Download a = x

Chapter 5
Integration
1
5.1
Estimating with Finite Sums
2
Riemann Sums
Approximating area bounded by the graph
between [a,b]
3
Area is approximately given by
f(c1)Dx1 + f(c2)Dx2+ f(c3)Dx3+ … + f(cn)Dxn





Partition of [a,b] is the set of
P = {x0, x1, x2, … xn-1, xn}
a = x0< x1< x2 …< xn-1 < xn=b
cn[xn-1, xn]
||P|| = norm of P = the largest
of all subinterval width
4
Riemann sum for f on [a,b]
Rn = f(c1)Dx1 + f(c2)Dx2+
f(c3)Dx3+ … +f(cn)Dxn
5
Figure 5.4




Let the true value of the
area is R
Two approximations to R:
cn= xn corresponds to case
(a). This under estimates
the true value of the area
R if n is finite.
cn= xn-1 corresponds to
case (b). This over
estimates the true value of
the area S if n is finite.
go back
6
Limits of finite sums


Example 5 The limit of finite approximation to
an area
Find the limiting value of lower sum
approximation to the area of the region R
below the graphs f(x) = 1 - x2 on the interval
[0,1] based on Figure 5.4(a)
7
Solution
Dxk = (1 - 0)/n= 1/n ≡Dx; k = 1,2,…n
 Partition on the x-axis: [0,1/n], [1/n, 2/n],…, [(n-1)/n,1].
 ck = xk = kDx = k/n
 The sum of the stripes is
Rn = Dx1 f(c1) + Dx2 f(c2) + Dx3 f(c3) + …+ Dxn f(cn)
= Dx f(1/n) + Dx f(2/n) + Dx f(3/n) + …+ Dxn f(1)
= ∑k=1n Dx f(kDx) = Dx ∑k=1n f (k/n)
= (1/ n) ∑k=1n [1 - (k/n)2]
= ∑k=1n 1/ n - k2/n3 = 1 – (∑k=1n k2)/ n3
= 1 – [(n) (n+1) (2n+1)/6]/ n3 = 1 – [2 n3 + 3 n2+n]/(6n3)

∑k=1n k2 = (n) (n+1) (2n+1)/6
8

Taking the limit of n → ∞
 2n3 + 3n2 + n 
lim Rn = R = 1 
= 1 2 / 6 = 2 / 3

3
n 
6n




The same limit is also obtained if cn = xn-1 is chosen
instead.
For all choice of cn  [xn-1,xn] and partition of P, the
same limit for S is obtained when n ∞
9
5.3
The Definite Integral
10
11
“The integral from a to b of f of x with
respect to x”
12

The limit of the Riemann sums of f on [a,b]
converges to the finite integral I
n
lim
|| P|| 0


 f (c )Dx
k =1
k
k
b
= I =  f ( x)dx
a
We say f is integrable over [a,b]
Can also write the definite integral as
b
b
b
a
a
a
I =  f ( x)dx =  f (t )dt =  f (u )du
b

=  f (what ever) d (what ever)
a
The variable of integration is what we call a
‘dummy variable’
13
Question: is a non continuous
function integrable?
14
Integral and nonintegrable functions


Example 1
A nonintegrable function on [0,1]
1, if x is rational
f ( x) = 
0, if x is irrational

Not integrable
15
Properties of definite integrals
16
17
18
Example 3 Finding bounds for an integral



Show that the value of
is less than 3/2

1
0
1 + cos xdx
Solution
Use rule 6 Max-Min Inequality
19
Area under the graphs of a nonnegative
function
20
Example 4 Area under the line y = x
b

Compute 0 xdx (the
Riemann sum)
and find the area A
under y = x over the
interval [0,b], b>0
21
By geometrical consideration:
Solution
A=(1/2)highwidth= (1/2)bb=
b2/2
Choose partition of n subinterval with equal width:
0 = x0 , x1, x2 ,
xn = b, Dxk = xk  xk 1 = Dx = b / n
Riemann sum:
n
n
lim  Dx f (ck ) = lim Dx  f ( xk )
n 
n 
k =1
n
k =1
n
= lim Dx  xk = lim Dx  k Dx
n 
n 
k =1
k =1
b
= lim ( Dx )  k = lim  
n 
n  n
 
k =1
2
n
2
n
k
k =1
 b  n ( n + 1)
 b  n ( n + 1)
= lim  
= lim  
n  n
n  n
2
2
 
 
b2  1  b2
= lim 1 +  =
n  2
 n 2
2
2
22
Using the additivity rule
for definite integration:
b
a
b
0
0
a
 xdx =  xdx +  xdx
b
b
a
b2 a 2
  xdx =  xdx   xdx =  , a  b
2 2
a
0
0
Using geometry, the area
is the area of a trapezium
A= (1/2)(b-a)(b+a)
= b2/2 - a2/2
Both approaches to
evaluate the area agree
23

