Transcript 4 1 Antidervatives

4-1: Antiderivatives
Objective:
1. To define and find
antiderivatives
Assignment:
• P. 255-258: 5-7, 15-41
odd, 47, 49, 51, 60, 67,
75, 87-92
Warm Up 1
Find each derivative.
1
2
1. 𝑦 = 𝑥 2 + 3𝑥 + 1
1
2
2. 𝑦 = 𝑥 2 + 3𝑥 − 1
Warm Up 2
Given 𝑦 ′ = 𝑥 + 3, what is 𝑦?
Objective 1
To define and find
antiderivatives
Differential Equations
A differential
equation in 𝑥 and
𝑦 is an equation
involving 𝑥, 𝑦, and
derivatives of 𝑦.
𝑦′ = 𝑥 + 3
𝑦 = 12𝑥 2 + 3𝑥 + 1
Solving differential equations involves working
backwards from 𝑦′ to find 𝑦. This process is
called integration or antidifferentiation.
Antiderivative
𝑓(𝑥) = 𝑥 + 3
Antiderivatives
A function 𝐹 is an antiderivative of 𝑓 on an
interval 𝐼 if 𝐹 ′ 𝑥 = 𝑓(𝑥) for all 𝑥 in 𝐼.
𝐹 𝑥 = 12𝑥 2 + 3𝑥 − 1
𝐹 𝑥 = 12𝑥 2 + 3𝑥
𝐹 𝑥 = 12𝑥 2 + 3𝑥 + 1
𝐹 𝑥 = 12𝑥 2 + 3𝑥 + 2
Antiderivatives
If 𝐹 is an antiderivative of 𝑓 on an interval 𝐼,
then 𝐺 is also an antiderivative of 𝑓 on 𝐼 iff
𝐺(𝑥) = 𝐹(𝑥) + 𝐶, for all 𝑥 in 𝐼, where 𝐶 is a
constant.
General
𝑓(𝑥) = 𝑥 + 3
𝐹 𝑥 = 12𝑥 2 + 3𝑥 − 1
𝐺 𝑥 = 12𝑥 2 + 3𝑥 + 𝐶
We denote the entire family of
antiderivatives of 𝑓 by adding a constant 𝐶.
antiderivative
of 𝑓
Antiderivatives
If 𝐹 is an antiderivative of 𝑓 on an interval 𝐼,
then 𝐺 is also an antiderivative of 𝑓 on 𝐼 iff
𝐺(𝑥) = 𝐹(𝑥) + 𝐶, for all 𝑥 in 𝐼, where 𝐶 is a
constant.
𝑓(𝑥) = 𝑥 + 3
𝐹 𝑥 = 12𝑥 2 + 3𝑥 − 1
𝐺 𝑥 = 12𝑥 2 + 3𝑥 + 𝐶
We denote the entire family of
antiderivatives of 𝑓 by adding a constant 𝐶.
General
solution to
the
differential
equation
𝐺′ 𝑥 = 𝑥 + 3
Antiderivatives
If 𝐹 is an antiderivative of 𝑓 on an interval 𝐼,
then 𝐺 is also an antiderivative of 𝑓 on 𝐼 iff
𝐺(𝑥) = 𝐹(𝑥) + 𝐶, for all 𝑥 in 𝐼, where 𝐶 is a
constant.
𝑓(𝑥) = 𝑥 + 3
𝐹 𝑥 = 12𝑥 2 + 3𝑥 − 1
𝐺 𝑥 =
1 2
2𝑥
+ 3𝑥 + 𝐶
Constant of
Integration
Exercise 1
Find the general solution to the differential
equation 𝑦′ = 2.
Differentials
The differential equation
𝑑𝑦
𝑑𝑥
= 𝑓(𝑥) can be written as
Differential Form
𝑑𝑦 = 𝑓 𝑥 𝑑𝑥
Differential
of 𝑦
Approximates
∆𝑦
Differential
of 𝑥
Approximates
∆𝑥
𝑑𝑦 is the amount of change in the
tangent line when 𝑥 changes by 𝑑𝑥
∆𝑦 is the amount of change in
𝒚 = 𝒇(𝒙) when 𝑥 changes by ∆𝑥
Indefinite Integration
To solve this differential equation, we apply the
process of antidifferentiation or indefinite
integration, which is denoted by ∫ .
