Figure 4 shows a sketch of the curve C with equation y = 5x^2 - 9x + 11, x > 0 The point P with coordinates (4, 15) lies on C - Edexcel - A-Level Maths Pure - Question 1 - 2017 - Paper 2

Substituting $x = 4$ into the derivative gives us:

$\frac{dy}{dx}\big|_{x=4} = 10(4) - 9 = 40 - 9 = 31$

Thus, the gradient at point P (4, 15) is 31.

Using the point-slope form of the equation of a line, we find:

$y - y_1 = m(x - x_1)$

Substituting $(x_1, y_1) = (4, 15)$ and $m = 31$ gives:

$y - 15 = 31(x - 4)$ $y = 31x - 124 + 15$ $$y = 31x - 109$$$$

The area of region R is bounded between the curve C and the line l from $x = 0$ to $x = 4$. Thus:

$\text{Area} = \int_0^4 \left( (5x^2 - 9x + 11) - (31x - 109) \right) dx$

The integrand simplifies to:

$5x^2 - 9x + 11 - 31x + 109 = 5x^2 - 40x + 120$

Now, compute the integral:

$\int_0^4 (5x^2 - 40x + 120) dx$

Calculating the integral yields:

$\left[ \frac{5}{3}x^3 - 20x^2 + 120x \right]_0^4$

Evaluating the bounds gives:

$= \left( \frac{5}{3} (4^3) - 20 (4^2) + 120(4) \right) - 0$ $= \left( \frac{5}{3} \cdot 64 - 320 + 480 \right)$ $= \frac{320}{3} - 320 + 480$ $= \frac{320 - 960 + 1440}{3} = \frac{800}{3}$

After simplification, we find:

$\text{Area} = 24$