Figure 1 shows a sketch of part of the curve with equation $y = \sqrt{x^2 + 1}, \; x > 0$ - Edexcel - A-Level Maths Pure - Question 5 - 2014 - Paper 1

Using the trapezium rule, the area $A$ can be approximated by:

$$A \approx \frac{1}{2} \times (b_1 + b_2) \times h$$

where $b_1$ and $b_2$ are the lengths of the bases, and $h$ is the width between the intervals. Here, we approximate:

• $y(1) = 1.414$
• $y(1.25) \approx 1.601$
• $y(1.5) = 1.803$
• $y(1.75) = 2.016$
• $y(2) = 2.236$

So, the width $h$ is $0.25$ for the interval $[1, 1.25]$, $0.25$ for $[1.25, 1.5]$, and so forth. Thus:

$$A \approx \frac{1}{2} \times (y_1 + 2y_2 + 2y_3 + 2y_4 + y_5) \times 0.25$$

Plugging in the values:

$$A \approx \frac{1}{2} \times (1.414 + 2(1.601) + 2(1.803) + 2(2.016) + 2.236) \times 0.25$$

Calculating this results in:

$$A \approx \frac{1}{2} \times (1.414 + 3.202 + 3.606 + 4.032 + 2.236) \times 0.25 \approx \frac{1}{2} \times 14.490 \times 0.25 \approx 1.8115$$

Thus, the approximate area of region $R$ is $1.81$ (to 2 decimal places).