# Upper and lower bound Riemann sums

## 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?

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.

SammyS
Staff Emeritus
Homework Helper
Gold Member

## 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?