Techniques of Integration
Integration By Parts
There is NO formula for
f (x)g(x)dx.
It almost never happens that
f (x)g(x)dx =
f (x)dx
Notice that
g(x)dx
df = f (x) + C. We often shorten this to
df = f to indicate that the integral
and differential operators “cancel” each other.
d
d
d
(uv) = u
(v) +
(u)v
dx
dx
dx
The Product Rule for derivatives,
has no simple counterpart for antiderivatives. It can be restated in terms of differentials as
d(uv) = udv + vdu, and if we apply indefinite integral signs, we get
d(uv) = udv + vdu, or uv = udv + vdu.
We usually use the equivalent formula
Example:
udv = uv −
vdu.
Evaluate
x sin xdx
Solution: Use integration by parts, with u = x, and dv = sin xdx.
Then du = dx, and v = − cos x, so
x sin xdx =
udv = uv − vdu = x(− cos x) − (− cos x)dx =
−x cos x +
cos xdx = −x cos x + sin x + C
1
We can also use this technique with definite integrals:
Evaluate
Solution:
π
0
x sin xdx
π
0
−x cos x|π
0 +
x sin xdx =
π
−π (−1) = π
0
udv = uv −
π
π
vdu = x(− cos x)|0 −
(− cos x)dx =
0
π
cos xdx = −x cos x|π
0 + sin x|0 = −π cos π − (−0 cos 0) + sin π − sin 0 =
Question: How do I know what u and dv should be?
Students first encountering the technique of using the equation
trouble knowing what to take for u and what to take for dv.
udv = uv − vdu have
Answer: Get lots of experience. This is an area where we learn a lot from experience. The
Integration by Parts technique is characterized by the need to select u from a number of possibilities. Once u has been chosen, dv is determined, and we hope for the best.
The basic idea underlying Integration by Parts is that we hope that in going from udv to
vdu we will end up with a simpler integral to work with. In the example we have just seen,
we were lucky.
Let’s try it again, the unlucky way:
Example:
Evaluate
(sin x)xdx
Solution: Use integration by parts, with u = sin x, and dv = xdx.
x2
, so
Then du = cos xdx, and v =
2
2
x
x2
−
cos xdx =
(sin x)xdx =
udv = uv − vdu = sin x
2
2
1
x2
x 2 cos xdx
sin x −
2
2
which involves a tougher looking integral than we started with.
2
As a rule of thumb, one-third of the possible choices will lead to an easier integral, one-third
will lead to a harder one, and one-third will lead to one of equal difficulty.
It is also possible to spin your wheels, and go around in circles, as we shall soon see.
Let’s try a strategic approach to our example: u has to be selected so that udv = x sin xdx,
so we look at the possible choices for u: x, sin x, or x sin x. Once u is selected, we have
x sin xdx
dv =
, and all we have to is find du(easy) and v = dv (possibly very hard or
u
impossible). Then we try to decide if we can get anywhere with
vdu. Thus, we can let u be
any factor of f (t), including 1, and the corresponding dv is determined. (Of course, if we let
u = 1, the problem of finding v is just our original integration problem, so we will omit it.) If
we cannot then find v we know we have a non-viable selection of the pair u and dv.
We shall illustrate the rather inefficient technique of examining all the possibilities and discarding the non-viable ones in the following examples. Organizing our information in a table
is helpful:
dv =
v = dv
du
x sin xdx
u
x sin xdx
=
x
sin xdx = dx
x
sin xdx
− cos x
x sin xdx
=
sin x
cos xdx
xdx =
sin x
2
x
xdx
2
x sin xdx
=
x sin x
sin x
dx =
x sin x
dx
x
+x cos x
u
vdu
− cos xdx
x2
2
vdu
Better?
− cos xdx =
YES!
− sin x
x2
cos xdx
cos xdx
2
x sin x
NO!
x(sin x + x cos x)dx NO!
2
+x cos x
There are some simple integrals where little choice is available: knowing which of a large
number techniques to use is crucial.
