Chapter 5: Integration
MAT186H1F Lec0107 Burbulla
Chapter 5 Lecture Notes
Five Lectures
Fall 2015
Chapter 5 Lecture Notes Five Lectures
MAT186H1F Lec0107 Burbulla
Chapter 5: Integration
Chapter 5: Integration
5.1 Approximating Areas Under Curves
5.2 The Definite Integral
5.3 The Fundamental Theorem of Calculus
5.4 Working With Integrals
5.5 Substitution Rule
Chapter 5 Lecture Notes Five Lectures
MAT186H1F Lec0107 Burbulla
Chapter 5: Integration
5.1
5.2
5.3
5.4
5.5
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
The Area Problem
The tangent line problem, see Section 2.1, is one of the geometric
problems that historically motivated calculus. The other problem is
the area problem:
y
y = f (x)
a
b x
Chapter 5 Lecture Notes Five Lectures
Chapter 5: Integration
What is the area of the region
under the graph of y = f (x),
above the x-axis, between
x = a and x = b? We shall
only assume that the area of a
rectangle is known: A = l ×w .
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Geometric Solution to the Area Problem
One way to approximate the area of the region under a curve is to
use lots of rectangles to approximate it. This is illustrated in the
figures below, in which first 10 rectangles and then 50 rectangles
are used to approximate the area under y = 6 + 3 sin x + 4 cos x on
the interval [0, 2π].
Chapter 5 Lecture Notes Five Lectures
MAT186H1F Lec0107 Burbulla
Chapter 5: Integration
5.1
5.2
5.3
5.4
5.5
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
What’s Calculus Got To Do With It?
The two previous figures suggest that more rectangles result in a
better approximation. We shall take this process to the limit, by
letting the number of rectangles used go to infinity. First we shall
execute this approach on a simple example, for which we know the
answer, to see if it works. The example will be y = x on [0, 10], for
which the region is a triangle with A = 50. The figures below show
approximations with 10 and 50 rectangles, respectively.
Chapter 5 Lecture Notes Five Lectures
Chapter 5: Integration
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Setting Things Up
Partition the interval [0, 10] into n subintervals, each of length
∆x = 10/n, by using the points
P = {x0 , x1 , x2 , . . . , xn }
with
xi = 0 + i∆x, for i = 0, 1, 2, 3, . . . n.
0
r
x0
r
x1
r
∆x
xi−1
r
xi
Chapter 5 Lecture Notes Five Lectures
r
xn−1
MAT186H1F Lec0107 Burbulla
10
r
xn
Chapter 5: Integration
5.1
5.2
5.3
5.4
5.5
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Determining the Height of the Rectangles
Figure: Using the right endpoint
of the subinterval to determine
height of the rectangle.
Chapter 5 Lecture Notes Five Lectures
Chapter 5: Integration
In each subinterval [xi−1 , xi ]
we could pick any point we
want to determine the corresponding y value that will
be the height of the rectangle. For this example we shall
pick the right endpoint of each
subinterval, so that the height
of the i-th rectangle will be xi ;
its area will be Ai = xi · ∆x.
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Adding Up the Areas of All The Rectangles
Let A be the area of the triangle determined by y = x on [0, 10].
A ' A1 + A2 + · · · + An =
n
X
Ai
i=1
n
X
n
X
10 10
=
xi · ∆x =
i
·
n n
i=1
i=1
2 X
n
10
100 n(n + 1)
=
i= 2 ·
n
n
2
i=1
100 (n2 + n)
1
⇒ A = lim 2 ·
= 50 lim 1 +
= 50
n→∞ n
n→∞
2
n
Chapter 5 Lecture Notes Five Lectures
MAT186H1F Lec0107 Burbulla
Chapter 5: Integration
5.1
5.2
5.3
5.4
5.5
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Example 1: More Calculus?
Now consider the region under y = x on the interval [a, b] with
area A.
We can calculate A as the difference of two triangles:
1
1
A = b 2 − a2 .
2
2
d 1 2
Hmmm . . .
x
= x;
dx 2
coincidence or connection?
