Let's solve the integral \(\int_{0}^{1} \sqrt{x^2 + 1}\, dx\) using the **trapezoidal rule** with \( n = 5 \). --- ### **Step 1: Find the width of each subinterval** The interval \([0, 1]\) is divided into \(n = 5\) subintervals. \[ h = \frac{b - a}{n} = \frac{1 - 0}{5} = 0.2 \] --- ### **Step 2: Find the \(x\)-values** \[ x_0 = 0, \quad x_1 = 0.2, \quad x_2 = 0.4, \quad x_3 = 0.6, \quad x_4 = 0.8, \quad x_5 = 1.0 \] --- ### **Step 3: Evaluate the function at these points** \[ f(x) = \sqrt{x^2 + 1} \] - \(f(x_0) = \sqrt{0^2 + 1} = 1\) - \(f(x_1) = \sqrt{0.2^2 + 1} = \sqrt{0.04 + 1} = \sqrt{1.04} \approx 1.0198\) - \(f(x_2) = \sqrt{0.4^2 + 1} = \sqrt{0.16 + 1} = \sqrt{1.16} \approx 1.0770\) - \(f(x_3) = \sqrt{0.6^2 + 1} = \sqrt{0.36 + 1} = \sqrt{1.36} \approx 1.1662\) - \(f(x_4) = \sqrt{0.8^2 + 1} = \sqrt{0.64 + 1} = \sqrt{1.64} \approx 1.2806\) - \(f(x_5) = \sqrt{1^2 + 1} = \sqrt{1 + 1} = \sqrt{2} \approx 1.4142\) --- ### **Step 4: Apply the trapezoidal rule formula** \[ \int_{a}^{b} f(x)\, dx \approx \frac{h}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + 2f(x_4) + f(x_5) \right] \] Plug in the values: \[ \int_{0}^{1} \sqrt{x^2 + 1}\, dx \approx \frac{0.2}{2} \left[ 1 + 2(1.0198) + 2(1.0770) + 2(1.1662) + 2(1.2806) + 1.4142 \right] \] --- ### **Step 5: Calculate the sum inside the brackets** \[ \begin{align*} &1 + 2(1.0198) + 2(1.0770) + 2(1.1662) + 2(1.2806) + 1.4142 \\ &= 1 + 2.0396 + 2.1540 + 2.3324 + 2.5612 + 1.4142 \\ &= 1 + 2.0396 = 3.0396 \\ &3.0396 + 2.1540 = 5.1936 \\ &5.1936 + 2.3324 = 7.5260 \\ &7.5260 + 2.5612 = 10.0872 \\ &10.0872 + 1.4142 = 11.5014 \end{align*} \] --- ### **Step 6: Finish the calculation** \[ \int_{0}^{1} \sqrt{x^2 + 1}\, dx \approx 0.1 \times 11.5014 = 1.15014 \] --- ## **Final Answer** \[ \boxed{1.15} \] (using three significant digits; you can round as needed). This is the approximate value of the integral using the trapezoidal rule with \(n = 5\).