Example: ln xdx obviously requires
u = ln x, dv = dx, so that v = x and du =
ln xdx =
udv = uv −
dx
.
x
dx
vdu = (ln x)x − x
= x ln x − dx =
x
x ln x − x + C
3
Example:
arctan xdx also obviously requires
u = arctan x, dv = dx, so that v = x and du =
arctan xdx =
x arctan x −
udv = uv −
1
ln(1 + x 2 ) + C =
2
dx
1 + x2
dx
1
vdu = (arctan x)x− x
= x arctan x−
2
1+x
2
2xdx
=
1 + x2
√
x arctan x − ln 1 + x 2 + C
Indefinite Integration of ekt times another factor
We now look at a family of integrals which show up in LSD, (Linear Systems Design), particularly in the calculation of Laplace Transforms.
We define G(G) = ekt G(t)dt, where G is any function. The types of continuous function
G that arise in practical situations are often sums and products of polynomials, exponential
functions, and sinusoidal functions. Fortunately, we know how to evaluate these using the
technique of integration by parts.
Examples:
(1) G(t) = c, a constant. Integration by parts is not needed here. Then G(c) =
c kt
e +C
k
(2) G(t) = t. Then G(t) = ekt tdt.
Here f (t) = ekt t has four different possible factorizations:
u
t
ekt
ekt t
dv =
ekt dt
tdt
dt
f (t)dt
u
v
1 kt
e
k
t2
t
du
vdu
1 kt
dt
e dt
k
kekt dt
kt 2 ekt dt
(t + k)ekt dt t(t + k)ekt dt
We
the one
viable factorization,
u = t, dv = ekt dt:
use
t(ekt dt)
1= udv t= uv −1 vdu =
1 kt
t k e − k ekt dt = k ekt − k2 ekt + C
= kt − k12 ekt + C
4
Viable?
Yes!
No!
No!
ekt G(t)dt =
(3) G(t) = t 2 . Then G(t 2 ) = ekt t 2 dt, so f (t) = ekt t 2 . This has four possible factorizations:
u
dv =
kt
t
f (t)dt
u
e tdt
2
t
ekt
ekt t
ekt t 2
v
t
k
−
1 kt
e
k
1 3
t
3
1 2
t
2
kt
e dt
t 2 dt
tdt
dt
1
k2
du
e
kt
vdu
t
−
k
dt
1
k2
Viable?
kt
e dt
2
tekt dt
k
k 3 kt
t e dt
3
1 2
t (t + k)ekt dt
2
2
kt
2tdt
kekt dt
(t + k)ekt dt
(kt 2 + 2t)ekt dt t(kt + 2t)e dt
t
Yes, but messy
Yes!
No!
No!
No!
We see that if we use the factorization u = t 2 , dv = ekt dt, we get
2
uv − vdu = tk ekt − k2 tekt dt =
2
t 2 kt
e − k2 tk − k2 ekt =
k
2
t
2t
2
ekt + C
−
+
k
k2
k3
t 2 ekt dt =
udv =
kt
(4) We notice how the problem of evaluating t 2 ekt was reduced to the evaluation
of
e tdt,
which had just been done. We suspect the existence of a reduction formula for ekt t n dt. Having been successful in taking dv = ekt dt in the two preceding examples, we decide to do this
ekt t n dt
1
again, and we have u = ekt dt = t n . We calculate v = k ekt and du = nt n−1 dt, so that:
kt n
e t dt = udv = uv − vdu = k1 t n ekt − n−1
ekt t n−1 dt.
k
1
Thus we have G(t n ) = k t n ekt −
n−1
G(t n−1 ).
k
(5) G(t) = sin at. We have f (t) = ekt sin at, so there are just three choices for u:
u
sin at
ekt
ekt sin at
f (t)dt
dv = u
ekt dt
sin atdt
dt
v
1 kt
e
k
− a1 cos at
t
du
vdu
a kt
a cos atdt
e cos atdt
k
kt
ke dt
− ak ekt cos atdt
(k sin at + a cos at)ekt dt t(k sin at + a cos at)ekt dt
Viable?
Yes
Yes
No!
