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

We are given the function \[ f(x)=x^3+1, \] and we want to approximate \[ \int_{0}^{8} \left(x^3+1\right)\,dx \] using a midpoint Riemann sum with 4 equal subdivisions. 1. **Determine the width of each subinterval:** The length of the interval is \[ 8-0=8. \] With 4 subdivisions, the width is \[ \Delta x=\frac{8}{4}=2. \] 2. **Identify the subintervals and their midpoints:** The 4 subintervals are: - \([0,2]\) with midpoint \( x_1= \frac{0+2}{2}=1 \). - \([2,4]\) with midpoint \( x_2= \frac{2+4}{2}=3 \). - \([4,6]\) with midpoint \( x_3= \frac{4+6}{2}=5 \). - \([6,8]\) with midpoint \( x_4= \frac{6+8}{2}=7 \). 3. **Evaluate \( f(x) \) at each midpoint:** - At \( x=1 \): \[ f(1)=1^3+1=1+1=2. \] - At \( x=3 \): \[ f(3)=3^3+1=27+1=28. \] - At \( x=5 \): \[ f(5)=5^3+1=125+1=126. \] - At \( x=7 \): \[ f(7)=7^3+1=343+1=344. \] 4. **Compute the midpoint Riemann sum:** The midpoint Riemann sum is given by: \[ \text{Sum}=\Delta x\left(f(1)+f(3)+f(5)+f(7)\right). \] Substitute the values: \[ \text{Sum}=2\left(2+28+126+344\right). \] First, calculate the sum inside: \[ 2+28=30,\quad 30+126=156,\quad 156+344=500. \] Now multiply by 2: \[ 2 \times 500=1000. \] Thus, the approximate value of the integral using the midpoint Riemann sum with 4 subdivisions is \[ 1000. \]