Question

Use the Left and Right Riemann Sums with 4 rectangles to estimate the (signed) area under the curve of y=x−1‾‾‾‾‾√3 on the interval of [3,6]. Round your answers to the second decimal place.

Answer 1

Given:

`\begin{gathered} y=\sqrt[3]{x-1} \\ on\text{ the interval \lbrack3,6\rbrack} \end{gathered}`

where a = 3, b = 6, n = 4.

This gives

`\triangle\text{x=}(6-3)/(4)=(3)/(4)`

Divide the interval into 4 subintervals of the length Δx with the following endpoints:

`a=3,\text{ }(15)/(4),(9)/(2),(21)/(4),6=b`

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

`4.349280826349355`

To the second decimal place, we have the Left Riemann sum to be

`4.35`

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

`4.686821998935723`

To the second decimal place, we have the Right Riemann sum to be

`4.69`