Using 4 equal subintervals for \( [0,12] \) and the trapezoidal rule, approximate: \[ \left.\begin{array}{l} \int_{0}^{12} f(x) d x=[?] \\ x \mid 036912 \\ \hline f(x) \end{array} \right\rvert\, 4581311.0 \] Round your answer to the nearest thousandth.

To apply the trapezoidal rule, we first determine the width of each subinterval. The interval \([0, 12]\) is divided into 4 equal subintervals, so each subinterval will be: \[ h = \frac{b - a}{n} = \frac{12 - 0}{4} = 3 \] The subintervals will be \([0, 3]\), \([3, 6]\), \([6, 9]\), and \([9, 12]\). Next, we can calculate the integral using the trapezoidal rule formula: \[ \int_a^b f(x) \, dx \approx \frac{h}{2} \left( f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + f(x_4) \right) \] Where \(x_0 = 0\), \(x_1 = 3\), \(x_2 = 6\), \(x_3 = 9\), and \(x_4 = 12\). Given the values of \(f(x)\): - \(f(0) = 4\) - \(f(3) = 5\) - \(f(6) = 8\) - \(f(9) = 13\) - \(f(12) = 11\) Now substituting the function values into our trapezoidal rule formula: \[ \int_0^{12} f(x) \, dx \approx \frac{3}{2} \left( f(0) + 2f(3) + 2f(6) + 2f(9) + f(12) \right) \] Calculating step-by-step: 1. \(f(0) = 4\) 2. \(f(3) = 5\), so \(2f(3) = 2 \times 5 = 10\) 3. \(f(6) = 8\), so \(2f(6) = 2 \times 8 = 16\) 4. \(f(9) = 13\), so \(2f(9) = 2 \times 13 = 26\) 5. \(f(12) = 11\) Now summing it up: \[ f(0) + 2f(3) + 2f(6) + 2f(9) + f(12) = 4 + 10 + 16 + 26 + 11 = 67 \] Substituting this back into our trapezoidal rule formula: \[ \int_0^{12} f(x) \, dx \approx \frac{3}{2} \times 67 = \frac{201}{2} = 100.5 \] Rounding it to the nearest thousandth, the approximate value is: \[ \int_{0}^{12} f(x) \, dx \approx 100.500 \]