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)

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

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

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)

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!

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.

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! 

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

Evaluate Definite Integrals: Example 2 Evaluate 0 𝜋 4 sec 2 𝑥 𝑑𝑥 tan 𝜋 4 − tan 0 =1

Evaluate Definite Integrals: Example 3 Find 0 1 𝑥 2 + 𝑥 𝑑𝑥 1 3 (1) 3 + 2 3 (1) 3 2 = 1 3 + 2 3 =1

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

Evaluate Definite Integrals: Example 5 initial Condition Given 𝑓 ′ 𝑥 = sin 𝑥 2 and 𝑓 2 =−5, find 𝑓 1 . 1 2 𝑓′(𝑥) 𝑑𝑥=𝑓 2 −𝑓 1 𝑓 1 =−5.495

Evaluate Definite Integrals: Example 6 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

Evaluate Definite Integrals Example 7 Initial Condition The graph of 𝑓′ on −2≤𝑥≤6 consists of two line segments and a semicircle as shown. Given that 𝑓 −2 =5. Find 𝑓(6) 𝑓 6 = −2 6 𝑓 𝑡 𝑑𝑡 +𝑓(−2) 𝑓 6 = 8+2𝜋 +5=13+2𝜋

Evaluate Definite Integrals Example 8 Absolute Value 0 1 2 −(2𝑥−1) 𝑑𝑥+ 1 2 2 2𝑥−1 𝑑𝑥=2.5

Evaluate Definite Integrals Example 9 Absolute Value −3 −2 − 3𝑥+6 𝑑𝑥 + −2 0 3𝑥+6 𝑑𝑥=7.5

Evaluate Definite Integrals Example 10 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 2 −6 6 − 2 2 −6 2 =8 Find the total distance from 𝑡=2 to 𝑡=6. 2 6 |2𝑡−6| 𝑑𝑡= 2 3 − 2𝑡−6 𝑑𝑡 + 3 6 2𝑡−6 𝑑𝑡 =10

Evaluate Definite Integrals Example 11 Motion The velocity of a particle moving on a line at time 𝑡 is 𝑣=5 𝑡 2 3 +6𝑡. How many meters did the particle travel from 𝑡=1 to 𝑡=8? 1 8 5 𝑡 2 3 +6𝑡 𝑑𝑡= 3 8 5 3 +3 8 2 − 3 1 5 3 +3 1 2 3 1 5 3 +3 1 2 =282m