The curve C has equation $y = x^2 - 5x + 4$ - Edexcel - A-Level Maths Pure - Question 9 - 2010 - Paper 4

To show that the point N (5, 4) lies on C, substitute x = 5 into the curve equation:

$y = (5)^2 - 5(5) + 4$

Calculating this gives:

$y = 25 - 25 + 4 = 4$

Since the coordinates match, N (5, 4) does indeed lie on the curve C.

To find the integral, we compute:

$\int (x^2 - 5x + 4) dx = \frac{x^3}{3} - \frac{5x^2}{2} + 4x + C$

where C is the constant of integration.

The area of region R can be calculated by finding the definite integral from x = 1 to x = 4:

$\text{Area} = \int_{1}^{4} (x^2 - 5x + 4) dx$

Using the result from part (c):

$= \left[ \frac{x^3}{3} - \frac{5x^2}{2} + 4x \right]_{1}^{4}$

Substituting the limits gives:

$= \left( \frac{4^3}{3} - \frac{5(4)^2}{2} + 4(4) \right) - \left( \frac{1^3}{3} - \frac{5(1)^2}{2} + 4(1) \right)$

Calculating both parts:

$= \left( \frac{64}{3} - 40 + 16 \right) - \left( \frac{1}{3} - \frac{5}{2} + 4 \right)$

This leads to:

$= \left( \frac{64 - 120 + 48}{3} \right) - \left( \frac{1 - 7.5 + 12}{3} \right) = \left( \frac{-8}{3} \right) - \left( \frac{5.5}{3} \right)$

Finally, this simplifies to:

$= \frac{-8 - 5.5}{3} = \frac{-13.5}{3} = -4.5$

Taking the absolute value of the area:

$= 4.5$