Figure 1 shows part of the curve C with equation $y = (1+x)(4-x)$ - Edexcel - A-Level Maths Pure - Question 4 - 2009 - Paper 2

The area A of region R can be found using the definite integral:

$A = ext{integral from } -1 ext{ to } 4 (4 + 3x - x^2) \, dx$

We first evaluate the integral:

$A = \int_{-1}^{4} (4 + 3x - x^2) \, dx$

Calculating each term separately:

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

Now we substitute the limits:

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

Calculating:

$A = \left[ 16 + \frac{3}{2}(16) - \frac{64}{3} \right] - \left[ -4 + \frac{3}{2} - \frac{1}{3} \right]$

Which simplifies to:

$A = \left[ 16 + 24 - \frac{64}{3} \right] - \left[ -4 + 1.5 + \frac{1}{3} \right]$

Now, simplifying further:

For the upper limit:

$A = \left[ 40 - \frac{64}{3} \right] = \frac{120 - 64}{3} = \frac{56}{3}$

For the lower limit:

$\left[ -2.5 + \frac{1}{3} \right] = -2.5 + 0.333 = -2.167 \text{ (approx)}$

Bringing both together:

$A = \frac{56}{3} + 2.167 = \frac{56 + 6.5}{3} = \frac{62.5}{3} = \frac{125}{6}$

Thus, the exact area of region R is:

$A = \frac{125}{6}$