Figure 1 shows a sketch of part of the curve with equation y = \frac{(x + 2)^{\frac{3}{2}}}{4} x > -2 The finite region R, shown shaded in Figure 1, is bounded by the curve, the x-axis and the line with equation x = 10 The table below shows corresponding values of x and y for y = \frac{(x + 2)^{\frac{3}{2}}}{4} - Edexcel - A-Level Maths Pure - Question 3 - 2018 - Paper 4

To use the trapezium rule, we first note the values for x and y:

x	-2	2	6	10
y	0	2	16	6\sqrt{3}

Now, we can calculate the area using:

$A \approx \frac{1}{2} \times h \times (y_0 + 2y_1 + 2y_2 + y_3)$

Where:

• $h = \frac{10 - (-2)}{3} = \frac{12}{3} = 4$
• $y_0 = 0, y_1 = 2, y_2 = 16, y_3 = 6\sqrt{3}$

Thus, calculating the area:

$A \approx \frac{1}{2} \times 4 \times (0 + 2 \times 2 + 2 \times 16 + 6\sqrt{3}) \approx 8 + 12\sqrt{3}$

Substituting the approximate value for $\sqrt{3} \approx 1.732$, we find:

$A \approx 8 + 12 \times 1.732 \approx 8 + 20.784 = 28.784$

Therefore, the approximate area for region R is 28.784, rounded to 3 decimal places: 28.784.