Figure 1 shows a sketch of the curve with equation $y = x ext{ln} x, ext{ for } x > 1$ - Edexcel - A-Level Maths Pure - Question 4 - 2010 - Paper 7

Using the trapezium rule, the area $A$ can be estimated as follows:

1. The values from the table are:

   • For $x = 1$: $y = 0$
   • For $x = 1.5$: $y = 0.608$
   • For $x = 2$: $y = 1.386$
   • For $x = 2.5$: $y = 2.291$
   • For $x = 3$: $y = 3.296$
   • For $x = 3.5$: $y = 4.385$
   • For $x = 4$: $y = 5.545$

2. The area estimate is given by:

   $A = \frac{1}{2} h (f(a) + f(b))$

   where $h$ is the width of the intervals ($0.5$ in this case) and $f(a)$ and $f(b)$ are the function values at the endpoints.

   Calculating:

   $A = 0.5 \times \left( 0 + 2 \times (0.608 + 1.386 + 2.291 + 3.296 + 4.385 + 5.545) \right) = 0.25 \times 29.477 \approx 7.37$

   Thus, the estimated area of region $R$ is approximately 7.37.

For this integral, we use the integration by parts formula:

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

Let us choose:

• $u = \text{ln} x$, hence $du = \frac{1}{x} \, dx$
• $dv = x \, dx$, thus $v = \frac{x^2}{2}$

Now applying the formula:

$\int x \text{ln} x \, dx = \frac{x^2}{2} \text{ln} x - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx$

This simplifies to:

$= \frac{x^2}{2} \text{ln} x - \int \frac{x}{2} \, dx$

Evaluating the last integral gives:

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

Using the result from part (c)(i):

The area $A$ from 1 to 4 for the specified region is given by:

$A = \left[ \frac{x^2}{2} \text{ln} x - \frac{x^2}{4} \right]_{1}^{4}$

Evaluating:

At $x = 4$: $= \left( \frac{16}{2} \text{ln}(4) - \frac{16}{4} \right) = 8 \text{ln}(4) - 4$

At $x = 1$: $= 0$

Therefore, the area $A = 8 \ln(4) - 4 ext{ (since log terms cancel out)}$$

Finally, writing the area in the required form:

$= \frac{1}{4} (16 \ln{2} - 15)$

Hence, the final answer for the area of $R$ in the specified form is $\frac{1}{4}(16 \ln 2 - 15)$.