ANTIDERIVATIVES AND INDEFINITE INTEGRATION
AB Calculus
ANTIDERIVATIVES AND INDEFINITE
INTEGRATION
y  x 2  3 x  12
y  2 x  3
y  2𝑥 + 3
y  x 2  3x
y 
y  x  3x  7
2
Rem:
2𝑥 + 3
DEFN: A function F is called an Antiderivative of the function f, if for
every x in f:
F /(x) = f(x)
If f (x) =
3x
2
then F(x) =
x
3
or
/
If f (x) =
3x
2
since
then f (x) =
x
3
d 3
( x )  3x 2
dx
ANTIDERIVATIVES
𝑎𝑑𝑑 𝑜𝑛𝑒 𝑡𝑜 𝑡ℎ𝑒 𝑒𝑥𝑝𝑜𝑛𝑒𝑛𝑡
𝑛𝑒𝑤 𝑒𝑥𝑝𝑜𝑛𝑒𝑛𝑡
Layman’s Idea:
A) What is the function that has f
.
(x) as its derivative?
n 1
x
-Power Rule: f ( x)  x n  F ( x) 
n 1
f ( x)  sin x  F ( x)   cos x
-Trig:
f ( x)  cos x  F ( x)  sin x
B) The antiderivative is never unique, all answers must include a
+C
(constant of integration)
The Family of Functions whose derivative is given.
Family of Graphs
All same
equations
with different
y intercept
+C
𝑑𝑦
= cos(𝑥) 𝑦 = − sin 𝑥 + 𝑐
𝑑𝑥
The Family of Functions whose derivative is given.
Notation:
dy
 f ( x)
dx
dy  f ( x )dx
y   f ( x)dx
Differential Equation
small
Differential Form (REM: A Quantity of change)
Integral symbol =
Integrand =

=sum
f ( x)
Variable of Integration =
dx
Summing a
bunch of little
changes
The Variable of Integration
y   f ( x )dx
Newton’s Law of gravitational attraction

NOW:
dr
Gm1m2
2
r
dr
tells which variable is being integrated r
Will have more meanings later!
The Family of Functions whose derivative
is given.
𝑦′ = 𝑥3
𝑦 =?
𝑦 ′ = 𝑥 −2
𝑥 −1
𝑦=
−1
𝑥4
𝑦=
4
𝑦′
=
1
𝑥2
3
𝑥2
2 3
𝑦=
= 𝑥2
3
3
2
𝑦′ = 𝑥
𝑦=
2
−3
1
𝑥3
1
3
=
1
3𝑥 3
Notation:
dy
 3x  4
dx
𝑑𝑦 =
3𝑥 + 4 𝑑𝑥
Differential Equation
Differential Form
( REM: A Quantity of change)
Increment of change
Antiderivative or Indefinite Integral
y
3𝑥 + 4 𝑑𝑥
Total (Net) change
General Solution
A) Indefinite Integration and the Antiderivative are the same thing.

General Solution
f ( x)dx  F ( x)  c
_________________________________________________________
ILL:
3𝑥 + 4 𝑑𝑥
3𝑥𝑑𝑥 +
4 𝑑𝑥
𝑥2
3
+ 4𝑥 + 𝑐
2
General Solution: EX 1.
General Solution: The Family of Functions
EX 1:
1
 x3 dx
𝑥 −3 𝑑𝑥
𝑥 −2
+𝑐
−2
 f ( x)dx  F ( x)  c
General Solution: EX 2.
General Solution: The Family of Functions
EX 2:
 (2sin x)dx
2 − cos 𝑥 + 𝑐
−2(cos 𝑥) + 𝑐
 f ( x)dx  F ( x)  c
General Solution: EX 3.
General Solution: The Family of Functions
EX 3:
1
f ( x) 
x
/
𝑥 −1 𝑑𝑥
𝑥0
0
𝑙𝑛 𝑥 + 𝑐
Careful !!!!!
= 𝑥 −1
 f ( x)dx  F ( x)  c
Verify the statement by showing the derivative of the right side
equals the integral of the left side.
9
3
− 4 𝑑𝑥 = 3 + 𝑐
𝑥
𝑥
𝑥 3 0 − 3(3𝑥 2 ) −9𝑥 2
=
=
3
2
𝑥
𝑥6
−9
= 4 +𝑐
𝑥
3
𝑡
𝑡 2 − sin 𝑡 𝑑𝑡 =
+ cos 𝑡 + 𝑐
3
General Solution
A) Indefinite Integration and the Antiderivative are the same thing.

