Figure 1 shows a sketch of part of the curve with equation $y = \frac{x}{1 + \sqrt{x}}$ - Edexcel - A-Level Maths Pure - Question 16 - 2013 - Paper 1

Using the trapezium rule, we can estimate the area as:

$\text{Area} \approx \frac{1}{2} \times (y_1 + y_n) + \sum_{i=1}^{n-1} y_i$

Here, we have:

• $y_1 = 0.5$
• $y_2 \approx 0.8284$
• $y_3 \approx 1.0981$
• $y_4 = 1.3333$

Calculating: $\text{Area} \approx \frac{1}{2} \times (0.5 + 1.3333) + (0.8284 + 1.0981)$

This gives us: $= \frac{1}{2} \times 1.8333 + 1.9265 = 0.91665 + 1.9265 \approx 2.84315$

Hence the estimated area of region $R$ is approximately $2.843$.

To find the exact area of region $R$, we first find the limits for $u$:

• When $x = 1$, $u = 1 + 1 = 2$.
• When $x = 4$, $u = 1 + 2 = 3$.

Thus the new limits for integration in terms of $u$ are from $2$ to $3$. Rewrite $x$ in terms of $u$: $x = (u - 1)^2$

Next, we find the area: $A = \int_{2}^{3} \frac{(u - 1)^2}{1 + (u - 1)} \cdot 2(u - 1) \, du$

Integrating this, we can obtain the exact area. The computation will involve polynomial expansion and standard integration techniques. After solving, the exact area will reflect the precise calculation of this setup, approximated to rac{1}{6}(11) - 2 and simplified appropriately.