Figure 4 shows a sketch of the curve C with equation $$ y = 5x^2 - 9x + 11, \, x \geq 0 $$ The point P with coordinates (4, 15) lies on C - Edexcel - A-Level Maths Pure - Question 16 - 2017 - Paper 2

Using the point-slope form of the line, we can find the equation of the tangent line at point P (4, 15):

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

This simplifies to:

$y = 31x - 109$

The area R can be determined by the integral of the difference between the curve C and the tangent line l from the y-axis to where they intersect. We first find the intersection points:

Set the equations equal:

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

Rearranging gives us:

$5x^2 - 40x + 120 = 0$

This simplifies to:

$x^2 - 8x + 24 = 0$

Using the quadratic formula:

$x = \frac{8 \pm \sqrt{(-8)^2 - 4(1)(24)}}{2(1)}$

We find out the limits of integration are from x = 0 to x = 4.

Now we compute the area R:

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

This becomes:

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

Calculating the integral:

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

Evaluating this from 0 to 4 gives:

$= \left( \frac{5}{3}(64) - 20(16) + 480 \right) - 0$

Calculating:

$= \left( \frac{320}{3} - 320 + 480 \right) = \frac{320}{3} + 480 - 320 = \frac{320 + 480\times 3 - 960}{3} = \frac{320 + 1440 - 960}{3} = \frac{800}{3} = 24$