A
a
b
Chapter 5 Lecture Notes Five Lectures
Chapter 5: Integration
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Example 2: y = x 2 on [0, 5]
0
r
x0
r
x1
r
∆x
xi−1
r
xi
r
xn−1
5
Take ∆x = 5/n; xi = i∆x, and again use the right endpoint of
each subinterval to determine the height of the approximating
rectangles: Ai = xi2 ∆x. Then
n
X
n 2
X
5
5
A = lim
Ai = lim
i
·
n→∞
n→∞
n
n
i=1
i=1
n
X
25
5
= lim
i2
·
n→∞
n2
n
i=1
Chapter 5 Lecture Notes Five Lectures
MAT186H1F Lec0107 Burbulla
r
xn
5.1
5.2
5.3
5.4
5.5
Chapter 5: Integration
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Some Summation Formulas
To finish these kinds of calculations you need to know formulas for
the summations that are involved. For instance, to get the area
under the line y = x we used the following formula for the sum of
the first n integers:
n
X
n(n + 1)
.
i=
2
i=1
To finish the calculation in Example 2 we need to know the sum of
the first n squares:
n
X
i2 =
i=1
n(n + 1)(2n + 1)
.
6
Chapter 5 Lecture Notes Five Lectures
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Chapter 5: Integration
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Example 2, Concluded
A =
n
53 X 2
i
n3
lim
n→∞
!
i=1
53 n(n + 1)(2n + 1)
= lim
·
n→∞ n3
6
53
n(n + 1)(2n + 1)
=
lim
3
6 n→∞
n
53
d x3
=
; again:
= x2 . . .
3
dx 3
Chapter 5 Lecture Notes Five Lectures
MAT186H1F Lec0107 Burbulla
Chapter 5: Integration
5.1
5.2
5.3
5.4
5.5
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Reviewing What We Have Done So Far To Find Area
To find the area under y = f (x) on the interval [a, b], partition the
interval [a, b] into n subintervals, each of length ∆x = (b − a)/n,
by using the points P = {x0 , x1 , x2 , . . . , xn } with
xi = a + i∆x = a + i
a
r
r
x0
x1
r
∆x
xi−1
(b − a)
, for i = 0, 1, 2, 3, . . . n.
n
r
r
xi
Chapter 5 Lecture Notes Five Lectures
Chapter 5: Integration
xn−1
b
r
xn
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Determining the Height of the Rectangles
y
y = f (x)
r
r
xi−1
r r
xi∗ xi
In each subinterval [xi−1 , xi ] we could pick
any point we want to determine the corresponding y value that will be the height of
the rectangle. In general, call xi∗ an arbitrary
point in [xi−1 , xi ]; then the area of the i-th
rectangle is Ai = f (xi∗ ) · ∆x. Usually, I will
pick the right endpoint of each subinterval,
so that the height of the i-th rectangle will
be f (xi ); and its area will be Ai = f (xi )·∆x.
Chapter 5 Lecture Notes Five Lectures
MAT186H1F Lec0107 Burbulla
Chapter 5: Integration
5.1
5.2
5.3
5.4
5.5
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Approximating Areas and Riemann Sums
Let xi∗ be any point in [xi−1 , xi ]. If f (x) ≥ 0 for x ∈ [a, b] and A is
the area of the region under the curve y = f (x), on the interval
[a, b], above the x-axis, then for any n,
A ≈ A1 + A2 + · · · + An =
n
X
Ai
i=1
=
n
X
f (xi∗ ) · ∆x
i=1
Any expression of the form
n
X
f (xi∗ ) · ∆x is called a Riemann Sum
i=1
of f on the interval [a, b], even if f is not always positive on [a, b].
Chapter 5 Lecture Notes Five Lectures
Chapter 5: Integration
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Example 3: Approximate Area under y = sin x on [0, π].
To form a Riemann sum we need to pick n and xi∗ . Let n = 4 and
let xi∗ = xi . Then ∆x = π/4, P = {0, π/4, π/2, 3π/4, π}, and
π
π
π
π
+ sin(π/2) · + sin(3π/4) · + sin π ·
4
4
4
4
1 π
π
1 π
= √ · +1· + √ · +0
4
2 4
2 4
√ π
= (1 + 2)
4
≈ 1.8961189
A ≈ sin(π/4) ·
Chapter 5 Lecture Notes Five Lectures
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Chapter 5: Integration
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Taking the Limit
To calculate the actual area A we take the limit as n → ∞. That is,
A = lim
n
X
n→∞
Ai = lim
n
X
n→∞
i=1
f (xi∗ ) · ∆x
i=1
If we choose xi∗ to be the right end point of each subinterval
[xi−1 , xi ], then xi∗ = xi = a + i∆x and
n
X
(b − a)
(b − a)
A = lim
f a+i
·
n→∞
n
n
i=1
In practice, this method of calculating areas is very difficult!
