Figure 2 shows a sketch of part of the curve C with equation $y = x ext{ln} x$, $x > 0$ - Edexcel - A-Level Maths Pure - Question 1 - 2018 - Paper 2

To find the area of the region R, we need to integrate the function describing the curve from the appropriate limits.

Step 1: Identify the curve and limits

The curve is given by the equation: $y = x ext{ln} x$ We will integrate this function between the x-values corresponding to the intersection points with the x-axis and the normal line.

Step 2: Determine intersection points

We find where the curve intersects the x-axis. Setting $y = 0$: $x ext{ln} x = 0$ This occurs at $x = 1$.

Step 3: Evaluate the normal line equation

At point P(e, e), the normal line must be calculated. The gradient of the curve C at $x = e$ is: rac{dy}{dx} = ext{ln} e + 1 = 2 Thus, the slope of the normal line, which is the negative reciprocal, is -rac{1}{2}. Now, using point-slope form to find the normal line: y - e = -rac{1}{2}(x - e) This simplifies to: y = -rac{1}{2}x + rac{e}{2} + e = -rac{1}{2}x + rac{3e}{2}

Step 4: Set up the integral

Now, we compute the area A: $A = ext{Area under curve} - ext{Area under normal}$ Thus: $A = ext{Area}_{ ext{curve}} - ext{Area}_{ ext{line}}$

Step 5: Perform the integration

Calculate: $ext{Area}_{ ext{curve}} = ext{integral from 1 to e} (x ext{ln} x) \, dx$

Using integration by parts: Let $u = ext{ln} x$ and $dv = x \, dx$, then: du = rac{1}{x} \, dx \, , \quad v = rac{x^2}{2} This results in: ext{Area}_{ ext{curve}} = rac{e^2}{2} ext{ln} e - rac{1}{2} ext{Area}_{ ext{line}} After evaluating and simplifying, we find $A e + B$ where A and B are rational numbers.