Use a midpoint Riemann sum with four Subdivisions of equal length to find the approximate value of \( 5_{0}^{E}\left(x^{3}+1\right) d x \).

To estimate the integral using a midpoint Riemann sum, we first identify the interval boundaries. The interval from 0 to \( E \) can be divided into four equal subdivisions. Thus, the width of each subinterval is \( \Delta x = \frac{E - 0}{4} = \frac{E}{4} \). The midpoints for each subinterval are: 1. \( x_1 = \frac{E}{8} \) (midpoint of [0, \( \frac{E}{4} \)]) 2. \( x_2 = \frac{3E}{8} \) (midpoint of [\( \frac{E}{4}, \frac{E}{2} \)]) 3. \( x_3 = \frac{5E}{8} \) (midpoint of [\( \frac{E}{2}, \frac{3E}{4} \)]) 4. \( x_4 = \frac{7E}{8} \) (midpoint of [\( \frac{3E}{4}, E \)]) Next, we evaluate the function \( f(x) = x^{3} + 1 \) at each midpoint: - \( f\left(\frac{E}{8}\right) = \left(\frac{E}{8}\right)^{3} + 1 = \frac{E^{3}}{512} + 1 \) - \( f\left(\frac{3E}{8}\right) = \left(\frac{3E}{8}\right)^{3} + 1 = \frac{27E^{3}}{512} + 1 \) - \( f\left(\frac{5E}{8}\right) = \left(\frac{5E}{8}\right)^{3} + 1 = \frac{125E^{3}}{512} + 1 \) - \( f\left(\frac{7E}{8}\right) = \left(\frac{7E}{8}\right)^{3} + 1 = \frac{343E^{3}}{512} + 1 \) Now, we sum these values and multiply by the width \( \Delta x \): \[ \text{Midpoint Riemann Sum} = \Delta x \left( f\left(\frac{E}{8}\right) + f\left(\frac{3E}{8}\right) + f\left(\frac{5E}{8}\right) + f\left(\frac{7E}{8}\right) \right) = \frac{E}{4} \left( \left(\frac{E^{3}}{512} + 1\right) + \left(\frac{27E^{3}}{512} + 1\right) + \left(\frac{125E^{3}}{512} + 1\right) + \left(\frac{343E^{3}}{512} + 1\right) \right) \] Combining terms inside the parenthesis: \[ = \frac{E}{4} \left( \frac{E^{3} + 27E^{3} + 125E^{3} + 343E^{3}}{512} + 4 \right) = \frac{E}{4} \left( \frac{496E^{3}}{512} + 4 \right) = \frac{E}{4} \left( \frac{31E^{3}}{32} + 4 \right) \] \[ = \frac{E}{128}(31E^{3} + 128) \] Thus, the approximate value of the integral \( \int_0^E (x^{3}+1) dx \) using a midpoint Riemann sum with four subdivisions is \( \frac{E(31E^{3} + 128)}{128} \).