Upper and lower bound Riemann sums

  • #1
248
3

Homework Statement


Find the upper, lower and midpoint sums for $$\displaystyle\int_{-3}^{3} (12-x^{2})dx$$
$$\rho = \Big\{-3,-1,3\Big\}$$

The Attempt at a Solution



For the upper:
(12-(-1)^2)(-1-(-3)) + (12-(-1))(3-(-1))
=74

For the lower:
(12-(-3)^2)(-1-(-3))+(12-3)(3-(-1))
=42

For midpoint sum:
I used the expression $$\frac{b+a}{n}$$

(12-(-2)^2)(2) + (12-(2)^2)(4)
=48

Are these values correct? Does the process seem good?

Please explain
Thank you
 

Answers and Replies

  • #2
Svein
Science Advisor
Insights Author
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I think you have misunderstood slightly. "Upper", "lower" and "midpoint" do not refer to the argument ("x") values, but to the function values.

In your case you need to calculate f(-3), f(-1) and f(3) and use those values to determine "upper" and "lower".
 
  • #3
248
3
@Svein I thought finding upper, lower and midpoints was essentially taking the partition {-3,-1,3} then breaking that into sub-intervals (-3,-1),(-1,3) then referencing the graph to see which is biggest or smallest in that interval. And in (-3,-1) if we are looking for upper that would be -1 then we would evaluate -1 by plugging it back into the original function and multiplying that by the distance between those two points and so on.
 
  • #5
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
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Homework Statement


Find the upper, lower and midpoint sums for $$\displaystyle\int_{-3}^{3} (12-x^{2})dx$$
$$\rho = \Big\{-3,-1,3\Big\}$$

The Attempt at a Solution



For the upper:
(12-(-1)^2)(-1-(-3)) + (12-(-1)^2)(3-(-1)) ##\ \ \ ## You didn't square the indicated -1 .
=74

For the lower:
(12-(-3)^2)(-1-(-3))+(12-(3^2))(3-(-1)) ##\ \ \ ## You didn't square the indicated 3 .
=42

For midpoint sum:
I used the expression $$\frac{b+a}{n}$$
(12-(-2)^2)(2) + (12-(2)^2)(4)
=48

Are these values correct? Does the process seem good?

Please explain
Thank you
You made errors in evaluating your function for the Upper and Lower sums.

How are upper and lower sums defined in your textbook/course?

The midpoint sum looks to be correct.
 

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