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 \[ f(x)=5 \tan x,\left[0, \frac{\pi}{3}\right] \]

To approximate the area of the region bounded by the graph of the function \( f(x) = 5 \tan x \) over the interval \([0, \frac{\pi}{3}]\) using the Midpoint Rule with \( n = 4 \), we first need to calculate the width of each subinterval. 1. **Calculate the width of each subinterval**: The width of each subinterval \( \Delta x \) is given by: \[ \Delta x = \frac{b - a}{n} = \frac{\frac{\pi}{3} - 0}{4} = \frac{\pi}{12}. \] 2. **Determine the midpoints**: Since \( a = 0 \) and \( b = \frac{\pi}{3} \), the midpoints of each subinterval can be found as follows: - For the first subinterval \([0, \frac{\pi}{12}]\): Midpoint = \( m_1 = 0 + \frac{\Delta x}{2} = \frac{\pi}{24} \). - For the second subinterval \([\frac{\pi}{12}, \frac{\pi}{6}]\): Midpoint = \( m_2 = \frac{\pi}{12} + \frac{\Delta x}{2} = \frac{\pi}{12} + \frac{\pi}{24} = \frac{\pi}{8} \). - For the third subinterval \([\frac{\pi}{6}, \frac{\pi}{4}]\): Midpoint = \( m_3 = \frac{\pi}{6} + \frac{\Delta x}{2} = \frac{\pi}{6} + \frac{\pi}{24} = \frac{5\pi}{24} \). - For the fourth subinterval \([\frac{\pi}{4}, \frac{\pi}{3}]\): Midpoint = \( m_4 = \frac{\pi}{4} + \frac{\Delta x}{2} = \frac{\pi}{4} + \frac{\pi}{24} = \frac{7\pi}{24} \). 3. **Evaluate the function at the midpoints**: Next, we calculate \( f(m_i) \) for each midpoint: - \( f(m_1) = f\left(\frac{\pi}{24}\right) = 5 \tan\left(\frac{\pi}{24}\right) \). - \( f(m_2) = f\left(\frac{\pi}{8}\right) = 5 \tan\left(\frac{\pi}{8}\right) \). - \( f(m_3) = f\left(\frac{5\pi}{24}\right) = 5 \tan\left(\frac{5\pi}{24}\right) \). - \( f(m_4) = f\left(\frac{7\pi}{24}\right) = 5 \tan\left(\frac{7\pi}{24}\right) \). 4. **Sum the areas of each rectangle**: We then use the Midpoint Rule formula: \[ A \approx \Delta x \left( f(m_1) + f(m_2) + f(m_3) + f(m_4) \right). \] 5. **Calculate the approximate area**: Thus, the approximate area can be computed as: \[ A \approx \frac{\pi}{12} \left( 5 \tan\left(\frac{\pi}{24}\right) + 5 \tan\left(\frac{\pi}{8}\right) + 5 \tan\left(\frac{5\pi}{24}\right) + 5 \tan\left(\frac{7\pi}{24}\right) \right). \] Factoring out the 5, we get: \[ A \approx \frac{5\pi}{12} \left( \tan\left(\frac{\pi}{24}\right) + \tan\left(\frac{\pi}{8}\right) + \tan\left(\frac{5\pi}{24}\right) + \tan\left(\frac{7\pi}{24}\right) \right). \] You can now compute the values of the tangents at the computed angles and plug them into the equation to get the final area approximation!