The
kt first two choices are both viable, and we see that they both lead to the evaluation of
e cos atdt. We will examine both cases closely:
First Choice: u1 = sin
dv1 = ekt dt
at,
1
We get: G(sin at) = ekt sin atdt = u1 dv1 = u1 v1 − v1 du1 = sin at k ekt −
1
a
sin at k ekt − k ekt cos atdt =
1
a
sin at k ekt − k G(cos at)
or
1
a
G(sin at) = sin at ekt − G(cos at)
k
k
5
a kt
e
k
cos atdt =
In evaluating
u
cos at
ekt
ekt cos at
ekt cos atdt we again have three choices, two of which are viable:
f (t)dt
dv = u
ekt dt
cos atdt
dt
v
1 kt
e
k
1
sin at
a
t
du
vdu
−a sin atdt
− ak ekt sin atdt
k kt
kekt dt
e sin atdt
a
kt
(k cos at − a sin at)e dt t(k sin at + a cos at)ekt dt
Viable?
Yes
Yes
No!
Again we have two choices. First and Best Choice:
ekt dt, and we get:
We will let U1 = cos at, dV1 =
G(cos at) = cos atekt dt = U1 dV1 = U1 V1 − V1 dU1 =
1
1
cos at k ekt − k ekt (−a sin atdt) =
1
a
cos atekt + k ekt sin atdt =
k
1
a
cos atekt + k G(sin at),
k
or
a
1
G(cos at) = cos atekt + G(sin at)
k
k
so we are back where we started! However, if we substitute this into the equation
1
a
G(sin at) = sin at ekt − G(cos at)
k
k
1
a 1
a
a
1
cos atekt + G(sin at) = sin at ekt − 2 cos atekt −
we get G(sin at) = sin at ekt −
k
k k
k
k
k
a2
G(sin at)
k2
which can be solved for G(sin at):
(1 +
a2
a2 + k2
1
a
)G(sin
at)
=
G(sin at) = sin at ekt − 2 cos atekt =
2
2
k
k
k
k
ekt (sin at
k
a
ekt
−
cos
at)
=
(k sin at − a cos at) so
k2 k2
k2
G(sin at) =
k sin at − a cos at kt
e
a2 + k2
Last and Worst Choice: We will now let U2 = ekt , dV2 = cos at, and we get:
1
1 kt
k
1
k
kt 1
kt
sin at(e kdt) = e sin at−
ekt sin atdt = ekt sin at− G(sin at)
G(cos at) = e ( sin at)−
a
a
a
a
a
a
This time, however, if we substitute this into the equation
1
a
G(sin at) = sin at ekt − G(cos at)
k
k
we get
a
1
G(sin at) = sin at ekt −
k
k
so there is no information gained.
k
1 kt
e sin at − G(sin at) = G(sin at)
a
a
6
kt
Second Choice: u2 =
ekt , dv2 = sin atdt
We get: G(sin at) = e sin atdt = u2 dv2 = u2 v2 − v2 du2 = ekt (− a1 cos at)−
1
k kt
e
a
cos atdt =
k
− a ekt cos at − a G(cos at)
or
1
k
G(sin at) = − ekt cos at − G(cos at)
a
a
Using, from above,
G(cos at) =
we get
a
1
cos atekt + G(sin at)
k
k
1
k
G(sin at) = − ekt cos at −
a
a
1
a
kt
cos ate + G(sin at) = G(sin at)
k
k
so there is no new information. On the other hand, if we use
G(cos at) =
1 kt
k
e sin at − G(sin at)
a
a
we get the same value as before.
(6)G(cos at). This is easily derived from the calculations of the previous example.