Chapter 5 Lecture Notes Five Lectures
Chapter 5: Integration
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Example 4: y = x 2 on [a, b].
a
r
x0
r
r
x1
∆x
xi−1
r
xi
r
xn−1
b
r
xn
Take ∆x = (b − a)/n; xi = a + i∆x, and again use the right
endpoint of each subinterval to determine the height of the
approximating rectangles: Ai = xi2 ∆x. Then
n
X
n X
(b − a) 2 (b − a)
A = lim
Ai = lim
a+i
·
n→∞
n→∞
n
n
i=1
i=1
!
2
n
X
(b − a)
(b − a)
2
2 (b − a)
= lim
a + 2ia
+i
·
n→∞
n
n
n
i=1
Chapter 5 Lecture Notes Five Lectures
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Chapter 5: Integration
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Summation Formulas
To finish this kind of calculations you need to know formulas for
the summations that are involved. To finish the calculation in
Example 2 we need to know the sum of the first n integers
n
X
i=
i=1
n(n + 1)
,
2
and the sum of the first n squares
n
X
i2 =
i=1
n(n + 1)(2n + 1)
.
6
Chapter 5 Lecture Notes Five Lectures
Chapter 5: Integration
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Example 4, Concluded
A =
a2 n ·
lim
n→∞
a)2
(b − a)
(b −
+ 2a
n
n2
·
n
X
i=1
i+
(b −
n3
=
=
=
n
X
!
i2
i=1
(b − a)2 n(n + 1) (b − a)3 n(n + 1)(2n + 1)
lim a2 b − a3 + 2a
+
·
n→∞
n2
2
n3
6
1
a2 b − a3 + a(b − a)2 + (b − a)3
3
1
a2 b − a3 + ab2 − 2a2 b + a3 + (b 3 − 3ab2 + 3ba2 − a3 )
3 3
1 3 1 3
d x
b − a ;
note:
= x2
3
3
dx 3
=
a)3
Chapter 5 Lecture Notes Five Lectures
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Chapter 5: Integration
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Riemann Sums
We shall generalize the procedure used in the previous examples. A
set of points
P = {x0 , x1 , . . . , xn }
is called a partition of [a, b] if
a = x0 < x1 < x2 < · · · < xn−1 < xn = b.
Let xi∗ be any point in the subinterval [xi−1 , xi ]. Then the
expression
n
X
f (xi∗ ) · (xi − xi−1 )
i=1
is called a general Riemann sum of f on [a, b].
Chapter 5 Lecture Notes Five Lectures
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Chapter 5: Integration
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
The Definite Integral
Let ∆ be size of the biggest subinterval [xi−1 , xi ]; that is,
∆ = maxni=1 (xi − xi−1 ).
∆ is called the norm of the partition P. If the limit
lim
∆→0
n
X
f (xi∗ ) · (xi − xi−1 )
i=1
exists, then it is called the definite integral of f on [a, b], and we
write
Z b
n
X
f (x) dx = lim
f (xi∗ ) · (xi − xi−1 ).
a
∆→0
i=1
We say f is integrable on [a, b].
Chapter 5 Lecture Notes Five Lectures
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Chapter 5: Integration
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Existence of Definite Integral
Z
b
If f is continuous on [a, b], then
f (x) dx always exists,
a
regardless of the choice of partition P, or the choice of the points
xi∗ , as long as ∆ → 0. Most commonly we use a regular partition,
in which
b−a
xi = a + i · ∆x, and ∆x =
.
n
Then
Z b
n
X
f (x) dx = lim
f (xi∗ ) · ∆x.
n→∞
a
i=1
The most common choices for xi∗ are the left or right endpoint, or
the midpoint, of the interval [xi−1 , xi ].
Chapter 5 Lecture Notes Five Lectures
Chapter 5: Integration
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Examples
Based on the calculations we did in Sections 5.1, we have
1.
Z
10
x dx = 50
0
2.
Z
a
b
1
1
x 2 dx = b 3 − a3
3
3
In both these examples we used regular partitions, and picked xi∗ to
be the right endpoint of the interval [xi−1 , xi ].
Chapter 5 Lecture Notes Five Lectures
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Chapter 5: Integration
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Basic Properties of the Definite Integral
b
Z
1.
