\( \int _ { 0 } ^ { 2 } 1 \sqrt { 1 + x } d x \quad n = 4 \)

To evaluate the integral \( \int_{0}^{2} \sqrt{1 + x} \, dx \) using numerical methods, let's apply the trapezoidal rule with \( n = 4 \). First, we divide the interval \([0, 2]\) into four equal parts, each of width \( \Delta x = \frac{2 - 0}{4} = 0.5\). The sample points will be: - \( x_0 = 0 \) - \( x_1 = 0.5 \) - \( x_2 = 1.0 \) - \( x_3 = 1.5 \) - \( x_4 = 2.0 \) Next, we need to evaluate the function: - \( f(x_0) = \sqrt{1 + 0} = 1 \) - \( f(x_1) = \sqrt{1 + 0.5} = \sqrt{1.5} \approx 1.2247 \) - \( f(x_2) = \sqrt{1 + 1} = \sqrt{2} \approx 1.4142 \) - \( f(x_3) = \sqrt{1 + 1.5} = \sqrt{2.5} \approx 1.5811 \) - \( f(x_4) = \sqrt{1 + 2} = \sqrt{3} \approx 1.7321 \) Now, we can apply the trapezoidal rule formula: \[ \int_{a}^{b} f(x) \, dx \approx \frac{\Delta x}{2} \left[ f(x_0) + 2 f(x_1) + 2 f(x_2) + 2 f(x_3) + f(x_4) \right] \] Plugging in the numbers, we calculate: \[ \int_{0}^{2} \sqrt{1+x} \, dx \approx \frac{0.5}{2} \left[ 1 + 2(1.2247) + 2(1.4142) + 2(1.5811) + 1.7321 \right] \] \[ = 0.25 \left[ 1 + 2.4494 + 2.8284 + 3.1622 + 1.7321 \right] \] \[ = 0.25 \left[ 11.1721 \right] \approx 2.7930 \] Thus, using the trapezoidal rule with \( n = 4 \), the approximate value of the integral \( \int_{0}^{2} \sqrt{1 + x} \, dx \) is roughly \( 2.7930 \).