𝑑𝑦 = 𝑓 𝑥 𝑑𝑥
𝑦=
𝑓 𝑥 𝑑𝑥 = 𝐹 𝑥 + 𝐶
Read as “the antiderivative of 𝑓 with respect to 𝑥”
Indefinite Integration
To solve this differential equation, we apply the
process of antidifferentiation or indefinite
integration, which is denoted by ∫ .
𝑑𝑦 = 𝑓 𝑥 𝑑𝑥
𝑦=
𝑓 𝑥 𝑑𝑥 = 𝐹 𝑥 + 𝐶
Read as “the integral of 𝑓 with respect to 𝑥”
Indefinite Integration
To solve this differential equation, we apply the
process of antidifferentiation or indefinite
integration, which is denoted by ∫ .
𝑑𝑦 = 𝑓 𝑥 𝑑𝑥
𝑦=
𝑓 𝑥 𝑑𝑥 = 𝐹 𝑥 + 𝐶
Integral sign
Integrand
Variable of
Integration
Constant of
Integration
Inverse Relationship
Differentiation and Integration are inverse
operations.
𝐹 ′ 𝑥 𝑑𝑥 = 𝐹 𝑥 + 𝐶
𝑑
𝑑𝑥
𝑓 𝑥 𝑑𝑥 = 𝑓(𝑥)
These equations can be use to build integration
rules from differentiation rules
Integration Rules
Differentiation Formula
Integration Formula
𝑑
𝐶 =0
𝑑𝑥
0 𝑑𝑥 = 𝐶
𝑑
𝑘𝑥 = 𝑘
𝑑𝑥
𝑘 𝑑𝑥 = 𝑘𝑥
𝑑
𝑘𝑓(𝑥) = 𝑘𝑓 ′ (𝑥)
𝑑𝑥
𝑘𝑓(𝑥) 𝑑𝑥 = 𝑘
𝑓(𝑥) 𝑑𝑥
Integration Rules
Differentiation Formula
𝑑 𝑛
𝑥 = 𝑛𝑥 𝑛−1
𝑑𝑥
Integration Formula
𝑥 𝑛 𝑑𝑥 =
Exercise 2
Describe the antiderivatives of 3𝑥.
Exercise 3
Integrate the following.
1.
1
∫ 𝑥 3 𝑑𝑥
2. ∫ 𝑥 𝑑𝑥
3. ∫ 2 sin 𝑥 𝑑𝑥
You can check your
answer to an
integration problem
by differentiating
Exercise 4
Integrate the following.
1. ∫ 𝑑𝑥
2. ∫ 𝑥 + 2 𝑑𝑥
3. ∫ 3𝑥 4 − 5𝑥 2 + 𝑥 𝑑𝑥
Exercise 5
𝑥+1
𝑑𝑥
𝑥
Exercise 6
sin 𝑥
𝑑𝑥
2
cos 𝑥
A Particular Solution
Up to this point, we
have only been finding
the general solution to
a differential equation.
Given an initial
condition, however,
we can find a
particular solution.
Differential Equation
𝑓 ′ (𝑥) = 3𝑥 2 − 1
General Solution
𝐹 𝑥 = 𝑥3 − 𝑥 + 𝐶
Initial Condition
𝐹 2 =4
Particular Solution
𝐹 𝑥 = 𝑥3 − 𝑥 − 2
Exercise 7
′
1
,
𝑥3
Find the general solution of 𝑓 𝑥 =
𝑥 > 0, and
find the particular solution that satisfies the initial
condition 𝐹(1) = 0.
Exercise 8
A ball is thrown upward with an initial velocity of 64
feet per second from an initial height of 80 feet.
Find the position function giving the height 𝑠 as a
function of the time 𝑡. When does the ball hit the
ground?
4-1: Antiderivatives
Objective:
1. To define and find
antiderivatives
Assignment:
• P. 255-258: 5-7, 15-41
odd, 47, 49, 51, 60, 67,
75, 87-92