Z
b
f (x) dx =
f (t) dt
a
a
b
Z
c dx = c · (b − a)
Z b
Z b
3.
cf (x) dx = c
f (x) dx
a
a
Z b
Z b
Z
4.
(f (x) + g (x)) dx =
f (x) dx +
2.
a
a
a
Chapter 5 Lecture Notes Five Lectures
Chapter 5: Integration
b
g (x) dx
a
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
5. If f (x) ≤ g (x) for all x ∈ [a, b], then
Z
b
Z
f (x) dx ≤
a
b
g (x) dx
a
6. If m ≤ f (x) ≤ M for all x ∈ [a, b], then
Z
m · (b − a) ≤
b
f (x) dx ≤ M · (b − a)
a
Rb
7. If f (x) ≥ 0 for all x ∈ [a, b], then a f (x) dx is the
area of the region under the curve y = f (x) on [a, b].
Chapter 5 Lecture Notes Five Lectures
MAT186H1F Lec0107 Burbulla
Chapter 5: Integration
5.1
5.2
5.3
5.4
5.5
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Extension of the Definition
If a = b, we agree to define
Z a
f (x) dx = 0.
a
If a > b, we agree to define
Z
b
a
Z
f (x) dx = −
a
f (x) dx.
b
Then for any c,
Z
b
Z
c
f (x) dx =
a
Z
f (x) dx +
a
Chapter 5 Lecture Notes Five Lectures
Chapter 5: Integration
b
f (x) dx
c
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
How To Calculate Definite Integrals?
All of the above properties of the definite integral follow from its
definition. The question remains:
Z b
Is there a simple way to calculate
f (x) dx?
a
Our previous numerical examples suggest the evaluation of definite
integrals has something to do with antiderivatives of f . In Section
5.3 we shall state and prove the formula for evaluating definite
integrals.
Chapter 5 Lecture Notes Five Lectures
MAT186H1F Lec0107 Burbulla
Chapter 5: Integration
5.1
5.2
5.3
5.4
5.5
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
The Evaluation Theorem
Let f be continuous on [a, b]. If F is an antiderivative of f on
[a, b], then
Z
a
b
f (x) dx = F (b) − F (a) ≡ [F (x)]ba .
Examples:
Z 10
1 2 10 1
1.
x dx =
x
= 100 − 0 = 50
2
2
0
0
3 5
Z 5
53
53
x
2
=
−0=
2.
x dx =
3
3
3
0
0
Chapter 5 Lecture Notes Five Lectures
Chapter 5: Integration
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Proof of the Evaluation Theorem
Apply the Mean Value Theorem to F on the subinterval [xi−1 , xi ] :
there is a number xi∗ ∈ [xi−1 , xi ] such that
F (xi ) − F (xi−1 ) = F 0 (xi∗ ) · (xi − xi−1 ) = f (xi∗ ) · (xi − xi−1 ).
Let ∆xi = xi − xi−1 . The Riemann sum can be calculated:
n
X
f (xi∗ )∆xi
=
n
X
(F (xi ) − F (xi−1 ))
i=1
i=1
= F (x1 ) − F (x0 ) + F (x2 ) − F (x1 ) + · · · + F (xn ) − F (xn−1 )
= F (xn ) − F (x0 ) = F (b) − F (a)
Z
⇒
b
f (x) dx
a
=
lim (F (b) − F (a)) = F (b) − F (a)
∆→0
Chapter 5 Lecture Notes Five Lectures
MAT186H1F Lec0107 Burbulla
Chapter 5: Integration
5.1
5.2
5.3
5.4
5.5
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Example 1
Find the area under y = cos x on the interval [0, π/2].
Z
π/2
A =
cos x dx
0
π/2
= [sin x]0
π = sin
− sin 0
2
= 1
Try finding this with Riemann sums!
Chapter 5 Lecture Notes Five Lectures
Chapter 5: Integration
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Example 2
Find the area under y = e x on the interval [−1, 2].
Z
2
A =
e x dx
=
−1
[e x ]2−1
2
−1
= e −e
e3 − 1
=
e
Chapter 5 Lecture Notes Five Lectures
MAT186H1F Lec0107 Burbulla
Chapter 5: Integration
5.1
5.2
5.3
5.4
5.5
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
The Fundamental Theorem of Calculus
Suppose f is continuous on [a, b]. Then:
1. If F 0 (x) = f (x) for all x ∈ (a, b), then
Z
b
f (x) dx = F (b) − F (a).
a
2. For any x ∈ (a, b),
d
dx
x
Z
f (t) dt
= f (x).
a
Part 1 is the Evaluation Theorem: the integral of a derivative.
