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 integral:

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

We now integrate:

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

Next, substituting the limits into the integral:

egin{align*} A &= \left[ 4(4) + \frac{3}{2}(4)^2 - \frac{1}{3}(4)^3 ight] - \left[ 4(-1) + \frac{3}{2}(-1)^2 - \frac{1}{3}(-1)^3 \right] \ &= [16 + 24 - \frac{64}{3}] - [-4 + \frac{3}{2} + \frac{1}{3}] \
&= \frac{48 - 64}{3} + 4 - \frac{3}{2} - \frac{1}{3} \ &= \frac{20}{3} + 4 - \frac{3}{2} - \frac{1}{3}. \end{align*}

Calculating the final area:

Combine the fraction values:

egin{align*} A &= \frac{20}{3} + \frac{12}{3} - \frac{45}{30} - \frac{10}{30} \
&= \frac{32}{3} - \frac{55}{30} \
&= \frac{20}{6} \
&= \frac{125}{6}. \end{align*} Thus, the exact area of region $R$ is rac{125}{6}.