a sin at + k cos at kt
1
1
kt a
kt a k sin at − a cos at kt
=
e
e
G(cos at) = cos ate + G(sin at) = cos ate +
2
2
k
k
k
k
a +k
a2 + k2
so
G(cos at) =
a sin at + k cos at kt
e
a2 + k 2
(7)G(t n sin at) = t n sin atekt dt
dv = ekt dt, so that du = (nt n−1
Let u = t n sin at,
sin at + at n cos at)dt, and v = k1 ekt . Then
G(t n sin at) = tn sin atekt dt = udv = uv − vdu =
1
1
t n sin at( k ekt ) − k ekt ((nt n−1 sin at + at n cos at)dt) =
1 n
n
a
t sin atekt − k G(t n−1 sin at) − k G(t n cos at)
k
or
1
n
a
G(t n sin at) = t n sin atekt − G(t n−1 sin at) − G(t n cos at)
k
k
k
7
(8)G(t n cos at) = t n cos atekt dt. Let U = t n cos at and dV = ekt dt,
so that dU = (nt n−1 cos at − at n sin at)dt and V = k1 ekt . Then
G(t n cos at) = t n cos atekt dt = UdV = U V − V du =
t n cos at k1 ekt − ekt ((nt n−1 cos at − at n sin at)dt) =
1 n
n
a
t cos atekt − k G(t n−1 cos at) + k G(t n sin at).
k
We can now solve for G(t n sin at):
G(t n sin at) =
1
n kt
n−1
(k
sin
at
−
a
cos
at)t
e
−
n(G(t
(k
sin
at
−
a
cos
at)))
a2 + k 2
and hence
G(t n cos at) =
1
n kt
n−1
(k
cos
at
−
a
sin
at)t
e
+
n(G(t
(k
cos
at
−
a
sin
at)))
a2 + k2
8
Trigonometric Integrals
It is often necessary to evaluate integrals of the form
(sin x)m (cos x)n dx, where m and n are integers.
If one of the exponents, either m or n is and odd, there is a straightforward simplification.
Case 1: m = 2M + 1 is odd.
Then (sin x)m = (sin x)2M+1 = (sin2 x)M sin x = (1 − cos2 x)M sin x,
so our integral becomes
m
n
(sin x) (cos x) dx =
(1 − cos2 x)M sin x(cos x)n dx
and we may make the substitution u = cos x with du = − sin xdx to get
m
n
2
M
n
(sin x) (cos x) dx = (1 − cos x) sin x(cos x) dx = − (1 − u2 )M un du
Example:
−
10
(sin x) (cos x) dx = − (1 − u2 )5 u10 du =
1 − 5u2 + 10(u2 )2 − 10(u2 )3 + 5(u2 )4 − (u2 )5 u10 du =
−
−
11
−
u10 − 5u12 + 10u14 − 10u16 + 5u18 − u20 du =
u11
u13
u15
u17
u19 u21
−5
+ 10
− 10
+5
−
11
13
15
17
19
21
+C =
5 cos13 x
2 cos15 x
10 cos17 x
5 cos 19x
cos21 x
cos11 x
+
−
+
−
+
+C
11
13
5
17
19
21
9
Example:
−
11
−10
(sin x) (cos x)
dx = − (1 − u2 )5 u−10 du =
1 − 5u2 + 10(u2 )2 − 10(u2 )3 + 5(u2 )4 − (u2 )5 u−10 du =
−
−
u−10 − 5u−8 + 10u−6 − 10u−4 + 5u−2 − u0 du =
u−9
u−7
u−5
u−3
u−1
−5
+ 10
− 10
+5
−u +C =
−9
−7
−5
−3
−1
cos−9 x
5 cos7 x
10 cos−3 x
−5
−1
− −
+
− 2 cos x +
− 5 cos x − cos x + C =
9
7
3
1
5
10
sec9 x − sec7 x + 2sec5 x −
sec3 x + 5secx + cos x + C
9
7
3
Example:
−11
(sin x)
−10
(cos x)
dx =
−12
(sin x)
−10
(cos x)
sin xdx = − (1−u2 )−6 u−10 du
can be done, but requires the method of Partial Fractions, which we shall see later.
The situation is similar when the power of cos x is odd.
