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

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

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

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.

Concept 4: The Mean Value Theorem of Integrals Info Only 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)

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

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

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)

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

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

Example 5: Average Value for a Function 1 2 1 3 4( 𝑥 2 +1) 𝑥 2 𝑑𝑥 =2 1 3 ( 𝑥 2 +1) 𝑥 2 𝑑𝑥 =2 1 3 𝑥 2 𝑥 2 + 1 𝑥 2 𝑑𝑥 =2 1 3 1+ 𝑥 −2 𝑑𝑥

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

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

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

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

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

Part a) Example 6: Application

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+5 3≤𝑥≤4 &1, 4≤𝑥,≤7 1 2 −2𝑥+5𝑑𝑥 = − 𝑥 2 +5𝑥 1 2 = −4+10 −[−1+5] =2

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

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+5 3≤𝑥≤4 &1, 4≤𝑥,≤7 =− 1 2 𝑥 2 +5𝑥 3 4 3 4 −𝑥+5𝑑𝑥 = −8+20 − −4.5+15 =1.5

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+5 3≤𝑥≤4 &1, 4≤𝑥,≤7 4 7 1𝑑𝑥 = 𝑥 4 7 = 7−4 =3

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

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

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

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