Solution:
Given:
[tex]\begin{gathered} y=\sqrt[3]{x-1} \\ on\text{ the interval \lbrack3,6\rbrack} \end{gathered}[/tex]
where a = 3, b = 6, n = 4.
This gives
[tex]\triangle\text{x=}\frac{6-3}{4}=\frac{3}{4}[/tex]
Divide the interval into 4 subintervals of the length Δx with the following endpoints:
[tex]a=3,\text{ }\frac{15}{4},\frac{9}{2},\frac{21}{4},6=b[/tex]
For the Left Riemann sum, we evaluate the function at the left endpoints of the subintervals. Thus, we have
sum up the values and multiply by Δx, we have
[tex]4.349280826349355[/tex]
To the second decimal place, we have the Left Riemann sum to be
[tex]4.35[/tex]
For the Right Riemann sum, we evaluate the function at the right endpoints of the subintervals. Thus, we have
sum up the values and multiply by Δx, we have
[tex]4.686821998935723[/tex]
To the second decimal place, we have the Right Riemann sum to be
[tex]4.69[/tex]
