Use a left and midpoint Rienman sum to approximate

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Homework Help Overview

The problem involves using left and midpoint Riemann sums to approximate the integral \(\int_{0}^{3} x^3 dx\). Participants are discussing the setup of the Riemann sums and the implications of choosing different endpoints for the approximation.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to clarify the choice of the left endpoint 'a' for the Riemann sum and how to calculate the values of \(x_i\). There are questions about the relationship between the left and right endpoints and whether they will yield the same results. Some participants are exploring the implications of choosing different values for \(n\) in the context of approximating the integral.

Discussion Status

The discussion is ongoing, with various interpretations being explored regarding the setup of the Riemann sums. Some participants have offered guidance on how to approach the problem, while others are questioning the assumptions being made about the endpoints and the values of \(n\). There is a recognition that approximating the integral involves summing the areas of rectangles based on chosen endpoints.

Contextual Notes

Participants note that the problem specifies using left and midpoint Riemann sums, which may impose constraints on how the approximation is approached. There is also mention of the need to define \(n\) for the calculations, which some participants find confusing.

  • #31
flyingpig said:
But the question said "right Reinman"
What's your point? All this means is that you use the right endpoint of each subinterval.

Pick a number n.
Divide the interval [0, 3] into that many subintervals.
Evaluate f at the [STRIKE]left end [/STRIKE] [STRIKE]midpoint[/STRIKE] right end of each subinterval.
Multiply by \Delta x.
Add the n terms.



BTW, it's Riemann.
 
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  • #32
What if you want to use left endpoint
 
  • #33
Try this:

Riemann Sum

Particularly the difference between right, midpoint, and left.
 
  • #34
flyingpig said:
What if you want to use left endpoint
See post #24.

I really hope that you have finished this problem by now. Approximating an integral by a Riemann sum is really nothing more than adding up the areas of a number of (n) rectangles. The rectangles are all \Delta x (= (b - a)/n) in width. For a left Riemann sum, you use the left endpoint of the subinterval to find the height of the rectangle. For a right Riemann sum, you use the right endpoint of the subinterval to find the height of the rectangle. For a midpoint Riemann sum, you use the midpoint of the subinterval to find the height of the rectangle.

To get your approximation, add the areas of the n rectangles. That's all there is to it.
 

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