General Solution
f ( x)dx  F ( x)  c
_________________________________________________________
4
ILL:
𝑥3 3 4
= 𝑥3 + 𝑐
3
4
4
𝑥𝑑𝑥
3
1
𝑥 𝑥
𝑑𝑥
1
𝑑𝑥
3
2𝑥
−1
−1
𝑥 𝑥2
=
3 2
−2+2
𝑥
1
−
2
=
1
−2
−2𝑥
+𝑐
1 −3 1 𝑥 −2
1 −2
𝑥 =
=− 𝑥
2
2 −2
4
Special Considerations
 3x dx
4
3𝑥 5
+𝑐
5
 ( x  3) dx
2
𝑥 2 − 6𝑥 + 9 𝑑𝑥
𝑥 2 𝑑𝑥 +
−6𝑥𝑑𝑥 +
𝑥3
𝑥2
+ −6
+ 9𝑥
3
2
𝑥3
− 3𝑥 2 + 9𝑥 + 𝑐
3
9𝑑𝑥
𝑥 2 − 2𝑥 + 1
𝑥−1
(𝑥 − 1)(𝑥 − 1)
(𝑥 − 1)
𝑥 − 1 𝑑𝑥
𝑥𝑑𝑥 −
1𝑑𝑥
𝑥2
−𝑥+𝑐
2
Initial Condition Problems:
B) Initial Condition Problems:
Particular solution < the single graph of the Family –
through a given point>
ILL:
dy
 2x 1
dx
through the point (1,1)
𝑑𝑦 = 2𝑥 − 1 𝑑𝑥
-Find General solution
𝑑𝑦 =
-Plug in Point < Initial Condition >
and solve for C
2𝑥 = 1 𝑑𝑥
𝑥2
𝑦=2
−𝑥+𝑐
2
𝑦 = 𝑥2 − 𝑥 + 𝑐
1 = 12 − 1 + c
1 = 1 −1+c
1=𝑐
𝑦 = 𝑥2 − 𝑥 + 1
through the point (1,1)
𝑦 = 𝑥2 − 𝑥 + 1
Initial Condition Problems: EX 4.
B) Initial Condition Problems:
Particular solution < the single graph of the Family –
through a given point.>
Ex 4:
1
f ( x) 
x
2
/
𝑦=
1
𝑥𝑑𝑥
2
1 𝑥2
𝑦=
+𝑐
2 2
1 2
𝑦 = 𝑥 +𝑐
4
1
f    1
2
2
1 1
−1 =
+𝑐
4 2
1
−1 =
+𝑐
16
17
−
=𝑐
16
1 2 17
𝑦= 𝑥 −
4
16
Initial Condition Problems: EX 5.
B) Initial Condition Problems:
Particular solution < the single graph of the Family –
through a given point.>
Ex 5:
f ( x)  cos( x)
/
𝑓 𝑥 =
cos 𝑥 𝑑𝑥
= sin 𝑥 + 𝑐
𝜋
1 = sin
+𝑐
3
3
1=
+𝑐
2
3
1−
=𝑐
2
f(

3
) 1
3
𝑓 𝑥 = sin 𝑥 + 1 −
2
Initial Condition Problems: EX 6.
B) Initial Condition Problems:
A particle is moving along the x - axis such that its
acceleration is a (t )  . 2
At t = 2 its velocity is 5 and its position is 10.
Find the function, x (t ), that models the particle’s motion.
𝑥(𝑡) = 𝑣(𝑡)
𝑥 ′′ 𝑡 = 𝑎(𝑡)
𝑣 𝑡 =
𝑎 𝑡 𝑑𝑡
𝑣 𝑡 =
2𝑑𝑡
𝑣 𝑡 = 2𝑡 + 𝑐
5=2 2 +𝑐
1=𝑐
𝑣 𝑡 = 2𝑡 + 1
𝑥 𝑡 =
𝑣 𝑡 𝑑𝑡
𝑥 𝑡 =
2𝑡 + 1 𝑑𝑡
𝑡2
𝑥 𝑡 =2
+𝑡+𝑐
2
10 = 22 + 2 + 𝑐
4=𝑐
𝑥 𝑡 = 𝑡2 + 𝑡 + 4
Initial Condition Problems: EX 7.
B) Initial Condition Problems:
EX 7:
If
no
Initial Conditions are given:
Find
𝑓 ′′
𝑥 =
f ( x)
1𝑑𝑥
𝑓 ′′ 𝑥 = 𝑥 + 𝑐
𝑓′ 𝑥 =
𝑥 + 𝑐 𝑑𝑥
2
𝑥
𝑓′ 𝑥 =
+ 𝑐1 𝑥 + 𝑐2
2
if
f
///
𝑓 𝑥 =
( x)  1
𝑥2
+ 𝑐1 𝑥 + 𝑐2
2
1 𝑥3
𝑥2
𝑓 𝑥 =
+ 𝑐1
+ 𝑐2 𝑥 + 𝑐3
2 3
2
Last Update:
• 12/17/10
• Assignment
– Xerox