Figure 2 shows a sketch of part of the curve C with equation y = x³ - 10x² + kx, where k is a constant - Edexcel - A-Level Maths Pure - Question 9 - 2010 - Paper 3

To find the area of region R, we will integrate the curve from x = 0 to x = 2 (the x-coordinate of point P).

First, substitute k = 28 into the curve equation:

y = x³ - 10x² + 28x.

Next, we compute the definite integral:

$\int_0^2 (x^3 - 10x^2 + 28x) \, dx.$

This can be evaluated as follows:

1. Find the antiderivative:

   $\int (x^3 - 10x^2 + 28x) \, dx = \frac{x^4}{4} - \frac{10x^3}{3} + 14x^2 + C.$

2. Evaluate from 0 to 2:

   \left[ \frac{(2)^4}{4} - \frac{10(2)^3}{3} + 14(2)^2 \right] - \left[ \frac(0)^4}{4} - \frac{10(0)^3}{3} + 14(0)^2 \right]

Calculating the values:

$= \left[ 4 - \frac{80}{3} + 56 \right] - 0 = 4 + 56 - \frac{80}{3} = 60 - \frac{80}{3} = \frac{180 - 80}{3} = \frac{100}{3}.$

Therefore, the exact area of region R is:

$\frac{100}{3}$.