Lecture2-slides-substitution.pdf - Lecture 2 Warm-Up...

Lecture 2
Warm-Up Problems
Suppose that f ( x ) =
Z
20
x
p
1 - 5 u
3
du .
Find f
0
( x ).
Find the following indefinite integrals
(Hint: simplify using the distributive property)
1.
Z
sec
2
( x )(1 + 2 cos
2
( x )) dx
2.
Z
( x
2
+ 3)
2
2 x dx
3.
Z
( x
2
+ 3)
100
2 x dx
Announcements:
I Quiz 1 on Tuesday, Sept 11: FTC and Substitution
I Homework 0 (optional) and Homework 1 are posted.
HW0 is due on Sept 12, HW1 is due on Sept 13, both at 8:45am.

Recall from last class
Theorem (The Fundamental Theorem of Calculus)
Suppose that f ( x ) is continuous on [ a , b ] . Then,
1. if g ( x ) =
R
x
a
f ( t ) dt for all x ∈ [ a , b ] , then g
0
( x ) = f ( x ) . In other words,
d
dx
Z
x
a
f ( t ) dt = f ( x ) .
2. Suppose F ( x ) is an antiderivative of f ( x ) (i.e. F
0
( x ) = f ( x ) ), then
Z
b
a
f ( x ) dx = F ( b ) - F ( a ) .

Recall from last class
Theorem (The Fundamental Theorem of Calculus)
Suppose that f ( x ) is continuous on [ a , b ] . Then,
1. if g ( x ) =
R
x
a
f ( t ) dt for all x ∈ [ a , b ] , then g
0
( x ) = f ( x ) . In other words,
d
dx
Z
x
a
f ( t ) dt = f ( x ) .
2. Suppose F ( x ) is an antiderivative of f ( x ) (i.e. F
0
( x ) = f ( x ) ), then
Z
b
a
f ( x ) dx = F ( b ) - F ( a ) .
“Integrals undo derivatives”

Recall from last class
Theorem (The Fundamental Theorem of Calculus)
Suppose that f ( x ) is continuous on [ a , b ] . Then,
1. if g ( x ) =
R
x
a
f ( t ) dt for all x ∈ [ a , b ] , then g
0
( x ) = f ( x ) . In other words,
d
dx
Z
x
a
f ( t ) dt = f ( x ) .
2. Suppose F ( x ) is an antiderivative of f ( x ) (i.e. F
0
( x ) = f ( x ) ), then
Z
b
a
f ( x ) dx = F ( b ) - F ( a ) .
“Integrals undo derivatives”
Properties of Definite Integrals
1.
Z
b
a
f ( x ) dx = -
Z
a
b
f ( x ) dx

Recall from last class
Theorem (The Fundamental Theorem of Calculus)
Suppose that f ( x ) is continuous on [ a , b ] . Then,
1. if g ( x ) =
R
x
a
f ( t ) dt for all x ∈ [ a , b ] , then g
0
( x ) = f ( x ) . In other words,
d
dx
Z
x
a
f ( t ) dt = f ( x ) .
2. Suppose F ( x ) is an antiderivative of f ( x ) (i.e. F
0
( x ) = f ( x ) ), then
Z
b
a
f ( x ) dx = F ( b ) - F ( a ) .
“Integrals undo derivatives”
Properties of Definite Integrals
1.
Z
b
a
f ( x ) dx = -
Z
a
b
f ( x ) dx
2.
Z
b
a
f ( x ) dx =
Z
c
a
f ( x ) dx +
Z
b
c
f ( x ) dx

Example
Worksheet 1, Problem 3b:
Find the derivative of h ( z ) =
Z
2
z
tan
1
u 2 + 1
du

Example
Worksheet 1, Problem 3b:
Find the derivative of h ( z ) =
Z
2
z
tan
1
u 2 + 1
du
h
0
( z ) =
d
dz
Z
2
z
tan
1
u 2 + 1
du
=
d
dz
-
Z
z
2
tan
1
u 2 + 1
du
= - tan
1
z
2
+ 1

FTC (part 1) + Chain Rule
Worksheet 1, Problem 3c:
Find the derivative of k ( x ) =
Z
x
- 2 x
sec
2
( s ) ds

FTC (part 1) + Chain Rule
Worksheet 1, Problem 3c:
Find the derivative of k ( x ) =
Z
x
- 2 x
sec
2
( s ) ds
k ( x ) =
Z
0
- 2 x
sec
2
( s ) ds +
Z
x
0
sec
2
( s ) ds

FTC (part 1) + Chain Rule
Worksheet 1, Problem 3c:
Find the derivative of k ( x ) =
Z
x
- 2 x
sec
2
( s ) ds
k ( x ) =
Z
0
- 2 x
sec
2
( s ) ds +
Z
x
0
sec
2
( s ) ds
= -
Z
- 2 x
0
sec
2
( s ) ds +
Z
x
0
sec
2
( s ) ds

FTC (part 1) + Chain Rule
Worksheet 1, Problem 3c:
Find the derivative of k ( x ) =
Z
x
- 2 x
sec
2
( s ) ds
k ( x ) =
Z
0
- 2 x
sec
2
( s ) ds +
Z
x
0
sec
2
( s ) ds
= -
Z
- 2 x
0
sec
2
( s ) ds +
Z
x
0
sec
2
( s ) ds
k
0
( x ) = - sec
2
( - 2 x )
d
dx
[ - 2 x ] + sec
2
( x )

FTC (part 1) + Chain Rule
Worksheet 1, Problem 3c:
Find the derivative of k ( x ) =
Z
x
- 2 x
sec
2
( s ) ds
k ( x ) =
Z
0
- 2 x
sec
2
( s ) ds +
Z
x
0
sec
2
( s ) ds
= -
Z
- 2 x
0
sec
2
( s ) ds +
Z
x
0
sec
2
( s ) ds
k
0
( x ) = - sec
2
( - 2 x )
d
dx
[ - 2 x ] + sec
2
( x )
= 2 sec
2
( - 2 x ) + sec
2
( x )

FTC (part 1) + Chain Rule
d
dx
"
Z
h ( x )
a
f ( t ) dt
#
= f ( h ( x ) ) h
0
( x )

FTC (part 1) + Chain Rule
Example:
Find the derivative of ‘ ( θ ) =
Z
sin( θ )
2
1
√
t 3 - t
dt .