Final Exam Review (Integration)

Review for Final Exam
Integration

Math 1a

January 13, 2008

Announcements
Oﬃce hours on the website (click “Exams”)
Email your TF, CA, or me with questions
Final: Thursday 9:15am in Hall B

Outline
The Riemann Integral
Estimating the integral
Properties of the integral
Comparison Properties of the Integral
The Fundamental Theorem of Calculus
Statement
Differentiation of functions defined by integrals
Properties of the area function
The Second FTC
Examples
Total Change
Indefinite Integrals
Integration by Substitution
Substitution for Indefinite Integrals
Substitution for Definite Integrals

The Riemann Integral
Learning Objectives

Compute the definite integral using a limit of Riemann sums
Estimate the definite integral using a Riemann sum (e.g.,
Midpoint Rule)
Reason with the definite integral using its elementary
properties.

The Area Problem

Given a function f deﬁned on [a, b], how can one ﬁnd the area
between y = 0, y = f (x), x = a, and x = b?
We divide and conquer.

Forming Riemann sums

We have many choices of how to approximate the area:

Ln = f (x0 )∆x + f (x1 )∆x + · · · + f (xn−1 )∆x
Rn = f (x1 )∆x + f (x2 )∆x + · · · + f (xn )∆x
x0 + x1 x1 + x2 xn−1 + xn
∆x + · · · + f
Mn = f ∆x + f ∆x
2 2 2

Forming Riemann sums

We have many choices of how to approximate the area:

Ln = f (x0 )∆x + f (x1 )∆x + · · · + f (xn−1 )∆x
Rn = f (x1 )∆x + f (x2 )∆x + · · · + f (xn )∆x
x0 + x1 x1 + x2 xn−1 + xn
∆x + · · · + f
Mn = f ∆x + f ∆x
2 2 2

In general, choose ci to be a point in the ith interval [xi−1 , xi ].
Form the Riemann sum
Sn = f (c1 )∆x + f (c2 )∆x + · · · + f (cn )∆x
n
= f (ci )∆x
i=1

Theorem

Theorem
If f is a continuous function on [a, b] or has ﬁnitely many jump
discontinuities, then

lim Sn = lim {f (c1 )∆x + f (c2 )∆x + · · · + f (cn )∆x}
n→∞ n→∞

exists and is the same value no matter what choice of ci we made.

Theorem

Theorem
If f is a continuous function on [a, b] or has finitely many jump
discontinuities, then

lim Sn = lim {f (c1 )∆x + f (c2 )∆x + · · · + f (cn )∆x}
n→∞ n→∞

exists and is the same value no matter what choice of ci we made.

Definition
The definite integral of f from a to b is the number
n
b
f (x) dx = lim f (ci ) ∆x
∆x→0
a i=1

Example (“Sample Exam”, Problem 6)
The rate at which the world’s oil is being consumed is increasing.
Suppose that the rate (measured in billions of barrels per year) is
given by the function r (t), where t is measured in years and t = 0
represents January 1, 2000.
(a) Write a definite integral that represents the total quantity of
oil used between the start of 2000 and the start of 2005.
(b) Suppose that r (t) = 32e 0.05t . Find the approximate value for
the definite integral from part (a) using a right-hand sum with
n = 5 subintervals.
(c) Interpret each of the five terms in the sum from part (b) in
terms of oil consumption.

Answers

5
(a) r (t) dt
0
(b)

1·32e 0.05(1) +1·32e 0.05(2) +1·32e 0.05(3) +1·32e 0.05(4) +1·32e 0.05(5)

(c) Each term stands for the approximate amount of oil used in
each year. For instance, the term 1 · 32e 0.05(3) is approximately
the amount of oil used between January 1, 2002 and January
1, 2003.

Example
1
4
Estimate dx using the midpoint rule and four divisions.
1 + x2
0

Example
1
4
1 + x2
0

Solution
1 1 3
The partition is 0 < < < < 1, so the estimate is
4 2 4
1 4 4 4 4
M4 = + + +
2 2 2 1 + (7/8)2
4 1 + (1/8) 1 + (3/8) 1 + (5/8)

Example
1
4
1 + x2
0

Solution
1 1 3
4 2 4
1 4 4 4 4
M4 = + + +
2 2 2 1 + (7/8)2
4 1 + (1/8) 1 + (3/8) 1 + (5/8)
1 4 4 4 4
= + + +
4 65/64 73/64 89/64 113/64

Example
1
4
1 + x2
0

Solution
1 1 3
4 2 4
1 4 4 4 4
M4 = + + +
2 2 2 1 + (7/8)2
4 1 + (1/8) 1 + (3/8) 1 + (5/8)
1 4 4 4 4
= + + +
4 65/64 73/64 89/64 113/64
150, 166, 784
≈ 3.1468
=
47, 720, 465


