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

Using the completed table:

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

The area can be estimated using the trapezium rule:

$A \approx \frac{h}{2} \left( y_0 + 2y_1 + 2y_2 + 2y_3 + 2y_4 + 2y_5 + y_6 \right)$

Where:

• $h = 0.5$ (the width between each x value)
• $y_0 = 0$, $y_1 = 0.608$, $y_2 = 1.386$, $y_3 = 2.291$, $y_4 = 3.296$, $y_5 = 4.385$, $y_6 = 5.545$

Substituting in these values results in:

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

Let:

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

Using integration by parts: $\int x \, \text{ln} \, x \, dx = uv - \int v \, du$ Substituting for $u$ and $v$ gives: $= \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx$

This simplifies to: $= \frac{x^2}{2} \ln x - \int \frac{x}{2} \, dx$

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

The exact area $A$ of $R$ can be calculated using the definite integral from $x=1$ to $x=4$:

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

Calculating the bounds:

• At $x = 4$: $= \frac{16}{2} \ln 4 - \frac{16}{4} = 8 \ln 4 - 4$

• At $x = 1$: $= \frac{1}{2} \ln 1 - \frac{1}{4} = 0 - \frac{1}{4} = -\frac{1}{4}$

Thus, the area: $A = \left( 8 \ln 4 - 4 \right) - \left( -\frac{1}{4} \right)$ $= 8 \ln 4 - \frac{15}{4}$

Rearranging gives: $A = \frac{1}{4} \left( 32 \ln 4 - 15 \right)$

So we can express it in the required form as: $A = \frac{1}{4} (a \ln 2 + b)$ Where $a = 64$ and $b = -15$.