Figure 1 shows part of a curve C with equation $y = 2x + \frac{8}{x^2} - 5$, $x > 0$ - Edexcel - A-Level Maths Pure - Question 1 - 2016 - Paper 2

To find the area of region R, we need to integrate the curve from x = 1 to x = 4 and subtract the area of the triangle formed by points P and Q.

1. Determine the points P and Q:

   • At $x = 1$:
     $y = 2(1) + \frac{8}{(1)^2} - 5 = 2 + 8 - 5 = 5.$
     Thus, $P(1, 5)$.
   • At $x = 4$:
     $y = 2(4) + \frac{8}{(4)^2} - 5 = 8 + \frac{8}{16} - 5 = 8 + 0.5 - 5 = 3.5.$
     Thus, $Q(4, 3.5)$.

2. Set up the integral:
   The area under curve C from $x = 1$ to $x = 4$:

   $A_c = \int_{1}^{4} \left( 2x + \frac{8}{x^2} - 5 \right)dx.$

3. Calculate the integral:

   • Integrate term by term:

   $A_c = \left[ x^2 + \frac{-8}{x} - 5x \right]_{1}^{4}.$

   • Evaluate at limits:

   At $x = 4$:
   $4^2 + \frac{-8}{4} - 5(4) = 16 - 2 - 20 = -6.$

   At $x = 1$:
   $1^2 + \frac{-8}{1} - 5(1) = 1 - 8 - 5 = -12.$

   So the area under the curve is:
   $A_c = -6 - (-12) = 6.$

4. Calculate the area of triangle PQR:
   The base PQ is given by:
   $\text{Base} = 4 - 1 = 3.$
   The height can be found from the y-coordinates of P and Q:
   $\text{Height} = 5 - 3.5 = 1.5.$
   Area of triangle:
   $A_t = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 3 \times 1.5 = 2.25.$

5. Calculate the area of region R:
   $\text{Area of R} = A_c - A_t = 6 - 2.25 = 3.75.$ The exact area of R is 3.75 square units.