Theorem (Additive Properties of the Integral)
Let f and g be integrable functions on [a, b] and c a constant.
Then
b
c dx = c(b − a)
1.
a


Then
b
c dx = c(b − a)
1.
a
b b b
2. [f (x) + g (x)] dx = f (x) dx + g (x) dx.
a a a


Then
b
c dx = c(b − a)
1.
a
b b b
2. [f (x) + g (x)] dx = f (x) dx + g (x) dx.
a a a
b b
3. cf (x) dx = c f (x) dx.
a a


Then
b
c dx = c(b − a)
1.
a
b b b
2. [f (x) + g (x)] dx = f (x) dx + g (x) dx.
a a a
b b
3. cf (x) dx = c f (x) dx.
a a
b b b
[f (x) − g (x)] dx = f (x) dx −
4. g (x) dx.
a a a

More Properties of the Integral

Conventions:
a b
f (x) dx = − f (x) dx
b a


Conventions:
a b
f (x) dx = − f (x) dx
b a
a
f (x) dx = 0
a


Conventions:
a b
f (x) dx = − f (x) dx
b a
a
f (x) dx = 0
a
This allows us to have
c b c
5. f (x) dx = f (x) dx + f (x) dx for all a, b, and c.
a a b

Example
Suppose f and g are functions with
4
f (x) dx = 4
0
5
f (x) dx = 7
0
5
g (x) dx = 3.
0
Find
5
[2f (x) − g (x)] dx
(a)
0
5
(b) f (x) dx.
4

Solution
We have
(a)
5 5 5
[2f (x) − g (x)] dx = 2 f (x) dx − g (x) dx
0 0 0
= 2 · 7 − 3 = 11

Solution
We have
(a)
5 5 5
[2f (x) − g (x)] dx = 2 f (x) dx − g (x) dx
0 0 0
= 2 · 7 − 3 = 11

(b)
5 5 4
f (x) dx −
f (x) dx = f (x) dx
4 0 0
=7−4=3

Theorem
Let f and g be integrable functions on [a, b].

Theorem
6. If f (x) ≥ 0 for all x in [a, b], then
b
f (x) dx ≥ 0
a

Theorem
b
f (x) dx ≥ 0
a

7. If f (x) ≥ g (x) for all x in [a, b], then
b b
f (x) dx ≥ g (x) dx
a a

Theorem
b
f (x) dx ≥ 0
a

7. If f (x) ≥ g (x) for all x in [a, b], then
b b
f (x) dx ≥ g (x) dx
a a

8. If m ≤ f (x) ≤ M for all x in [a, b], then
b
m(b − a) ≤ f (x) dx ≤ M(b − a)
a

Example
4
1
Estimate dx using the comparison properties.
x + sin2 πx
1

The Fundamental Theorem of Calculus
Learning Objectives

State and use both fundamental theorems of calculus
Understand the relationship between integration and
antidiﬀerentiation
Use FTC to compute derivatives of integrals with functions in
the limits
Use FTC to compute areas or other accumulations

Theorem (The First Fundamental Theorem of Calculus)
Let f be an integrable function on [a, b] and deﬁne
x
g (x) = f (t) dt.
a

If f is continuous at x in (a, b), then g is diﬀerentiable at x and

g (x) = f (x).

Example (Spring 2000 Final, Problem 7c)
100
dy
p 2 − p dp
Find if y =
dx x 3 +x

Example (Spring 2000 Final, Problem 7c)
100
dy
p 2 − p dp
Find if y =
dx x 3 +x

Solution u
p 2 − p dp. By the Fundamental Theorem of
Let A(u) =
1
u 2 − u. We have
Calculus, A (u) =
100
d
p 2 − p dp
y =
dx x 3 +x
x 3 +x
100
d
p2 p 2 − p dp
− p dp −
=
dx 1 1
d
A(100) − A(x 3 + x)
=
dx
= −A (x 3 + x) · (3x 2 + 1)
= −(3x 2 + 1) (x 3 + x)2 − (x 3 + x).

Facts about g from f

Let f be the function whose graph is given below.
Suppose the the position at time t seconds of a particle moving
t
along a coordinate axis is s(t) = f (x) dx meters. Use the
0
graph to answer the following questions.
4
3 •
(3,3)
2 • •
(2,2) (5,2)
1 •
(1,1)

1 2 3 4 5 6 7 8 9


t
0
4
What is the particle’s velocity
3 •
(3,3) at time t = 5?
2 • •
(2,2) (5,2)
1 •
(1,1)

1 2 3 4 5 6 7 8 9


t
0
4
What is the particle’s velocity
3 •
(3,3) at time t = 5?
2 • •
(2,2) (5,2) Solution
1 •
(1,1) Recall that by the FTC we
have
1 2 3 4 5 6 7 8 9
s (t) = f (t).

