Split up the interval [0, 3] into 6 subintervals. Each will have length
[tex]\Delta x_i = \dfrac{3-0}6 = \dfrac12[/tex]
So the partition will be
[0, 1/2], [1/2, 1], [1, 3/2], [3/2, 2], [2, 5/2], [5/2, 6]
The right endpoints form an arithmetic sequence given by
[tex]x_i = \dfrac i2[/tex]
where 1 ≤ i ≤ 6.
Then the Riemann sum is
[tex]\displaystyle \sum_{i=1}^6 f(x_i) \Delta x_i = \frac12 \sum_{i=1}^6 \left(\left(\frac i2\right)^3 - 6\left(\frac i2\right)\right)[/tex]
[tex]\displaystyle \sum_{i=1}^6 f(x_i) \Delta x_i = \frac12 \sum_{i=1}^6 \left(\frac{i^3}8 - 3i\right)[/tex]
[tex]\displaystyle \sum_{i=1}^6 f(x_i) \Delta x_i = \frac1{16} \sum_{i=1}^6 i^3 - \frac32 \sum_{i=1}^6 i[/tex]
Recall the formulas for sums of powers:
[tex]\displaystyle \sum_{i=1}^n i = \frac{n(n+1)}2[/tex]
[tex]\displaystyle \sum_{i=1}^n i^3 = \frac{n^2(n+1)^2}4[/tex]
It follows that
[tex]\displaystyle \sum_{i=1}^6 f(x_i) \Delta x_i = \frac1{16} \cdot \frac{6^2\cdot7^2}4 - \frac32 \cdot \frac{6\cdot7}2 = -\frac{63}{16} \approx \boxed{-3.938}[/tex]
