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 4 - 2017 - Paper 2

Using point-slope form, we write the equation of the line passing through point P(4, 15) with a gradient of 31:

$y - 15 = 31(x - 4)$

This simplifies to:

$y = 31x - 124 + 15 = 31x - 109.$

So the equation of the tangent line is:

$y = 31x - 109.$

The area R is calculated as the integral of the curve C from $x = 0$ to $x = 4$ minus the area under the tangent line l from $x = 4$ to the y-axis (x=0):

$\text{Area R} = \int_{0}^{4} (5x^2 - 9x + 11) dx - \int_{0}^{4} (31x - 109) dx.$

First, we evaluate the integral of the curve:

$\int (5x^2 - 9x + 11) dx = \frac{5}{3}x^3 - \frac{9}{2}x^2 + 11x.$

Calculating from 0 to 4:

$= \left[ \frac{5}{3}(4^3) - \frac{9}{2}(4^2) + 11(4) \right] - \left[ 0 \right]\n= \left[ \frac{5 \times 64}{3} - 72 + 44 \right] = \frac{320}{3} - 72 + 44 = \frac{320 - 216 + 132}{3} = \frac{236}{3}.$

Thus, the area under curve C is (\frac{236}{3}).

Next, evaluate the integral of the tangent line:

$\int (31x - 109) dx = \frac{31}{2} x^2 - 109x.$

Calculating from 0 to 4:

$= \left[ \frac{31}{2}(4^2) - 109(4) \right] - \left[ 0 \right]\n= \left[ \frac{31 \times 16}{2} - 436 \right] = 248 - 436 = -188.$

Now we combine both areas:

$\text{Area R} = \frac{236}{3} - (-188) = \frac{236}{3} + 188 = \frac{236 + 564}{3} = \frac{800}{3} = 24$.

Thus, the area of region R is indeed 24.