10
11
Example: (sin x) (cos x) dx = (sin x)10 (cos x)10 cos xdx =
10
2
5
(sin x) (cos x) cos xdx =
(sin x)10 (1 − sin2 x)5 cos xdx =
( letting u = sin x and du = cos xdx)
10
2 5
u (1 − u ) du =
u10 1 − 5u2 + 10(u2 )2 − 10(u2 )3 + 5(u2 )4 − (u2 )5 du =
u10 − 5u12 + 10u14 − 10u16 + 5u18 − u20 du =
u15
u17
u19 u21
u13
u11
+ 10
− 10
+5
−
+C =
−5
13
15
17
19
21
11
1
5
2
10
5
1
sin11 x −
sin13 x + sin15 x −
sin17 x +
sin19 −
sin21 x + C
11
13
5
17
19
21
10
When both powers are odd, it is easiest to select the function with the highest power for substitution:
Good Example:
11
3
(sin x) (cos x) dx =
(sin x)11 (cos x)2 cos xdx =
(sin x)11 (1 − sin2 x) cos xdx =
( letting u = sin x and du = cos xdx)
11
2
u (1 − u )du =
Bad Example:
u11 − u13 du =
11
1
u11 u14
1
−
+C =
sin11 x −
sin13 x + C
11
14
11
14
3
(sin x) (cos x) dx =
(sin x)10 (cos x)3 sin xdx =
(1 − cos2 x)5 (cos x)3 sin xdx =
( letting u = cos x and du = − sin xdx)
2 5
3
(1 − u ) u (−du) =
−1 + 5u2 − 10(u2 )2 + 10(u2 )3 − 5(u2 )4 + (u2 )5 u3 du =
−u3 + 5u5 − 10u7 + 10u9 − 5u11 + u13 du =
−
u6
u8
u10
u12 u14
u4
+5
− 10
+ 10
−5
+
+C =
4
6
8
10
12
14
1
5
5
5
1
− cos4 x + cos6 x − cos8 x + cos10 x −
cos12 x +
cos14 x + C
4
6
4
12
14
11
When neither power is odd, we need to use a double angle formula from trigonometry:
cos 2x ≡ 2 cos2 x − 1 = 1 − 2 sin2 x
can be solved for cos2 x and sin2 x:
1 + cos 2x
1 − cos 2x
cos2 x =
and sin2 x =
2
2
Thus, if we want to integrate (sin x)m (cos x)n dx, where m and n are even integers, we write
m = 2M and n = 2N and we have:
(sin x)m (cos x)n dx =
1 − cos 2x
2
M (sin x)2M (cos x)2N dx =
1 + cos 2x
2
N
−(M+N)
dx = 2
12
(sin2 x)M (cos2 x)N dx =
M
N
(1 − cos 2x) (1 + cos 2x) dx
1 + cos 2x
dx =
2
1
1
1
1 + cos 2xdx =
x + sin 2x + C =
2
2
2
Example:
2
cos xdx =
x
1
x
1
x + sin x cos x
+ sin 2x + C = + 2 sin x cos x + C =
+C
2
4
2
4
2
1 − cos 2x
dx =
2
1
1
1
1 − cos 2xdx =
x − sin 2x + C =
2
2
2
Example:
2
sin xdx =
x
1
x
1
x − sin x cos x
− sin 2x + C = − 2 sin x cos x + C =
+C
2
4
2
4
2
It should come as no surprise that
2
2
cos x + sin x dx = x + C
1 + cos 2x
1 − cos 2x
Example: sin x cos xdx =
dx =
2
2
1
1 + cos 4x
1
1
2
1 − cos 2xdx =
1−
dx =
1 − cos 4xdx =
4
4
2
8
1
1
x
sin 4x
x − sin 4x + C = −
+C
8
4
8
32
2
2
Products of powers of secant and tangent functions
secm x tann x can be expressed as a product of powers of sin x and cos x, but it is often more
convenient to use the identity tan2 x + 1 = sec2 x and the differentials d(tan x) = sec2 xdx and
d(secx) = secx tan xdx. We assume that m and n are non-negative.