One can prove the following Riemannian sum
of the functions f(x)=c and f(x)= x2:
24
Average value of a continuous function
revisited

Average value of nonnegative continuous
function f over an interval [a,b] is
f (c1 ) + f (c2 ) +
n
f (cn )
1 n
=  f (ck )
n k =1
Dx n
1 n
=
f (ck ) =
Dxf (ck )


b  a k =1
b  a k =1

In the limit of n ∞, the average =
b
1
f ( x)dx

ba a
25
26
27
Example 5 Finding average value

Find the average value
of f ( x) = 4  x 2
over [-2,2]
28
5.4
The Fundamental Theorem of Calculus
29
Mean value theorem for definite integrals
30
31
32
Example 1 Applying the mean value
theorem for integrals

Find the average value of
f(x)=4-x on [0,3] and where
f actually takes on this value
as some point in the given
domain.

Solution
Average = 5/2
Happens at x=3/2


33
Fundamental theorem Part 1
x



Define a function F(x): F ( x ) =  f (t )dt
a
x,a  I, an interval over which f(t) > 0 is
integrable.
The function F(x) is the area under the
graph of f(t) over [a,x], x > a
34
35
Fundamental theorem Part 1 (cont.)
F ( x + h)  F ( x) =
x +h
 f ( t ) dt
x
F ( x + h )  F ( x ) 1 x +h
=  f ( t ) dt = f ( c ) ; x  c  x + h
h
h x
mean value theorem
F ( x + h)  F ( x)
lim
= F ( x ) = f ( x )
h 0
h
The above result holds true
even if f is not positive
definite over [a,b]
36
37
Note: Convince yourself that
(i) F(x) is an antiderivative of f(x)
(ii)f(x) is a derivative of F(x)
d/dx
f(x) is a derivative
of F(x) because
F'(x)= f(x)
x
F ( x ) =  f (t )dt
a
(
f(x) = F'(x)
)dx
F(x) is an antiderivative
x
of f(x) because F ( x) = a f (t )dt
38
The main use of theorem 4 is …


d
It tells us that
dx
(
x
a
)
f (t )dt = f ( x )
In pragmatic terms, if a function is expressed
in terms of an integral of the form
x
F ( x ) =  f (t )dt
a
d
then the derivative of F(x), F ( x ) , is simply
dx
f(x)
Example 3 Applying the fundamental
theorem

Use the fundamental theorem to find
x
x
d
(a )  cos tdt
dx a
d
1
(b) 
dt
2
dx a 1 + t
5
x2
dy
(c )
if y =  3t sin tdt
dx
x
dy
(d )
if y =  cos tdt
dx
1
40
Solution for (d): you have to invoke chain
rule
x
2
d
F ( x ), where F ( x )   cos tdt
dx
1

Chain rule says if F(x)= (f◦u)(x)= f [u(x)],
d
d
d
d
d
F ( x) = ( f u ) ( x) =
f [u( x )] =
f (u )  u( x )
dx
dx
dx
du
dx
41
Solution for (d): you have to invoke chain
rule
x2
F ( x ) =  cos t dt is a composite function of the form F(x)=f [u(x)]
1
F ( x ) = f [u ( x )], where
u
f (u ) =  cos t dt , u ( x ) = x 2
so that
1
u
 d 2
d
d
d
d 
F ( x) =
f (u )  u ( x ) =
  cos t dt   ( x )
dx
du
dx
du  1
 dx
u

d 
2
=
cos
t
dt

2
x
=
cos
u

2
x
=
2
x
cos
x
(
)


du  1

42
Example 4 Constructing a function with a
given derivative and value




Find a function y = f(x) on the domain (-p /2, p/2) with
derivative dy/dx = tan x that satisfies f(3)=5.
The strategy:
Use the fundamental theorem of calculus.
Think along this line: find
a function F(x) of the form
x
F ( x ) =  q(t )dt
such that
a
d
F ( x ) = q( x ), with q( x ) = tan x
dx
Example 4 (Cont. 1)
Solution
x
dF
= tan x.
 Stage 1: If F ( x ) =  tan tdt , then
a


dx
Stage 2: construct the function f(x) using F(x),
and then try to make f(x) so constructed
fulfills the condition of f(3)=5.
The way to construct f(x) from F(x) is
obviously
dy
y = f ( x ) = F ( x ) + constant (so that
= tan x )
dx
x
=  tan tdt + constant
a
44
Example 4 (Cont. 2)
x
f ( x ) =  tan tdt + constant
a



