The curve C has equation $y = x^2 - 5x + 4$ - Edexcel - A-Level Maths Pure - Question 7 - 2010 - Paper 4

To show that the point N(5, 4) lies on the curve C, we substitute $x = 5$ into the equation of the curve: $y = 5^2 - 5(5) + 4 = 25 - 25 + 4 = 4$ Since the calculated y-coordinate is 4, we have confirmed that the point N(5, 4) is indeed on the curve.

To find the integral, we can break it down as follows: $\int (x^2 - 5x + 4) \, dx = \int x^2 \, dx - 5\int x \, dx + 4\int 1 \, dx$ Calculating each integral gives:

1. $\int x^2 \, dx = \frac{x^3}{3}$
2. $\int x \, dx = \frac{x^2}{2}$
3. $\int 1 \, dx = x$ Putting it all together: $\int (x^2 - 5x + 4) \, dx = \frac{x^3}{3} - \frac{5x^2}{2} + 4x + C$

To find the area of the region R, we use the definite integral between the limits of L and M, which are 1 and 4: $\text{Area} = \int_{1}^{4} (x^2 - 5x + 4) \, dx$ Calculating this integral: $= \left[ \frac{x^3}{3} - \frac{5x^2}{2} + 4x \right]_{1}^{4}$ Evaluating at the bounds gives: $\left( \frac{4^3}{3} - \frac{5(4)^2}{2} + 4(4) \right) - \left( \frac{1^3}{3} - \frac{5(1)^2}{2} + 4(1) \right)$ After simplifying results in: $= \left( \frac{64}{3} - 40 + 16 \right) - \left( \frac{1}{3} - \frac{5}{2} + 4 \right)$ Solving further provides the exact area of region R.