Figure 1 shows part of the curve $y = \frac{3}{\sqrt{1+4x}}$ - Edexcel - A-Level Maths Pure - Question 4 - 2009 - Paper 3

To find the area of region $R$, we need to set up the definite integral with respect to $x$:

$\text{Area}(R) = \int_{0}^{2} \frac{3}{\sqrt{1+4x}} \, dx$

Now, we perform integration:

1. We can rewrite the integral as follows:

   $\int \frac{3}{\sqrt{1+4x}} \, dx = \frac{3}{4} \int (1+4x)^{-1/2} \, d(1+4x)$

2. The integral evaluates to:

   $= \frac{3}{4} \left[ (1+4x)^{1/2} \right]_{0}^{2}$

3. Substituting the limits into the equation gives us:

   $= \frac{3}{4} \left( (1+8)^{1/2} - (1)^{1/2} \right)$

4. Calculating this result:

   $= \frac{3}{4} \left(\sqrt{9} - \sqrt{1}\right) = \frac{3}{4} (3 - 1) = \frac{3}{4} \times 2 = \frac{3}{2}$

Thus, the area of region $R$ is $\frac{3}{2}$ square units.