Figure 2 shows a sketch of part of the curve C with equation $y = x \ln x, \, x > 0$ The line I is the normal to C at the point $P(e, e)$ - Edexcel - A-Level Maths Pure - Question 3 - 2018 - Paper 2

To determine the area of region R, we begin by identifying the relevant boundaries. The area under the curve can be assessed using integration.

1. Determine the area bounded by curve C and the x-axis:

   • The equation of the curve is given as $y = x \ln x$.
   • Find the area from $x = 1$ to $x = e$:

   $ext{Area}_R = \int_{1}^{e} (x \ln x) \, dx$

   Using integration by parts, let:

   • $u = \ln x$
   • $dv = x \, dx$
   • Therefore, $du = \frac{1}{x} \, dx$ and $v = \frac{x^2}{2}$.

   By applying integration by parts:

   $\int (x \ln x) \, dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C$

   • Evaluating the definite integral gives:

   $[ \frac{e^2}{2} \ln e - \frac{e^2}{4} ] - [ \frac{1^2}{2} \ln 1 - \frac{1^2}{4} ] = [ \frac{e^2}{2} - \frac{e^2}{4} ] - [ 0 - \frac{1}{4} ]\n = \frac{e^2}{4} + \frac{1}{4}$

   Therefore, the area between the curve and the x-axis from $x = 1$ to $x = e$ is:

   $\text{Area}_R = \frac{e^2 + 1}{4}$

2. Include the area under the line I:

   • The line I is the normal to the curve at point $P(e,e)$,
   • The slope of the curve at this point is given by:

   $\frac{dy}{dx} = \ln e + 1 = 2\n$

   • Thus, the slope of the normal is $-\frac{1}{2}$.
   • The equation of the normal line is thus:

   $y - e = -\frac{1}{2}(x - e)\n y = -\frac{1}{2}x + \frac{e}{2} + e = -\frac{1}{2}x + \frac{3e}{2}\n$

3. Final Area Calculation:

   • We then compute the area between $x = 1$ and $x = e$ beneath line I,

   $\text{Area} = \int_{1}^{e} [-\frac{1}{2}x + \frac{3e}{2}] \, dx = [-\frac{1}{4}x^2 + \frac{3e}{2}x]_{1}^{e}\n$

   • This evaluates to:

   $[-\frac{1}{4}e^2 + \frac{3e^2}{2}] - [-\frac{1}{4} + \frac{3e}{2}] = [\frac{5e^2}{4} - \frac{3e}{2} + \frac{1}{4}]$

   • Bringing it all together yields:

   $\text{Total Area} = \frac{e^2 + 1}{4} + \frac{5e^2 - 6e + 1}{4} = \frac{6e^2 - 6e + 2}{4} = \frac{3(e^2 - e + \frac{1}{3})}{2}$

Thus, the area can be expressed as:$\frac{3}{4}(e^2 - e + \frac{1}{3})$ where $A = \frac{3}{4}$, and $B$ is the rational number corresponding to the non-$e$ term.