Figure 2 shows a sketch of part of the curve with equation y = 4x³ + 9x² - 30x - 8, -0.5 ≤ x ≤ 2.2 The curve has a turning point at the point A - Edexcel - A-Level Maths Pure - Question 3 - 2016 - Paper 2

First, we find the x-coordinates at which the curve intersects the x-axis. This is already given as B(2, 0) and C(-1/4, 0).

The area can be found through integration of the curve between these two points:

$Area = \int_{-1/4}^{2} (4x^3 + 9x^2 - 30x - 8) \, dx$

Calculating the integral:

1. Find the antiderivative:

   • The antiderivative of $4x^3$ is $x^4$.
   • The antiderivative of $9x^2$ is $3x^3$.
   • The antiderivative of $-30x$ is $-15x^2$.
   • The antiderivative of $-8$ is $-8x$.

   Thus,

   $\int (4x^3 + 9x^2 - 30x - 8) \, dx = x^4 + 3x^3 - 15x^2 - 8x + C$

2. Evaluate from -1/4 to 2: $\left[ x^4 + 3x^3 - 15x^2 - 8x \right]_{-1/4}^{2}$

Calculating for $x=2$ yields: $2^4 + 3(2^3) - 15(2^2) - 8(2) = 16 + 24 - 60 - 16 = -36$

Now for $x=-1/4$: $\left(-\frac{1}{4}\right)^4 + 3\left(-\frac{1}{4}\right)^3 - 15\left(-\frac{1}{4}\right)^2 - 8\left(-\frac{1}{4}\right)$

• Calculating each term:
  • $\left(-\frac{1}{4}\right)^4 = \frac{1}{256}$
  • $3\left(-\frac{1}{4}\right)^3 = -\frac{3}{64}$
  • $-15\left(-\frac{1}{4}\right)^2 = -15\cdot \frac{1}{16} = -\frac{15}{16}$
  • $-8\left(-\frac{1}{4}\right) = 2$
• Combine these terms to find the value at $x = -1/4$: $\frac{1}{256} - \frac{3}{64} - \frac{15}{16} + 2$ Finally finding a common denominator and simplifying gives us the value:

After evaluating both sides, the area is given by:

$Area = 32.52$

Thus, to two decimal places, the area of region R is:

32.52.