Question Find the relative error when using the midpoint rule to calculate \( \int_{0}^{2}\left(4 x^{2}+10\right) d x \) 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 include the symbol "\%" in your answer.

To find the relative error when using the midpoint rule to calculate the integral \( \int_{0}^{2}\left(4 x^{2}+10\right) d x \) with 4 subintervals, we will follow these steps: 1. **Calculate the exact value of the integral.** 2. **Apply the midpoint rule to estimate the integral.** 3. **Calculate the relative error.** ### Step 1: Calculate the Exact Value of the Integral The exact value of the integral can be calculated as follows: \[ \int_{0}^{2} (4x^2 + 10) \, dx \] First, we find the antiderivative: \[ F(x) = \frac{4}{3}x^3 + 10x \] Now, we evaluate it from 0 to 2: \[ F(2) = \frac{4}{3}(2^3) + 10(2) = \frac{4}{3}(8) + 20 = \frac{32}{3} + 20 = \frac{32}{3} + \frac{60}{3} = \frac{92}{3} \] \[ F(0) = 0 \] Thus, the exact value of the integral is: \[ \int_{0}^{2} (4x^2 + 10) \, dx = F(2) - F(0) = \frac{92}{3} \] ### Step 2: Apply the Midpoint Rule We will use 4 subintervals. The width of each subinterval \( \Delta x \) is: \[ \Delta x = \frac{b - a}{n} = \frac{2 - 0}{4} = \frac{1}{2} \] The midpoints of each subinterval are: - For the first interval \([0, \frac{1}{2}]\): midpoint \( m_1 = \frac{0 + \frac{1}{2}}{2} = \frac{1}{4} \) - For the second interval \([\frac{1}{2}, 1]\): midpoint \( m_2 = \frac{\frac{1}{2} + 1}{2} = \frac{3}{4} \) - For the third interval \([1, \frac{3}{2}]\): midpoint \( m_3 = \frac{1 + \frac{3}{2}}{2} = \frac{5}{4} \) - For the fourth interval \([\frac{3}{2}, 2]\): midpoint \( m_4 = \frac{\frac{3}{2} + 2}{2} = \frac{7}{4} \) Now we evaluate the function at these midpoints: \[ f(m_1) = f\left(\frac{1}{4}\right) = 4\left(\frac{1}{4}\right)^2 + 10 = 4\left(\frac{1}{16}\right) + 10 = \frac{1}{4} + 10 = \frac{41}{4} \] \[ f(m_2) = f\left(\frac{3}{4}\right) = 4\left(\frac{3}{4}\right)^2 + 10 = 4\left(\frac{9}{16}\right) + 10 = \frac{36}{16} + 10 = \frac{36}{16} + \frac{160}{16} = \frac{196}{16} = \frac{49}{4} \] \[ f(m_3) = f\left(\frac{5}{4}\right) = 4\left(\frac{5}{4}\right)^2 + 10 = 4\left(\frac{25}{16}\right) + 10 = \frac{100}{16} + 10 = \frac{100}{16} + \frac{160}{16} = \frac{260}{16} = \frac{65}{4} \] \[ f(m_4) = f\left(\frac{7}{4}\right) = 4\left(\frac{7}{4}\right)^2 + 10 = 4\left(\frac{49}{16}\right) + 10 = \frac{196}{16} + 10 = \frac{196}{16} + \frac{160}{16} = \frac{356}{16} = \frac{89}{4} \] Now we can calculate the estimate using the midpoint rule: \[ \text{Midpoint Estimate} = \Delta x \left( f(m_1) + f(m_2) + f(m_3) + f(m_4) \right) \] \[ = \frac{1}{2} \left( \frac{41}{4} + \frac{49}{4} + \frac{65}{4} + \frac{89}{4} \right) \] \[ = \frac{1}{2} \left( \frac{41 + 49 + 65 + 89}{4} \right) = \frac{1}{2} \left( \frac{244}{4} \right) = \frac{1}{2} \cdot 61 = 30.5 \] ### Step 3: Calculate the Relative Error The relative error is given by: \[ \text{Relative Error} = \left| \frac{\text{Exact Value} - \text{Estimate}}{\text{Exact Value}} \right| \] Substituting the values we have: \[ \text{Exact Value} = \frac{92}{3} \quad \text{and} \quad \text{Estimate} = 30.5 \] Calculating the relative error: \[ \text{Relative Error} = \left| \frac{\frac{92}{3} - 30.5}{\frac{92}{3}} \right| \] First, we convert \( 30.5 \) to a fraction: \[ 30.5 = \frac{61}{2} \] Now we find a common denominator to subtract: \[ \frac{92}{3} - \frac{61}{2} = \frac{184}{6} - \frac{183}{6} = \frac{1}{6} \] Now substituting back into the relative error formula: \[ \text{Relative Error} = \left| \frac{\frac{1}{6}}{\frac{92}{3}} \right| = \left| \frac{1}{6} \cdot \frac{3}{92} \right| = \frac{3}{552} = \frac{1}{184} \] Now, converting this to a percentage: \[ \text{Relative Error as a percent} = \left( \frac{1}{184} \right) \times 100