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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))(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