Find the values of a and constant so that f(3)=5
This can be done by choosing a = 3, constant =5.
Verify this:
x =3
f (3) =  tan tdt + 5=0+5=5
a =3

So, finally, the function we are seeking is
x
f ( x ) =  tan tdt + 5
3
45
Fundamental theorem, part 2 (The
evaluation theorem)
46
To calculate the definite integral of f over
[a,b], do the following


1. Find an antiderivative F of f, and
2. Calculate the number
b
 f ( x)dx = F (b)  F (a)
=
a
47
To summarise
x
d
dF ( x)
f (t )dt =
= f ( x)

dx a
dx
 dF (t ) 
a  dt  dt = a f (t )dt = F ( x)  F (a)
x
x
48
Example 5 Evaluating integrals
p
( a )  cos xdx
0
0
( b)
p
sec x tan xdx
 /4
4
4 
3
x  2  dx
(c)  
x 
2
1
49
Example 7 Canceling areas



Compute
(a) the definite integral
of f(x) over [0,2p]
(b) the area between
the graph of f(x) and
the x-axis over [0,2p]
50
Example 8 Finding area using
antiderivative

Find the area of the region between the xaxis and the graph of f(x) = x3 - x2 – 2x,
-1 ≤ x ≤ 2.

Solution
First find the zeros of f.
f(x) = x(x+1) (x-2)


51
52
5.5
Indefinite Integrals and the Substitution
Rule
53
Note

The indefinite integral of f with respect to x,
 f ( x)dx
is a function plus an arbitrary constant
b

A definite integral  f ( x)dx is a number.
a
54
Antiderivative and indefinite integral in
terms of variable x

If F(x) is an antiderivative of f(x),
d
F ( x) = f ( x)
dx


the indefinite integral of f(x) is

f ( x )dx = F ( x ) + C
55
A useful mnemonic
d
(
dx

Example:
+ constant ) =
dx =
+ constant
d
(tan x + constant) = sec 2 x
dx
2
sec
 x dx = tan x + constant
56
Antiderivative and indefinite integral with
chain rule
d
F ( x ) = f ( x ) , i.e., F ( x ) antiderivative of f ( x ),
dx
d

F [u ] = f ( u ) , where u = u ( x ) .
du
d
Applying chain rule to
F u ] :
dx
du ( x ) dF ( u ) du
d
d
du
F u ] =

=
 f (u ) 
F u ] =
 f (u )
dx
dx
du
dx
dx
dx
du
In other words, F  u ] is an antiderivative of
 f ( u ) , so that we can write
dx
 du

  dx  f ( u )  dx = F u ] + C
57
The power rule in integral form
 u n +1 
d  u n +1 
 n du 
n du
 u

=u
+C
 dx = 
dx  n + 1 
dx
 dx 
 n +1
 n du 

n  du
n
u
dx
=
u
dx
=
u
  dx    dx   du
du
differential of u ( x), du is du =
dx
dx
58
Example 1 Using the power rule
The strategy is to convert the integral into
the form

1 + y  2 y dy = ?
2
du
Let u = 1 + y , du =
dy = 2 ydy.
dy
2

1 + y 2  2 y dy =  u  du = ...
59
Example 2 Adjusting the integrand by a
constant

4t  1 dt = ?
Let u = 4t  1, du = 4dt ,

1
1
4t  1 dt =  u dt =  u 4dt =  u du = ...
4
4
60
Substitution: Running the chain rule
backwards
du
dx =  f (u )du
dx
Used to find the integration with the integrand in the
form of the product of f [ g ( x )]  g '( x )
let u = g ( x);  f [ g ( x)]  g ( x)dx =  f (u ) 
 f [ g ( x)]  g '( x)dx =  f (u)du
f (u )
du
61
Example 3 Using substitution
 cos(7 x + 5) dx =  cos u 
u
1
du
7
du 1
1
= sin u + C = sin ( 7 x + 5 ) + C
7 7
7
62
Example 4 Using substitution
?
=
dx
x
sin
x

