v = x - 1 \quad \Rightarrow \quad dv = dx$

Then,

$I = (x - 1) \text{ln}x - \int (1) \text{ln}x \, dx$

Carrying out this integration results in:

$I = \frac{x^2}{2} \text{ln}x - \frac{x^2}{4} + C$

Evaluating from $1$ to $3$ gives:

$I = \frac{1}{2} \ln^3 3$.