Lecture 2
Warm-Up Problems
Suppose thatf(x) =1-5u3du.Findf0(x).Find the following indefinite integrals(Hint: simplify using the distributive property)1.Zsec2(x)(1 + 2 cos2(x))dx2.Z(x2+ 3)22x dx3.Z(x2+ 3)1002x dx
Z
20
x
p
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
.