So s (5) = f (5) = 2.


t
0
4
Is the acceleration of the par-
3 •
(3,3) ticle at time t = 5 positive or
2 negative?
• •
(2,2) (5,2)
1 •
(1,1)

1 2 3 4 5 6 7 8 9


t
0
4
Is the acceleration of the par-
3 •
(3,3) ticle at time t = 5 positive or
2 negative?
• •
(2,2) (5,2)
1 •
(1,1) Solution
We have s (5) = f (5), which
1 2 3 4 5 6 7 8 9
looks negative from the
graph.


t
0
4
What is the particle’s position
3 •
(3,3) at time t = 3?
2 • •
(2,2) (5,2)
1 •
(1,1)

1 2 3 4 5 6 7 8 9


t
0
4
What is the particle’s position
3 •
(3,3) at time t = 3?
2 • •
1 •
(1,1) Since on [0, 3], f (x) = x, we
have
1 2 3 4 5 6 7 8 9
3
9
s(3) = x dx = .
2
0


t
0
4
At what time during the ﬁrst 9
3 •
(3,3) seconds does s have its largest
2 value?
• •
(2,2) (5,2)
1 •
(1,1)

1 2 3 4 5 6 7 8 9


t
0
4
3 •
2 value?
• •
(2,2) (5,2)
1 •
(1,1) Solution
1 2 3 4 5 6 7 8 9


t
0
4
3 •
2 value?
• •
(2,2) (5,2)
1 •
(1,1) Solution
The critical points of s are
1 2 3 4 5 6 7 8 9
the zeros of s = f .


t
0
4
3 •
2 value?
• •
(2,2) (5,2)
1 •
(1,1) Solution
By looking at the graph, we
1 2 3 4 5 6 7 8 9
see that f is positive from
t = 0 to t = 6, then negative
from t = 6 to t = 9.


t
0
4
3 •
2 value?
• •
(2,2) (5,2)
1 •
(1,1) Solution
Therefore s is increasing on
1 2 3 4 5 6 7 8 9
[0, 6], then decreasing on
[6, 9]. So its largest value is
at t = 6.


t
0
4
Approximately when is the ac-
3 •
(3,3) celeration zero?
2 • •
(2,2) (5,2)
1 •
(1,1)

1 2 3 4 5 6 7 8 9


t
0
4
Approximately when is the ac-
3 •
(3,3) celeration zero?
2 • •
1 •
(1,1) s = 0 when f = 0, which
happens at t = 4 and t = 7.5
1 2 3 4 5 6 7 8 9
(approximately)


t
0
4
When is the particle moving
3 •
(3,3) toward the origin? Away from
2 the origin?
• •
(2,2) (5,2)
1 •
(1,1)

1 2 3 4 5 6 7 8 9


t
0
4
3 •
2 the origin?
• •
(2,2) (5,2)
1 •
(1,1) Solution
The particle is moving away
1 2 3 4 5 6 7 8 9
from the origin when s > 0
and s > 0.


t
0
4
3 •
2 the origin?
• •
(2,2) (5,2)
1 •
(1,1) Solution
Since s(0) = 0 and s > 0 on
1 2 3 4 5 6 7 8 9
(0, 6), we know the particle is
moving away from the origin
then.


t
0
4
3 •
2 the origin?
• •
(2,2) (5,2)
1 •
(1,1) Solution
After t = 6, s < 0, so the
1 2 3 4 5 6 7 8 9
particle is moving toward the
origin.


t
0
4
On which side (positive or neg-
3 •
(3,3) ative) of the origin does the
2 • •
particle lie at time t = 9?
(2,2) (5,2)
1 •
(1,1)

1 2 3 4 5 6 7 8 9


t
0
4
3 •
2 • •
(2,2) (5,2)
1 •
(1,1) Solution
We have s(9) =
1 2 3 4 5 6 7 8 9 6 9
f (x) dx + f (x) dx,
0 6
where the left integral is
positive and the right integral
is negative.


t
0
4
3 •
2 • •
(2,2) (5,2)
1 •
(1,1) Solution
In order to decide whether
1 2 3 4 5 6 7 8 9
s(9) is positive or negative,
we need to decide if the ﬁrst
area is more positive than the
second area is negative.


t
0
4
3 •
2 • •
(2,2) (5,2)
1 •
(1,1) Solution
This appears to be the case,
1 2 3 4 5 6 7 8 9
so s(9) is positive.

