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 formula:

$I \approx \frac{h}{2} (y_0 + 2y_1 + y_2)$

where $h$ is the width between x values (which is 1 in this case), $y_0 = 0$ (for $x=1$), $y_1 = 2\text{ln}2 \approx 1.3863$, and $y_2 = 2\text{ln}3 \approx 2.1972$.

Calculating:

$I \approx\frac{1}{2} (0 + 2(1.3863) + 2.1972) = \frac{1}{2} (0 + 2.7726 + 2.1972) = \frac{1}{2} (4.9698) \approx 2.4849\approx 2.485$
Thus, approximating $I$ gives $I \approx 2.485$.

Using the trapezium rule again with updated points:

Here, $y_0 = 0.2027$ (for $x=1.5$), $y_1 = 2\text{ln}2 \approx 1.3863$, $y_2 = 1.3743$ (for $x = 2.5$), and $y_3 = 2\text{ln}3 \approx 2.1972$.

Using the same formula:

$I \approx \frac{h}{2} (y_0 + 2y_1 + 2y_2 + y_3)$

Thus:

$I \approx \frac{0.5}{2} (0.2027 + 2(1.3863) + 1.3743 + 2.1972) \approx 1.6891$
So, approximating $I$ gives $I \approx 1.6891$.

Increasing the number of points used in the trapezium rule allows for a better approximation of the area under the curve, leading to reduced errors in estimation. In Figure 3, as the number of segments increases, the trapezoids used to approximate the area align more closely with the curve. This behavior reduces the overestimation and underestimation effects typical of using fewer segments, resulting in a more accurate overall estimation of the integral.

To solve the integral:

Using integration by parts where we let:

$u = \text{ln} x, \quad dv = (x - 1) dx$
Then we find:

$du = \frac{1}{x}dx, \quad v = \frac{x^2}{2} - x$

Now substituting these into the integration by parts formula:

$\int u \, dv = uv - \int v \, du$

Therefore:

$\int (x - 1) \text{ln} x \, dx = \left( \frac{x^2}{2} - x \right) \text{ln} x - \int \left( \frac{x^2}{2} - x \right) \frac{1}{x}dx$

Continuing with the integrals and evaluating at limits from 1 to 3, we arrive at the exact value of $\frac{1}{2} \text{ln} 3$.