1. Concept 3: Finding the Area
The Fundamental Theorem of Calculus is the primary method to find the area “under the curve”
“Under the curve” is a loosely used phrase. More accurately, it is the area bound between a curve and an axis

2. THERE IS NO EXAMPLE OR SLE FOR THIS CONCEPT
Examples
THERE IS NO EXAMPLE OR SLE FOR THIS CONCEPT

3. Concept 4: The Mean Value Theorem of Integrals
In Derivatives:
𝑓 ′ 𝑐 = 𝑓 𝑏 −𝑓(𝑎) 𝑏−𝑎
That’s…
That’s Slope

4. Concept 4: The Mean Value Theorem of Integrals
Info Only
If you see “info only”, don’t write anything from the slide. If your OCD can’t handle that, leave space in your notebook and download this presentation from my website. I will give you NO time beyond my talking to write.

5. Concept 4: The Mean Value Theorem of Integrals
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If 𝑓 is continuous on a closed interval [𝑎, 𝑏], then there exists a number 𝑐 in the closed interval [𝑎, 𝑏] such that:
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 =𝑓(𝑐)(𝑏−𝑎)

7. Concept 4: The Mean Value Theorem of Integrals
Info Only
Let 𝑓(𝑚) represent the lower approximation
Let 𝑓(𝑀) represent the greater approximation
Let 𝑓(𝑥) represent the actual value

8. Concept 4: The Mean Value Theorem of Integrals
Info Only
If 𝑓 𝑚 ≤𝑓 𝑥 ≤𝑓(𝑀)
Then
𝑎 𝑏 𝑓 𝑚 𝑑𝑥≤ 𝑎 𝑏 𝑓 𝑥 𝑑𝑥≤ 𝑎 𝑏 𝑓 𝑀 𝑑𝑥

9. Concept 4: The Mean Value Theorem of Integrals
Info Only
𝑎 𝑏 𝑓 𝑚 𝑑𝑥≤ 𝑎 𝑏 𝑓 𝑥 𝑑𝑥≤ 𝑎 𝑏 𝑓 𝑀 𝑑𝑥
𝑎 𝑏 𝑓(𝑚)(𝑏−𝑎) ≤ 𝑎 𝑏 𝑓 𝑥 𝑑𝑥 ≤ 𝑎 𝑏 𝑓(𝑀)(𝑏−𝑎)
𝑓 𝑚 ≤ 1 𝑏−𝑎 𝑎 𝑏 𝑓 𝑥 𝑑𝑥 ≤𝑓(𝑀)

10. Concept 4: The Mean Value Theorem of Integrals
So, it’s an average?
1 𝑏−𝑎 𝑎 𝑏 𝑓 𝑥 𝑑𝑥

11. Example 5: Average Value for a Function
Find the average value of:
𝑓 𝑥 = 4 𝑥 𝑥 2 ;[1, 3]
1 𝑏−𝑎 𝑎 𝑏 𝑓 𝑥 𝑑𝑥
= 1 3− ( 𝑥 2 +1) 𝑥 2 𝑑𝑥

12. Example 5: Average Value for a Function
( 𝑥 2 +1) 𝑥 2 𝑑𝑥
= ( 𝑥 2 +1) 𝑥 2 𝑑𝑥
= 𝑥 2 𝑥 𝑥 2 𝑑𝑥
= 𝑥 −2 𝑑𝑥

13. Example 5: Average Value for a Function
2 𝑥− 1 𝑥 1 3
=2 3− 1 3 −2 1− 1 1
= −2 0
=2 8 3
= 16 3

14. Student Led Example 5: Average Value for a Function
Find the AVERAGE VALUE of:
𝑔 𝑥 =3 𝑥 2 −2𝑥;[1, 4] 16

15. Given a piecewise function:
Example 6: Application
Given a piecewise function:
Evaluate the function over the given interval
Determine the average value of 𝑓 for that interval
Determine the answers to parts (a) and (b) if the function were translated two units up

16. Part a) Given: 𝑓 𝑥 = −&2𝑥+5, 1≤𝑥≤2 &𝑥−1, 2≤𝑥,≤3 −&𝑥+5 3≤𝑥≤4 &1, 4≤𝑥,≤7
Example 6: Application
Given: 𝑓 𝑥 = −&2𝑥+5, 1≤𝑥≤2 &𝑥−1, 2≤𝑥,≤3 −&𝑥+5 3≤𝑥≤4 &1, 4≤𝑥,≤7

17. Part a)
Example 6: Application
Evalute:
1 7 𝑓 𝑥 𝑑𝑥

18. Part a)
Example 6: Application

19. 1 2 −2𝑥+5𝑑𝑥 = − 𝑥 2 +5𝑥 1 2 = −4+10 −[−1+5] =2 Part a)
Example 6: Application
𝑓 𝑥 = −&2𝑥+5, 1≤𝑥≤2 &𝑥−1, 2≤𝑥,≤3 −&𝑥 ≤𝑥≤4 &1, ≤𝑥,≤7
1 2 −2𝑥+5𝑑𝑥
= − 𝑥 2 +5𝑥 1 2
= −4+10 −[−1+5] =2

20. Part a)
Example 6: Application 2
1.5
𝑓 𝑥 = −&2𝑥+5, 1≤𝑥≤2 &𝑥−1, 2≤𝑥,≤3 −&𝑥 ≤𝑥≤4 &1, ≤𝑥,≤7
= 𝑥 2 −𝑥 2 3
2 3 𝑥−1𝑑𝑥
= 4.5−3 − 2−2
=1.5

21. 1.5 1.5 2 =− 1 2 𝑥 2 +5𝑥 3 4 3 4 −𝑥+5𝑑𝑥 = −8+20 − −4.5+15 =1.5 Part a)
Example 6: Application 2
1.5
1.5
𝑓 𝑥 = −&2𝑥+5, 1≤𝑥≤2 &𝑥−1, 2≤𝑥,≤3 −&𝑥 ≤𝑥≤4 &1, ≤𝑥,≤7
=− 𝑥 2 +5𝑥 3 4
3 4 −𝑥+5𝑑𝑥
= −8+20 − −4.5+15
=1.5

22. 1.5 1.5 3 2 4 7 1𝑑𝑥 = 𝑥 4 7 = 7−4 =3 Part a) Example 6: Application
𝑓 𝑥 = −&2𝑥+5, 1≤𝑥≤2 &𝑥−1, 2≤𝑥,≤3 −&𝑥 ≤𝑥≤4 &1, ≤𝑥,≤7
4 7 1𝑑𝑥
= 𝑥 4 7
= 7−4 =3

23. Part a)
Example 6: Application
1.5 2
1.5 3 8 + = + +

24. Find the average area 1 𝑏−𝑎 𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = 8 6 = 1 7−1 ⋅8 Part b)
Example 6: Application
Find the average area
1 𝑏−𝑎 𝑎 𝑏 𝑓 𝑥 𝑑𝑥
= 1 7−1 ⋅8
= 8 6

25. Area avg = 20 6 = 10 3 Area=8+12=20 Part c)
Example 6: Application
Find the area if 𝑓(𝑥) was translated 2 units up
𝑓(𝑥) spans 6 units wide and is raised 2 units up for 12 units of additional area
Area avg = 20 6 = 10 3
Area=8+12=20

26. Student Led Example 6: Application
Example 5 on p.282