Use a midpoint Riemann sum with four subdivisions of equal length to find the approximate value of \( S_{0}^{8}\left(x^{3}+1\right) d x \).

To approximate the integral \( S_{0}^{8} (x^{3} + 1) \, dx \) using a midpoint Riemann sum with four subdivisions, we first divide the interval \([0, 8]\) into four equal parts. Each subinterval will have a width \( \Delta x = \frac{8 - 0}{4} = 2 \). The subintervals are: 1. \([0, 2]\) 2. \([2, 4]\) 3. \([4, 6]\) 4. \([6, 8]\) Next, we find the midpoints of each subinterval: 1. Midpoint of \([0, 2]\): \( m_1 = 1 \) 2. Midpoint of \([2, 4]\): \( m_2 = 3 \) 3. Midpoint of \([4, 6]\): \( m_3 = 5 \) 4. Midpoint of \([6, 8]\): \( m_4 = 7 \) Now we evaluate the function \( f(x) = x^3 + 1 \) at each midpoint: 1. \( f(m_1) = f(1) = 1^3 + 1 = 2 \) 2. \( f(m_2) = f(3) = 3^3 + 1 = 27 + 1 = 28 \) 3. \( f(m_3) = f(5) = 5^3 + 1 = 125 + 1 = 126 \) 4. \( f(m_4) = f(7) = 7^3 + 1 = 343 + 1 = 344 \) Finally, we calculate the midpoint Riemann sum: \[ S \approx \Delta x \cdot (f(m_1) + f(m_2) + f(m_3) + f(m_4)) = 2 \cdot (2 + 28 + 126 + 344) = 2 \cdot 500 = 1000 \] Thus, the approximate value of the integral is \( 1000 \).