Theorem (The Second Fundamental Theorem of Calculus)
Suppose f is integrable on [a, b] and f = F for another function f ,
then
b
f (x) dx = F (b) − F (a).
a

Examples

Find the following integrals:
1 1 2 2
1
x 2 dx, x 3 dx, x n dx (n = −1), dx
x
0 0 1 1
π 1
e x dx
sin θ dθ,
0 0

The Integral as Total Change

Another way to state this theorem is:
b
F (x) dx = F (b) − F (a),
a

or the integral of a derivative along an interval is the total change
between the sides of that interval. This has many ramiﬁcations:


b
F (x) dx = F (b) − F (a),
a


Theorem
If v (t) represents the velocity of a particle moving rectilinearly,
then
t1
v (t) dt = s(t1 ) − s(t0 ).
t0


b
F (x) dx = F (b) − F (a),
a


Theorem
If MC (x) represents the marginal cost of making x units of a
product, then
x
C (x) = C (0) + MC (q) dq.
0


b
F (x) dx = F (b) − F (a),
a


Theorem
If ρ(x) represents the density of a thin rod at a distance of x from
its end, then the mass of the rod up to x is
x
m(x) = ρ(s) ds.
0

A new notation for antiderivatives

To emphasize the relationship between antidiﬀerentiation and
integration, we use the indeﬁnite integral notation

f (x) dx

for any function whose derivative is f (x).

A new notation for antiderivatives

To emphasize the relationship between antidiﬀerentiation and
integration, we use the indeﬁnite integral notation

f (x) dx

for any function whose derivative is f (x). Thus

x 2 dx = 3 x 3 + C .
1

My ﬁrst table of integrals

[f (x) + g (x)] dx = f (x) dx + g (x) dx

x n+1
x n dx = cf (x) dx = c f (x) dx
+ C (n = −1)
n+1
1
e x dx = e x + C dx = ln |x| + C
x
ax
ax dx = +C
sin x dx = − cos x + C
ln a
csc2 x dx = − cot x + C
cos x dx = sin x + C

sec2 x dx = tan x + C csc x cot x dx = − csc x + C
1
√ dx = arcsin x + C
sec x tan x dx = sec x + C
1 − x2
1
dx = arctan x + C
1 + x2

Integration by Substitution
Learning Objectives

Given an integral and a specific substitution, perform that
substitution
Use the substitution method to evaluate definite and
indefinite integrals

Theorem (The Substitution Rule)
If u = g (x) is a diﬀerentiable function whose range is an interval I
and f is continuous on I , then

f (g (x))g (x) dx = f (u) du

or
du
f (u) dx = f (u) du
dx

This is the “anti” version of the chain rule.

Example
2
xe x dx
Find

Example
2
xe x dx
Find

Solution
Let u = x 2 . Then du = 2x dx and x dx = 1 du. So
2

2
xe x dx = e u du
1
2

= 1 eu + C
2
2
= 1 ex + C
2

Theorem (The Substitution Rule for Deﬁnite Integrals)
If g is continuous and f is continuous on the range of u = g (x),
then
b g (b)
f (g (x))g (x) dx = f (u) du.
a g (a)

Example
π
cos2 x sin x dx.
Compute
0

Example
π
cos2 x sin x dx.
Compute
0

Solution (Slow Way)
cos2 x sin x dx and then
First compute the indeﬁnite integral
evaluate.

Example
π
cos2 x sin x dx.
Compute
0

Solution (Slow Way)
cos2 x sin x dx and then
First compute the indeﬁnite integral
evaluate. Let u = cos x. Then du = − sin x dx and

cos2 x sin x dx = − u 2 du

= − 1 u 3 + C = − 1 cos3 x + C .
3 3

Therefore π
π
cos2 x sin x dx = − 1 cos3 x = 2.
3 3
0
0

Solution (Fast Way)
Do both the substitution and the evaluation at the same time.

Solution (Fast Way)
Do both the substitution and the evaluation at the same time. Let
u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1.

Solution (Fast Way)
Do both the substitution and the evaluation at the same time. Let
u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1. So
−1
π
cos2 x sin x dx = −u 2 du
0 1
1
u 2 du
=
−1
2
1
= 3 u3
1
=.
−1 3

Final Exam Review (Integration)

Recommended

Recommended

More Related Content

What's hot

What's hot (18)

Viewers also liked

Viewers also liked (20)

Similar to Final Exam Review (Integration)

Similar to Final Exam Review (Integration) (20)

More from Matthew Leingang

More from Matthew Leingang (20)

Recently uploaded

Recently uploaded (20)

Final Exam Review (Integration)