Use a left and midpoint Rienman sum to approximate

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SUMMARY

The discussion focuses on using left and midpoint Riemann sums to approximate the integral of the function \( f(x) = x^3 \) over the interval [0, 3]. Participants clarify the calculation of \( \Delta x \) as \( \frac{3}{n} \) and emphasize the importance of selecting appropriate endpoints for evaluation. The conversation highlights the distinction between left, right, and midpoint Riemann sums, with specific instructions on how to compute these approximations by evaluating the function at the designated points and summing the areas of rectangles formed by these evaluations.

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  • Understanding of Riemann sums
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  • #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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