Figure 1 shows a sketch of the curve with equation $y = x \, ext{ln} \, x$, $x > 1$ - Edexcel - A-Level Maths Pure - Question 3 - 2010 - Paper 7

To use the trapezium rule, we apply the formula:

$A \approx \frac{h}{2} \left( y_0 + 2 \sum_{i=1}^{n-1} y_i + y_n \right),$

where:

• $h$ is the distance between the $x$ values, and
• $y_i$ are the corresponding values of $y$.

Here, we calculate:
$A \approx \frac{0.25}{2} \left( 0 + 2(0.608 + 1.386 + 2.291 + 3.296 + 4.385 + 5.545)\right) \approx (a).$
Giving the answer to 2 decimal places, the area of $R$ is approximately $\approx 7.37$.

Here, we can use the integration by parts formula:

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

Let:

• $u = \text{ln} \, x$
• $dv = x \, dx$

Thus, we find:

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

Now applying integration by parts:

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

This simplifies to:

$= \frac{x^2}{2} \text{ln} \, x - \frac{1}{2} \int x \, dx = \frac{x^2}{2} \text{ln} \, x - \frac{1}{4} x^2 + C.$

Using the integral from 1 to 4:

$\int_1^4 x \, \text{ln} \, x \, dx = \left[ \frac{x^2}{2} \text{ln} \, x - \frac{1}{4} x^2 \right]_1^4.$

Calculating this:

1. At $x=4$:

   $\frac{4^2}{2} \text{ln} \, 4 - \frac{1}{4} (4^2) = 8 \text{ln} \, 4 - 4.$

2. At $x=1$:

   $\frac{1^2}{2} \text{ln} \, 1 - \frac{1}{4} (1^2) = 0 - \frac{1}{4} = -\frac{1}{4}.$

So, the area of $R$ is: $= \left( 8 \text{ln} \, 4 - 4 \right) - ( -\frac{1}{4}) = 8 \text{ln} \, 4 - \frac{15}{4}.$

This can be summarized as: $\text{Area} = \frac{1}{4}(a \text{ln} \, 2 + b),$ where $a = 64$ and $b = -15.$