3
2
u = x ; du = 3x dx
2
3
 sin x
3
u
1
1
x dx =  sin u  du =  cos u + C
3
3
1
2
3
du
1
3
= cos x + C
3
63
Example 5 Using Identities and
substitution
1
2
2
dx
=
sec
2
x
dx
=
sec
=
 cos2 2 x

 2ux dx
1
2
du
1
1
1
2
sec u du = tan u + C = tan 2 x + C

2 d
2
2
du
tan u
64
Example 6 Using different
substitutions

2z
3
z2 +1
dz =  ( z + 1)
2
u
1/ 3
1/ 3
2 zdz =  u 1/ 3du =...
du
65
The integrals of

2
sin x
and
2
cos x
Example 7
1
 sin x dx = 2 1  cos 2 x dx
x 1
=   cos 2 x dx
2 2
u 1
2
2
du
x 1
=   cos udu =...
2 4
66
The integrals of

2
sin x
and
2
cos x
Example 7(b)
1
 cos x dx = 2  cos 2 x + 1 dx = ...
2
67
Example 8 Area beneath the curve
y=sin2 x



For Figure 5.24, find
(a) the definite integral
of y(x) over [0,2p].
(b) the area between
the graph and the xaxis over [0,2p].
68
5.6
Substitution and
Area Between Curves
69
Substitution formula
x =b
let u = g ( x);

x =a
f [ g ( x)]  g ( x)dx =
x =b

x =a
u = g (b )
du
f [u ]  dx =  f (u )du
dx
u=g (a)
70
Example 1 Substitution

Evaluate
1
2
3
3
x
x
+ 1 dx

1
u ( x =1)
x =1

x =1
x 3 + 1  3 x 2 dx =
u1/ 2
du

u1/ 2  du =...
u ( x =1)
71
Example 2 Using the substitution formula
x =p / 2
p
cot x csc 2 xdx = ?
x= / 4
u2
 cot x csc xdx = cotu x  csc duxdx =   udu =  2 + c
2
2
cot 2 x
=
+c
2
 1
cot x
cot x
1 2
2
=
= cot (p / 4 )  cot (p / 2 )  =
p / 4 cot x csc xdx =  2
2 p /2 2 
 2
p /4
1
0


p /2
2
p /2
2
p /4
2
72
Definite integrals of symmetric
functions
73
74
Example 3 Integral of an even
function
2
Evaluate
4
2
x

4
x
+ 6 ) dx
(
2
Solution:
f ( x) = x 4  4 x 2 + 6;
f ( x) = (  x )  4 (  x ) + 6 = x 4  4 x 2 + 6 = f ( x)
4
2
even function
How about integration of the same
function from x=-1 to x=2
75
Area between curves
76
n
n
k =1
k =1
A   DAk =  Dxk ( f (ck )  g (ck ) 
A = lim
|| P|| 0
n
 Dx ( f (c )  g (c ) =   f ( x)  g ( x)] dx
k =1
b
k
k
k
a
77
78
Example 4 Area between intersecting
curves

Find the area of the region
enclosed by the parabola
y = 2 – x2 and the line y = -x.
DA = ( f ( x)  g ( x) )  Dx
n
A = lim  DAk =  dA;
n 
A=
A
0
k =1
b2
a 1
[ f ( x)  g ( x)]dx
2 x2
=
2
1
(2  x
2
x
 x ) dx = ...
79
Example 5 Changing the integral to
match a boundary change

Find the area of the
shaded region
Area = A + B
A=
2
B=
4
0
2
xdx;
x  ( x  2)dx
80
n
n
k =1
k =1
A   DAk =  Dyk ( f (ck )  g (ck ) 
A = lim
|| P|| 0
n
 Dy ( f (c )  g (c ) =   f ( y)  g ( y)] dy
k =1
d
k
k
k
c
DAk
81
Example 6 Find the area of the region in
Example 5 by integrating with respect to y
DA = ( f ( y)  g ( y))  Dy
n
A = lim  DAk = 
n 
k =1
y =4
y =0
[ f ( y )  g ( y )]dy
=  ( y + 2 )  ( y 2 ) dy = ...
2
0
82