Question Find the relative error when using the midpoint rule to calculate \( \int_{0}^{2}\left(3 x^{2}+2\right) d x \) using 4 subintervals. Find the relative error as a percent rounded to the nearest hundredth if necessary. Do not round until your final answer and do not inclu the symbol "\%" in your answer.

To find the relative error when using the midpoint rule for the integral \(\int_{0}^{2}\left(3 x^{2}+2\right) d x\) with 4 subintervals, first calculate the exact value of the integral. The integral computes as follows: \[ \int_{0}^{2}\left(3 x^{2}+2\right) d x = \left[ x^3 + 2x \right]_{0}^{2} = (2^3 + 2 \times 2) - (0 + 0) = 8 + 4 = 12. \] Now, using the midpoint rule with 4 subintervals (each of length \(\Delta x = \frac{2-0}{4} = 0.5\)), the midpoints \(x_i\) for each subinterval are \(0.25\), \(0.75\), \(1.25\), and \(1.75\). Now evaluate the function at each of these midpoints: \[ f(0.25) = 3(0.25)^2 + 2 = 3(0.0625) + 2 = 2.1875, \] \[ f(0.75) = 3(0.75)^2 + 2 = 3(0.5625) + 2 = 3.6875, \] \[ f(1.25) = 3(1.25)^2 + 2 = 3(1.5625) + 2 = 6.6875, \] \[ f(1.75) = 3(1.75)^2 + 2 = 3(3.0625) + 2 = 11.1875. \] Now, calculate the approximate integral using the midpoint rule: \[ M \approx \Delta x \left( f(0.25) + f(0.75) + f(1.25) + f(1.75) \right) = 0.5 \left( 2.1875 + 3.6875 + 6.6875 + 11.1875 \right). \] Calculating the sum: \[ M \approx 0.5 \times 23.75 = 11.875. \] Now, we find the absolute error: \[ \text{Absolute Error} = | \text{Exact Value} - \text{Approximate Value} | = | 12 - 11.875 | = 0.125. \] Next, calculate the relative error: \[ \text{Relative Error} = \frac{\text{Absolute Error}}{\text{Exact Value}} = \frac{0.125}{12} \approx 0.01041666667. \] Finally, convert the relative error to a percent: \[ \text{Relative Error as a Percent} = 0.01041666667 \times 100 \approx 1.041666667. \] Rounding to the nearest hundredth: **1.04**