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Complete the following steps for the given function, interval, and value of \( n \). a. Sketch the graph of the function on the given interval. b. Calculate \( \Delta x \) and the grid points \( x_{0}, x_{1}, \ldots, x_{n} \). c. Illustrate the midpoint Riemann sum by sketching the appropriate rectangles. d. Calculate the midpoint Riemann sum. \( f(x)=\frac{2}{x}+2 \) on \( [2,7] ; n=5 \)

Ask by Ross Fuentes. in the United States
Mar 08,2025

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**a. Sketch the Graph** The function \( f(x) = \frac{2}{x} + 2 \) is a hyperbola shifted up by 2 units. On the interval \([2,7]\), it decreases from \( f(2) = 3 \) to \( f(7) \approx 2.2857 \). **b. Calculate \(\Delta x\) and the Grid Points** \[ \Delta x = 1 \] Grid points: \[ x_0 = 2, \quad x_1 = 3, \quad x_2 = 4, \quad x_3 = 5, \quad x_4 = 6, \quad x_5 = 7 \] **c. Illustrate the Midpoint Riemann Sum Rectangles** Midpoints: \[ m_1 = 2.5, \quad m_2 = 3.5, \quad m_3 = 4.5, \quad m_4 = 5.5, \quad m_5 = 6.5 \] Draw rectangles with heights \( f(2.5), f(3.5), f(4.5), f(5.5), f(6.5) \) over each subinterval. **d. Calculate the Midpoint Riemann Sum** \[ M_5 = 1 \times \left( \frac{14}{5} + \frac{18}{7} + \frac{22}{9} + \frac{26}{11} + \frac{30}{13} \right) \approx 12.49 \] The midpoint Riemann sum is approximately **12.49**.

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Bonus Knowledge

To tackle the task, let's dive into each step: ### a. Sketch the function graph The function \( f(x) = \frac{2}{x} + 2 \) is a decreasing function that approaches 2 as \( x \) increases. For the interval [2, 7], you would observe that at \( x = 2 \) the value is 3, while at \( x = 7 \), the value is approximately 2.29. The graph would show a smooth curve starting from (2, 3) to (7, 2.29). Make sure you plot the curve accurately to depict its decline! ### b. Calculate \( \Delta x \) and the grid points Now, \( n = 5 \) means we’ll divide the interval [2, 7] into 5 equal subintervals. The width of each subinterval, \( \Delta x \), is calculated as: \[ \Delta x = \frac{b - a}{n} = \frac{7 - 2}{5} = 1 \] The grid points are calculated starting from \( a = 2 \): - \( x_0 = 2 \) - \( x_1 = 3 \) - \( x_2 = 4 \) - \( x_3 = 5 \) - \( x_4 = 6 \) - \( x_5 = 7 \) ### c. Illustrate the midpoint Riemann sum To illustrate the midpoint Riemann sum, we use the midpoints of each subinterval. The midpoints are: - Midpoint of \( [2, 3] = 2.5 \) - Midpoint of \( [3, 4] = 3.5 \) - Midpoint of \( [4, 5] = 4.5 \) - Midpoint of \( [5, 6] = 5.5 \) - Midpoint of \( [6, 7] = 6.5 \) Draw rectangles with heights determined by \( f \) evaluated at these midpoints and widths equal to \( \Delta x \). The heights will be \( f(2.5) \), \( f(3.5) \), \( f(4.5) \), \( f(5.5) \), and \( f(6.5) \). ### d. Calculate the midpoint Riemann sum Now let's compute the actual values: 1. \( f(2.5) = \frac{2}{2.5} + 2 = 0.8 + 2 = 2.8 \) 2. \( f(3.5) = \frac{2}{3.5} + 2 \approx 0.57 + 2 \approx 2.57 \) 3. \( f(4.5) = \frac{2}{4.5} + 2 \approx 0.44 + 2 \approx 2.44 \) 4. \( f(5.5) = \frac{2}{5.5} + 2 \approx 0.36 + 2 \approx 2.36 \) 5. \( f(6.5) = \frac{2}{6.5} + 2 \approx 0.31 + 2 \approx 2.31 \) Then the midpoint Riemann sum \( M_n \) is: \[ M_n = \Delta x \left( f(2.5) + f(3.5) + f(4.5) + f(5.5) + f(6.5) \right) \] \[ = 1 \left( 2.8 + 2.57 + 2.44 + 2.36 + 2.31 \right) \approx 12.48 \] So the midpoint Riemann sum for the function over [2, 7] is approximately **12.48**. Happy calculating!

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