Part 2 is a new differentiation rule: the derivative of an integral.
Chapter 5 Lecture Notes Five Lectures
Chapter 5: Integration
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Proof of Part 2
Rx
For x ∈ [a, b], let A(x) = a f (t) dt. A is called an area function.
Let mh and Mh be the min and max of f on [x, x + h],
respectively. Assume h > 0.
mh · h ≤
y
⇒ mh ≤
≤ Mh · h
≤ Mh
As h → 0, both mh , Mh → f (x). By
the squeeze law,
A(x)
a
A(x + h) − A(x)
A(x + h) − A(x)
h
x x +h b t
A(x + h) − A(x)
= f (x).
h→0
h
A0 (x) = lim
Chapter 5 Lecture Notes Five Lectures
MAT186H1F Lec0107 Burbulla
Chapter 5: Integration
5.1
5.2
5.3
5.4
5.5
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Example 3
Since we have already done some examples of Part 1, I’ll do a
couple of examples of Part 2.
!
Z x2 p
q
d
t 4 + t 2 + 1 dt
=
(x 2 )4 + (x 2 )2 + 1 · 2x
dx
3
p
= 2x x 8 + x 4 + 1
Note: the lower limit, a = 3, has nothing do with the answer. Also,
it
√ would be extremely difficult to find an antiderivative of
t 4 + t 2 + 1 with respect to t. The power of Part 2 of the
Fundamental Theorem is that you don’t need the antiderivative.
Chapter 5 Lecture Notes Five Lectures
Chapter 5: Integration
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Example 4
Z
d
dx
10
t2
e dt
=
sin x
d
dx
Z
−
sin x
t2
e dt
10
2
=
−e sin
x
=
− cos x e sin
· cos x
2
x
In general,
d
dx
Z
!
h(x)
f (t) dt
= f (h(x))h0 (x) − f (g (x))g 0 (x).
g (x)
Chapter 5 Lecture Notes Five Lectures
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Chapter 5: Integration
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
n
X
i2
Example 5: Find lim
n→∞
n3
i=1
Solution: this limit can be evaluated by switching to an integral.
But first we have to guess, or figure out, which function on which
interval has such a Riemann sum. If we take a regular partition of
[a, b] = [0, 1] and pick xi∗ = xi , then
n
n
X
X
i2
i2 1
=
n3
n2 n
i=1
=
i=1
n
X
i2
⇒ lim
n→∞
i=1
n
X
i=1
1
Z
=
n3
0
Chapter 5 Lecture Notes Five Lectures
1 3
x 2 dx =
x
3
1
=
0
1
3
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Chapter 5: Integration
Example 6: Find lim
f (xi∗ )∆x, if f (x) = x 2
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
n
X
e 3i/n+2
n→∞
n
i=1
Solution: If we take a regular partition of [a, b] = [0, 1] and pick
xi∗ = xi , then
n
X
e 3i/n+2
⇒ lim
i=1
n
X
n→∞
i=1
n
e 3i/n+2
n
=
n
X
i=1
1
Z
=
0
f (xi∗ )∆x, if f (x) = e 3x+2
1 3x+2
e 3x+2 dx =
e
3
Chapter 5 Lecture Notes Five Lectures
1
=
0
MAT186H1F Lec0107 Burbulla
1 5
e − e2
3
5.1
5.2
5.3
5.4
5.5
Chapter 5: Integration
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Net Signed Area
If f (x) is not always positive for x ∈ [a, b], we can still calculate
the expression
Z b
A=
f (x) dx.
a
The value of A then represents the net signed area; the area above
the x-axis and under the curve y = f (x), minus the area below the
x-axis and above the curve y = f (x).
Chapter 5 Lecture Notes Five Lectures
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Chapter 5: Integration
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Example 7
For y = x on [−2, 1],
1
x2
x dx =
2
−2
Z
1
−2
1 4
3
= − =− .
2 2
2
This is the net signed area, the
area of the triangle above the xaxis minus the area of the triangle
under the x-axis:
Figure: y = x on [−2, 1]
Chapter 5 Lecture Notes Five Lectures
1
1
3
·1·1− ·2·2=− .
