Use the Midpoint Rule with \( n=4 \) to approximate the area of the region bounded by the graph of the function and the \( x \)-axis over the given interval. \[ f(x)=x^{2}+5,[0,2] \] Need Help? Rwadit

Sure, I'd be happy to help you apply the **Midpoint Rule** to approximate the area under the curve of the function \( f(x) = x^2 + 5 \) over the interval \([0, 2]\) with \( n = 4 \) subintervals. ### **Step 1: Determine the Width of Each Subinterval (\( \Delta x \))** The interval \([a, b] = [0, 2]\) is divided into \( n = 4 \) subintervals. \[ \Delta x = \frac{b - a}{n} = \frac{2 - 0}{4} = 0.5 \] ### **Step 2: Identify the Midpoints of Each Subinterval** Since each subinterval has a width of \( 0.5 \), the subintervals and their midpoints are: 1. **First Subinterval:** \([0, 0.5]\) **Midpoint:** \( m_1 = 0 + \frac{0.5}{2} = 0.25 \) 2. **Second Subinterval:** \([0.5, 1.0]\) **Midpoint:** \( m_2 = 0.5 + \frac{0.5}{2} = 0.75 \) 3. **Third Subinterval:** \([1.0, 1.5]\) **Midpoint:** \( m_3 = 1.0 + \frac{0.5}{2} = 1.25 \) 4. **Fourth Subinterval:** \([1.5, 2.0]\) **Midpoint:** \( m_4 = 1.5 + \frac{0.5}{2} = 1.75 \) ### **Step 3: Evaluate the Function at Each Midpoint** Calculate \( f(m_i) \) for each midpoint: 1. \( f(0.25) = (0.25)^2 + 5 = 0.0625 + 5 = 5.0625 \) 2. \( f(0.75) = (0.75)^2 + 5 = 0.5625 + 5 = 5.5625 \) 3. \( f(1.25) = (1.25)^2 + 5 = 1.5625 + 5 = 6.5625 \) 4. \( f(1.75) = (1.75)^2 + 5 = 3.0625 + 5 = 8.0625 \) ### **Step 4: Apply the Midpoint Rule Formula** The Midpoint Rule formula is: \[ M_n = \Delta x \left[ f(m_1) + f(m_2) + f(m_3) + f(m_4) \right] \] Plugging in the values: \[ M_4 = 0.5 \times \left(5.0625 + 5.5625 + 6.5625 + 8.0625\right) \] \[ M_4 = 0.5 \times 25.25 = 12.625 \] ### **Result** Using the Midpoint Rule with \( n = 4 \), the approximate area under the curve \( f(x) = x^2 + 5 \) from \( x = 0 \) to \( x = 2 \) is **12.625**. ### **Verification (Optional)** For comparison, let's calculate the exact area using the definite integral: \[ \int_{0}^{2} (x^2 + 5) \, dx = \left[ \frac{x^3}{3} + 5x \right]_0^2 = \left( \frac{8}{3} + 10 \right) - (0 + 0) = \frac{8}{3} + 10 = \frac{38}{3} \approx 12.666... \] The Midpoint Rule approximation of **12.625** is very close to the exact value of approximately **12.666...**, demonstrating the effectiveness of the Midpoint Rule with \( n = 4 \) for this function and interval.