Lesson 19 : Section 4.9
Antiderivatives
Dr. Robinson
BC Calculus
Sections 1 and 2: November 11, 2015
Section 3: November 13, 2015
 Objective: Students will be able to find antiderivatives of functions. Students will be able to evaluate
indefinite integrals. Students will be able to apply anitderivatives to solve application problems
involving simple differential equations.
 Key Concepts
 Antiderivative
 General Antiderivative
 Common Antiderivatives
 Initial Value Problems and Differential Equations
 Indefinite Integrals
 Procedures
 Presentation of material
 Review of homework/in-class work on problem set
 Problem Set
 13, 19, 23, 33, 37, 43, 47, 49, 53, 55, 65, 67, 69, 71, 75, 77, 79, 83, 85, 89, 93, 95
 Assessments
 Quiz 8: 4.7 - 5.1
 Test 3
2 | Lesson 19: 4.9
BC Calculus
Dr. Robinson
Antiderivative
Definition: A function F(x) is an antiderivative of f (x) on an interval I, iff F' (x) = f (x) ∀ x ∈ I.
Examples:
f (x) = 2 x
g(x) = cos x
F(x) =
G(x) =
Dr. Robinson
BC Calculus
Lesson 19: 4.9 | 3
F(x) = x2
G(x) = sin x
f (x) = 2 x
g(x) = cos x
F(x) = x2 is not the only function that is an antiderivative of f (x) = 2 x. In fact, ∀ C ∈ , F(x) = x 2 + C is
an antiderivative of f (x). We say that the function f (x) has a family of functions F(x) + C that are antideriva
tives.
Definition: If F(x) is an antiderivative of f (x), then the most general antiderivative of f (x) is
F(x) + C.
Theorem: Power Rule for Indefinite Integrals
p
 x ⅆx =
Antiderivatives you already know.
xp+1
p+1
+C
4 | Lesson 19: 4.9
BC Calculus
Algebraic Rules for Antiderivatives
Theorem: If F(x) is an antiderivative of f (x), and G(x) is an antiderivative of g(x), then
1. k f (x)
has antiderivative
k F(x) + C
2. - f (x)
has antiderivative
-F(x) + C
3. f (x) ± g(x)
has antiderivative
F(x) ± G(x) + C
These statements are true because of the linearity of the derivative operator.
Example:
f (x) =
3
x
+ sin 2 x
Dr. Robinson
Dr. Robinson
BC Calculus
F(x) = 3 · 2
x -
1
2
Lesson 19: 4.9 | 5
cos 2 x + C
Differential Equations and Initial Value Problems - IVPs
Finding an antiderivative for a function f (x) is the same as finding a function y = f (x) that will solve the
equation
ⅆy
= f (x)
ⅆx
called a differential equation. The function y = f (x) is often referred to as y(x).
The general solution is the family of functions
y = F(x) + C
for any constant C.
If we are given the additional information F(x0 ) = y0 , then we have what is called an initial value and an
initial value problem or IVP. This means that we choose the single function out of the family of general
solutions such that the graph of the function passes throught the point (x0 , y0 ). Given the initial value, we
can plug it into the equation
y = F(x) + C
and solve for the constant C, i.e., y0 = F(x0 ) + C ⟹ C = y0 - F(x0 ) We can find the particular function
that is in the family of anitderivatives that satisfies the additional requirement of the initial value.
1. The general solution to the differential equation
ⅆy
= f (x)
ⅆx
is the family of antiderivates
y = F(x) + C.
2. The particular solution to the IVP
ⅆy
= f (x)
ⅆx
F(x0 ) = y0
is the single function
y = F(x) + (y0 - F(x0 )).
6 | Lesson 19: 4.9
BC Calculus
Example: Solve the IVP
ⅆy
= 3 x2
given
y(1) = -1,
ⅆx
i.e., the graph of y passes through the point (1, -1).
Dr. Robinson
Dr. Robinson
BC Calculus
ⅆy
ⅆx
Lesson 19: 4.9 | 7
= 3 x2 ⟹ y = x 3 + C
y(1) = -1 ⟹ -1 = 13 + C ⟹ C = -2
y = x3 - 2
is the particular solution to the IVP.
y
y  x3 + C
C2
C1
C0
2
C  -1
1
C  -2
1
-1
x
(1,-1)
-2
Concept of Antiderivative Applied to Motion
If we know an object’s acceleration, and the initial position and the initial velocity, then we can find the
particular function that describes the object’s position.
Example: A balloon ascending at the rate of 12 ft/sec is at a height of 80 feet above the ground when a
package is dropped. How long does it take for the package to reach the ground?
8 | Lesson 19: 4.9
BC Calculus
Dr. Robinson
The acceleration near the surface of the earth is due to gravity pulling down on the package.
a(t) = -32 ft  sec2
Acceleration is negative because gravity is acting in the direction of decreasing height s(t).
At the time the package is released, the velocity is given by 12 ft/sec, so we know the initial condition
v(0) = 12. This velocity is positive because the balloon is traveling in the direction of increasing height.
ⅆv
a(t) =
so v(t) is the antiderivative of a(t)
ⅆt
v(t) = -32 t + C
v(0) = 12 ⟹ C = 12
So the particular velocity function is
v(t) = -32 t + 12
At the time the package is released, the position of the balloon is 80 feet, so we know the initial condition
s(0) = 80.
ⅆs
v(t) =
so s(t) is the antiderivative of
v(t)
ⅆt
t2
s(t) = -32
+ 12 t + C
2
s(0) = 80 ⟹ C = 80
So the particular function the describes the height of the package at time t is
s(t) = -16 t2 + 12 t + 80
To answer the question, we solve s(t) = 0.
0 = -16 t2 + 12 t + 80 ⟹ t = 2.64
So, the package reaches the ground 2.64 seconds after being released.
Dr. Robinson
BC Calculus
Lesson 19: 4.9 | 9
Indefinite Integral
Definition: The set of all antiderivates of f (x) is called the indefinite integral of f w.r.t. x and is denoted by
 f (x) ⅆx
 is the integral sign
f (x) is called the integrand
ⅆx is a differential and is required in the integral
ⅆx indicates which variable the integral is with repsect to
Examples:
∫ 2 x ⅆx
∫ cos x ⅆx
10 | Lesson 19: 4.9
BC Calculus
Dr. Robinson
2
∫ 2 x ⅆx = x + C
∫ cos x ⅆx = sin x + C
Note: When evaluating an indefinite integral, you must have the arbitrary constant C in your answer in
order to indicate the entire set of antiderivatives. You will get points off if you do not include the constant of integration.
Example:
2
2
 x - 2 x + 5 ⅆx =  x ⅆx -  2 x ⅆx +  5 ⅆx
x3
=
3
=
x3
3
+ C1 - x2 + C2  + (5 x + C3 )
- x2 + 5 x + (C1 + C2 + C3 )
C
We can just integrate term-by-term and then add the constant.
2
 x - 2 x + 5 ⅆx =
x3
3
- x2 + 5 x + C
Back to Equations of Motion
Because we have that
a(t) = v' (t)
we see that v(t) =  a(t) ⅆt
and since v(t) = s' (t)
we have that s(t) =  v(t) ⅆt