How Do You Estimate a Double Integral Using Riemann Sums?

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SUMMARY

The discussion focuses on estimating a double integral using Riemann sums, specifically for the function ∫∫(1-xy²)dA over the region R = [0,4] x [-1,2]. The participants confirm the setup of the Riemann sum with m=2 and n=3, using lower right corners for sample points. An alternative approach using upper left corners is also explored, leading to a corrected calculation resulting in -8. The participants validate each other's methods and calculations throughout the discussion.

PREREQUISITES
  • Understanding of double integrals and Riemann sums
  • Familiarity with the concept of sample points in integration
  • Basic knowledge of function evaluation at specific points
  • Ability to perform arithmetic operations with sums
NEXT STEPS
  • Study the properties of Riemann sums in multivariable calculus
  • Learn about the differences between lower and upper Riemann sums
  • Explore numerical integration techniques beyond Riemann sums
  • Investigate the application of double integrals in real-world scenarios
USEFUL FOR

Students studying calculus, particularly those focusing on multivariable integration, as well as educators looking for practical examples of Riemann sums in action.

AATroop
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Homework Statement


If R = [0,4]x[-1,2], use a Riemann sum with m=2, n=3 to estimate the value of ∫∫(1-xy^2)dA. Take the sample points to be the lower right corners.

Homework Equations


None

The Attempt at a Solution


2*1[f(2,-1) + f(2,0) + f(2,1) + f(4,-1) + f(4,0) + f(4,1)] = some value

Just wondering if the setup is correct, particularly the 2 and the 1 at the beginning.

Edit:

Would the solution for the upper left corners of the rectangles be

2*1[f(0,-1)+ f(0,0)+ f(0,1) +f(2,-1) + f(2,0) + f(2,1)]
 
Last edited:
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AATroop said:

Homework Statement


If R = [0,4]x[-1,2], use a Riemann sum with m=2, n=3 to estimate the value of ∫∫(1-xy^2)dA. Take the sample points to be the lower right corners.


Homework Equations


None


The Attempt at a Solution


2*1[f(2,-1) + f(2,0) + f(2,1) + f(4,-1) + f(4,0) + f(4,1)] = some value

Just wondering if the setup is correct, particularly the 2 and the 1 at the beginning.

Yes, that looks OK

Edit:

Would the solution for the upper left corners of the rectangles be

2*1[f(0,-1)+ f(0,0)+ f(0,1) +f(2,-1) + f(2,0) + f(2,1)]

(0,-1) isn't the upper left corner of any rectangle. Did you accidentally do lower left?
 
Yes, I think I did do lower left. I changed it and it came out to -8. Sounds close enough to me haha. Thanks a bunch for your help.
 

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