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 formula:

$\text{Area} \approx \frac{h}{2} (y_0 + 2y_1 + 2y_2 + y_3)$

where $h$ is the width between each $x$ value, which equals $1$ in this case (since $[1, 2, 3, 4]$ are our values).

Filling in the table values we have:

• $y_0 = 0.5$
• $y_1 = 0.8284$
• $y_2 = 1.0977$
• $y_3 = 1.3333$

Thus, the area is:

$\text{Area} \approx \frac{1}{2}(0.5 + 2 \cdot 0.8284 + 2 \cdot 1.0977 + 1.3333)$

Calculating this gives:

$\text{Area} \approx \frac{1}{2}(0.5 + 1.6568 + 2.1954 + 1.3333) \approx \frac{1}{2}(5.6855) \approx 2.84275$

Rounded to three decimal places, we find:

$\text{Area} \approx 2.843$.

We make the substitution $u = 1 + \sqrt{x}$, which implies:

$\sqrt{x} = u - 1$ $x = (u - 1)^2$

Thus, the differential becomes:

$dx = 2(u - 1) du.$

The limits change as follows:

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

Now we need to set up the integral:

$\text{Area}(R) = \int_{2}^{3} (u - 1) imes 2(u - 1) du = 2 \int_{2}^{3} (u - 1)^2 du$

Expanding the integrand gives:

$= 2 \int_{2}^{3} (u^2 - 2u + 1) du$

Now, integrating term by term:

$= 2 \left[ \frac{u^3}{3} - u^2 + u \right]_{2}^{3}$

Evaluating this from $2$ to $3$ yields:

$= 2 \left( \left[ \frac{27}{3} - 9 + 3 \right] - \left[ \frac{8}{3} - 4 + 2 \right] \right)$

This simplifies to:

$= 2 \left( 9 - 9 + 3 - \left( \frac{8}{3} - 2 \right) \right) \approx 2 \left( 3 - \frac{2}{3} \right) = 2 \cdot \frac{7}{3} = \frac{14}{3}$

Thus, the exact area of $R$ is:

$\text{Area}(R) = \frac{14}{3}$.