1. 6-2 definite integrals

2. *Yesterday, we explored Riemann Sums using both a set width for the rectangles as well as varying widths.
We found that if we averaged our individual answers, we were able to come close to “the real thing.”
*Imagine if we were to make the widths super small – creating LOTS of rectangles. We would be practically finding the real area instead of estimating the area.
Definite Integral is a limit of Riemann Sums.

3. Thm: All Continuous Functions are Integrable
Definite Integral:
height
base
partitions
Thm: All Continuous Functions are Integrable
If f is continuous on [a, b], then the definite integral exists on [a, b].
(watch how the notation morphs…)
*Because the limits are all the same, we don’t need the partitions…
* f is continuous on [a, b] * n subintervals
upper limit
variable of integration
function/integrand
lower limit

4. Ex 1) The interval [–1, 3] is partitioned into n subintervals
of equal length Let mk denote the midpoint of
the kth subinterval. Express the limit
Since the points were chosen from the subintervals of the partition, it’s a limit of Riemann sums. (didn’t “have” to be midpoints)

5. Def: Definite Integral and Area Under a Curve
If y = f (x) is nonnegative and integrable on [a, b], then the area under the curve as defined using a definite integral is
Integral Area
Ex 2) Evaluate the integral
Means area

6. *Functions can be above or below the x-axis…
 below the x-axis gives us a negative value for area because the “height” is a negative value
If we want TOTAL AREA (above AND below) =
= area above axis – area below axis
Thm: Integral of a Constant
If f (x) = c is a constant function on [a, b], then
Why does this make sense?
It’s a rectangle!

7. Calculator Time!! MATH  9: fnInt
(Enter whichever applies to your calculator)
fnInt (f (x), x, a, b) OR
Ex) Evaluate the following using your calculator. a) b)
 (Let’s go back now and finish the last page of yesterday’s activity!)

8. Give a geometric explanation why some of the answers are negative and some are zero.
Direction
L  R +
Graph
above + 1 + 6)
answer 7) –2
L  R +
below – – 00

9. – 8) Area above = area below  cancel each other 9) –2 R  L – above +
Area above = area below
 cancel each other 9) –2
R  L –
above + – 00

10. 10) 2
R  L –
below – +
Same
11) 2

11. homework
Pg. 286 #1–7, 10, 13, 14, 15, 22, 25, 31–34