1. Warm Up 1. Find 2 6 2𝑥+1 𝑑𝑥 without using a calculator
2. Let 𝐴 𝑥 = 0 𝑥 𝑓(𝑡) 𝑑𝑡. Represent 2 6 𝑓(𝑡) 𝑑𝑡 in terms of 𝐴(𝑥)

2. Section 4.4 Day 1 Fundamental Theorem of Calculus
AP Calculus AB

3. Learning Targets
Define the Fundamental Theorem of Calculus parts I and II
Apply the Fundamental Theorem of Calculus parts I and II
Evaluate an integrals with an absolute value function as the integrand
Define the Mean Value Theorem/Average Value for Integrals
Apply the Mean Value Theorem/Average Value for Integrals

4. Fundamental Theorem of Calculus Part I Definition
If a function 𝑓 is continuous on [𝑎, 𝑏] and 𝐹 is an antiderivative of 𝑓 on [𝑎, 𝑏], then
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 =𝐹 𝑏 −𝐹(𝑎)
𝑜𝑟 𝑎 𝑏 𝑓 𝑥 𝑑𝑥 +𝐹 𝑎 =𝐹(𝑏)

5. Fundamental Theorem of Calculus Part I Why does this work?
What does this mean for us?
The definite integral can be solved without having to use Riemann Sums or areas!

6. Fundamental Theorem of Calculus Part I Why does this work?
Let’s look at each piece individually:
1. 𝑎 𝑏 𝑓(𝑥) 𝑑𝑥 represents the area under the derivative curve from [𝑎, 𝑏]
2. 𝐹 𝑏 −𝐹(𝑎) represents the difference between the function values of 𝐹 at 𝑎 and 𝑏.

7. Fundamental Theorem of Calculus Part I Why does this work?
Now, let’s look at each piece individually in the context of position (m) and velocity (m/s):
1. 𝑎 𝑏 𝑓(𝑥) 𝑑𝑥 represents the area under the velocity curve from 𝑎, 𝑏 . Recall, that the time units will cancel resulting in just meters.
2. 𝐹 𝑏 −𝐹(𝑎) represents the distance traveled between 𝑎 and 𝑏.
This is the exact same thing! 

8. Evaluate Definite Integrals: Example 1
−3 2 − −3 1 =− 2 3

9. Evaluate Definite Integrals: Example 2
Find 𝑥 2 + 𝑥 𝑑𝑥
1 3 (1) (1) = =1

10. Evaluate Definite Integrals: Example 3 Initial Condition
Given 𝑑𝑦 𝑑𝑥 =3 𝑥 2 +4𝑥−5 with the initial condition 𝑦 2 =−1. Find 𝑦(3).
𝑦 3 = 𝑥 2 +4𝑥−5 𝑑𝑥+𝑦(2)
𝑦 3 = −5 3 − −5 2 = 24
𝑦 3 =24−1=23

11. Evaluate Definite Integrals: Example 4 Initial Condition
A pizza with a temperature of 95°𝐶 is put into a 25°𝐶 room when 𝑡=0. The pizza’s temperature is decreasing at a rate of 𝑟 𝑡 =6 𝑒 −0.1𝑡 °𝐶 per minute. Estimate the pizza’s temperature when 𝑡=5 minutes.
𝑟 5 = 0 5 −6 𝑒 −0.1𝑡 𝑑𝑡 +𝑟 0 =71.391°𝐶

12. Evaluate Definite Integrals Example 5 Absolute Value
−(2𝑥−1) 𝑑𝑥 𝑥−1 𝑑𝑥=2.5

13. Evaluate Definite Integrals Example 6 Motion
An object has a velocity function of 𝑣 𝑡 =2𝑡−6
Find the distance from time 𝑡=2 to 𝑡=6.
2 6 2𝑡−6 𝑑𝑡= −6 6 − −6 2 =8
Find the total distance from 𝑡=2 to 𝑡=6.
2 6 |2𝑡−6| 𝑑𝑡= 2 3 − 2𝑡−6 𝑑𝑡 𝑡−6 𝑑𝑡 =10