Upper and lower bound Riemann sums

[Niaboc67]

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

 

[Svein]

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".

 

[Niaboc67]

@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.

 

[Svein]

For a thorough explanation of Riemann integral and Riemann sums, see: http://math.feld.cvut.cz/mt/txtd/1/txe3da1a.htm.

 

[SammyS]

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.