2
2
2
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Chapter 5: Integration
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Integrating Even and Odd Functions
Recall:
I
f is even if f (−x) = f (x), for all x.
I
f is odd if f (−x) = −f (x), for all x.
The graph of an even function is symmetric with respect to the
y -axis; the graph of an odd function is symmetric with respect to
the origin. Consequently, for a > 0:
Z a
Z a
I If f is even, then
f (x) dx = 2
f (x) dx.
0
Z −a
a
I If f is odd, then
f (x) dx = 0.
−a
Chapter 5 Lecture Notes Five Lectures
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Chapter 5: Integration
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Example 1
Z
2
4
3
(x − x ) dx
Z
2
−2
x5
4
x dx − 0 = 2
5
= 2
0
π/2
(cos x − sin x) dx
Z
π/2
Z
2
−π/2
Z
= 2
0
Chapter 5 Lecture Notes Five Lectures
=
0
64
;
5
π/2
cos x dx −
=
−π/2
x 3 dx
−2
2
Z
5
2
x dx −
=
−2
Z
Z
4
sin5 x dx
−π/2
π/2
π/2
cos x dx − 0 = 2[sin x]0
MAT186H1F Lec0107 Burbulla
=2
Chapter 5: Integration
5.1
5.2
5.3
5.4
5.5
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Average Value of a Function
For any finite list of numbers, y1 , y2 , . . . , yn , the average value is
simply
y1 + y2 + · · · + y n
y=
.
n
We shall now use the definition of the definite integral to develop
the average value of a function f on [a, b]. Take a regular partition
of [a, b] and let xi∗ be any point in [xi−1 , xi ]. The average value of
the n sample points from the graph of f (x) is
f (x1∗ ) + f (x2∗ ) + · · · + f (xn∗ )
.
n
Chapter 5 Lecture Notes Five Lectures
Chapter 5: Integration
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Since the above partition is regular we have
∆x =
1
b−a
∆x
⇔ =
.
n
n
b−a
n
f (x1∗ ) + f (x2∗ ) + · · · + f (xn∗ )
1 X
So
=
f (xi∗ ) · ∆x and
n
b−a
i=1
lim
n→∞
f (x1∗ ) + f (x2∗ ) + · · · + f (xn∗ )
n
n
=
=
X
1
lim
f (xi∗ ) · ∆x
b − a n→∞
i=1
Z b
1
f (x) dx
b−a a
This is the average value of f on [a, b]; it is denoted by f , or fave .
Chapter 5 Lecture Notes Five Lectures
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Chapter 5: Integration
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Example 2
Find the average value of f (x) = x 2 on [0, 3].
fave =
1
3
Z
3
x 2 dx
0
1 1 3
=
x
3 3
27
=
9
= 3
Chapter 5 Lecture Notes Five Lectures
3
0
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Chapter 5: Integration
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Example 3
Suppose x is the position of a particle at time t, and v its velocity
at time t. Then according to the above definition
1
b−a
Z
b
v dt
a
is the average velocity of the particle for a ≤ t ≤ b. This matches
with what we think of as average velocity since
1
b−a
Z
b
v dt =
a
x(b) − x(a)
,
b−a
because x is an antiderivative of v .
Chapter 5 Lecture Notes Five Lectures
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Chapter 5: Integration
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Mean Value Theorem for Integrals
If f is continuous on the closed interval [a, b], then there
is a number c ∈ [a, b] such that
b
Z
f (x) dx = f (c) · (b − a).
a
Proof: Let m and M be the min and max values of f on [a, b].
Z
b
m ≤ f (x) ≤ M ⇒ m(b − a) ≤
f (x) dx ≤ M(b − a)
a
Z b
1
⇒ m≤
f (x) dx ≤ M
b−a a
Chapter 5 Lecture Notes Five Lectures
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Chapter 5: Integration
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Suppose f (a1 ) = m; f (b1 ) = M. If a1 < b1 , apply the Intermediate
Value Theorem to f on [a1 , b1 ] : There is a number c ∈ [a1 , b1 ]
⊂ [a, b] such that
1
f (c) =
b−a
If b1 < a1 , use [b1 , a1 ]. ¶
Z
1
4
x3
2
x dx =
3
Z
b
f (x) dx.
a
For example:
4
= 21;
1
3f (c) = 21 ⇔ c 2 = 7 ⇒ c =
Chapter 5 Lecture Notes Five Lectures
√
7.
MAT186H1F Lec0107 Burbulla
Chapter 5: Integration
5.1
5.2
5.3
5.4
5.5
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Techniques of Integration
I
I
I
The full story on techniques of integration appears in
Chapter 7. Right now we will just cover the most basic
integration technique, known as substitution, or change of
variable, or transformation.
For this method the old-fashioned differential notation is very
useful, and takes advantage of the integral symbol, which
includes the differential dx.
So if u is a function of x, then we can write
du =
I
du
dx.
dx
Hence the general rule for picking u : look for a function
whose derivative is multiplying dx.
Chapter 5 Lecture Notes Five Lectures
Chapter 5: Integration
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Example 1
Z
3x 2
p
x 3 + 9 dx
( let u = x 3 + 9)
Z p
=
x 3 + 9 (3x 2 dx)
Z
√
=
u du
Z
=
u 1/2 du
=
=
Chapter 5 Lecture Notes Five Lectures
2 3/2
u +C
3
2 3
(x + 9)3/2 + C
3
MAT186H1F Lec0107 Burbulla
Chapter 5: Integration
5.1
5.2
5.3
5.4
5.5
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Example 2
Z
sin3 (2x) cos(2x) dx
=
( let u = sin(2x))
=
=
=
Chapter 5 Lecture Notes Five Lectures
Chapter 5: Integration
Z
1
sin3 (2x) 2 cos(2x) dx
2
Z
1
u 3 du
2
11 4
u +C
24
1 4
sin (2x) + C
8
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Example 3
Z
3
p
x 2 − 1 dx
=
( let u = x 2 − 1)
=
x
=
=
=
Z
p
1
2
x
x 2 − 1 (2x dx)
2
Z
√
1
(u + 1) u du
2
Z
1 3/2
1/2
u +u
du
2
1 2 5/2 2 3/2
u + u
+C
2 5
3
1 2
1
(x − 1)5/2 + (x 2 − 1)3/2 + C
5
3
Chapter 5 Lecture Notes Five Lectures
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Chapter 5: Integration
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Example 4
When you make a substitution in a definite integral you should also
change the limits of integration.
3
Z
p
3x 2 x 3 + 9 dx
3p
Z
x 3 + 9 (3x 2 dx)
=
0
0
( let u = x 3 + 9)
Z
=
=
=
Chapter 5 Lecture Notes Five Lectures
Chapter 5: Integration
36 √
2 3/2
u du =
u
3
9
2 3/2 2 3/2
36 − 9
3
3
2
(216 − 27) = 126
3
36
9
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Example 5
Z
π/4
sin3 (2x) cos(2x) dx
=
( let u = sin(2x))
=
0
=
=
=
Chapter 5 Lecture Notes Five Lectures
Z
1 π/4 3
sin (2x) 2 cos(2x) dx
2 0
Z
1 1 3
u du
2 0
1 1 4 1
u
2 4
0
1 1
−0
2 4
1
8
MAT186H1F Lec0107 Burbulla
Chapter 5: Integration
5.1
5.2
5.3
5.4
5.5
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
Examples 2 and 5: Before and After. u = sin(2x)
Figure: A =
R π/4
0
sin3 (2x) cos(2x) dx
Chapter 5 Lecture Notes Five Lectures
Chapter 5: Integration
Figure: A =
1
2
R1
0
u 3 du
MAT186H1F Lec0107 Burbulla
5.1
5.2
5.3
5.4
5.5
Approximating Areas Under Curves
The Definite Integral
The Fundamental Theorem of Calculus
Working With Integrals
Substitution Rule
General Formula for Change of Variables
Suppose u = f (x); then du = f 0 (x) dx and
b
Z
0
Z
f (b)
g (f (x)) f (x) dx =
a
g (u) du.
f (a)
Proof: Suppose G is an antiderivative of g . Then G ◦ f (x) is an
antiderivative of g (f (x)) f 0 (x), by the Chain Rule, so
Z
b
g (f (x)) f 0 (x) dx
a
Z
= [G (f (x))]ba = G (f (b)) − G (f (a))
f (b)
and
f (a)
f (b)
g (u) du = [G (u)]f (a) = G (f (b)) − G (f (a))
Chapter 5 Lecture Notes Five Lectures
MAT186H1F Lec0107 Burbulla