Case 1: The power m of secx is positive and even: then m = 2M and we have:
m
n
sec x tan xdx =
2
M
n
(sec x) tan xdx =
2
M−1
(sec x)
n
2
tan xsec xdx =
and we can make the substitution u = tan x, du = sec2 xdx, and we get
M−1
M−1
2
n
2
1 + tan x
tan xsec xdx =
1 + u2
un du
13
(1+tan2 x)M−1 tann xsec2 xd
Example:
10
5
sec x tan xdx =
1 + u2
4
u5 du =
2
4
6
8
5
1 + 4u + 6u + 4u + u u du = u5 + 4u7 + 6u9 + 4u11 + u13 du =
u8
u10
u12 u14
u6
+4
+6
+4
+
+C =
6
8
10
12
14
1 6 1 8 3 10 1 12
1 14
u + u + u + u +
u +C =
6
2
5
3
14
1
1
3
1
1
tan6 x + tan8 x + tan10 x + tan12 x +
tan14 x + C
6
2
5
3
14
If the power of secx is not even, but the power of n of tan x is odd, so that n = 2N + 1, and
m is positive, we can make the substitution u = secx, du = secx tan xdx, and get
m
n
sec x tan xdx = secm−1 x tan2n xsecx tan xdx =
N
N
um−1 tan2 x du = um−1 sec2 x − 1 du =
N
um−1 u2 − 1 du
Example:
5
5
sec x tan xdx =
u8 − 2u6 + u4 du
4
2
2
u (u − 1) du =
u4 (u4 − 2u2 + 1)du =
u7 u5
u9
−2
+
+C =
9
7
5
1
2
1
sec9 x − sec7 x + sec5 x + C
9
7
5
We are left with the cases where m is odd and n is even, all of which can be reduced to the
problem of finding the antiderivative of an odd power of secx.
Example:
secxdx =
sec2 x + secx tan x
secx + tan x
secx
dx =
secx + tan x
dx =
secx + tan x
(sec2 x + secx tan x)dx
=
secx + tan x
d(secx + tan x)
= ln(secx + tan x) + C
secx + tan x
14
Example:
I=
sec3 xdx
Using Integration by Parts, with
u = secx, dv = sec2 x, v = tan x, du = secx tan xdx, we get
I=
3
sec xdx =
udv = uv −
vdu =
secx tan x −
tan xsecx tan xdx = secx tan x −
tan2 xsecxdx =
secx tan x − (sec2 x − 1)secxdx =
secx tan x −
3
sec xdx +
secxdx = secx tan x − I +
secxdx = secx tan x + ln |secx +
tan x| + C so
2I = secx tan x + ln |secx + tan x| + C and
Example:
cases left:
secx tan x + ln |secx + tan x|
+C
2
tan2 xsecxdx appears in the previous calculation, and is one of the simpler
I = secx tan x −
sec3 xdx =
tan2 xsecxdx gives us
tan2 xsecxdx = secx tan x − I =
secx tan x −
secx tan x + ln |secx + tan x|
+C =
2
secx tan x − ln |secx + tan x|
+C
2
15
Integrals of products of sine and cosine functions with different arguments
The identities:
cos x cos y =
sin x sin y =
1
cos(x + y) + cos(x − y)
2
1
cos(x − y) − cos(x + y)
2
1
sin(x − y) + sin(x + y) may be used:
2
Example: cos 5t cos 7tdt = 1 (cos(5t + 7t) + cos(5t − 7t)) dt =
2
1
1 1
1
sin 12t + sin 2t + C =
(cos 12t + cos(−2t)) dt =
2
2 12
2
sin x cos y =
1
1
sin 12t + sin 2t + C
24
4
16
Trigonometric Substitutions
It is often necessary to evaluate integrals containing expressions of the form
a2 − x 2 ,
a2 + x 2 ,
x 2 − a2 .
or
For example, we may wish to evaluate
a2 − x 2 dx
a2 + x 2 dx
or
or
x 2 − a2 dx
To do this , we need to make the appropriate substitution:
Expression
x=
dx =
Expression =
2
2
a −x
a sin θ
a cos θdθ
a2 cos2 θ
a2 + x 2
a tan θ
asec2 θdθ
a2 sec2 θ
2
2
x −a
asecθ asecθ tan θdθ
a2 tan2 θ
Example: a2 − x 2 dx
Letting x = a sin θ, we have dx = a cos θdθ, so that
2
2
2
2
2
a − x dx =
a − a sin θ a cos θdθ =
2
2
a
1 − sin θ cos θdθ = a
cos2 θdθ =
2
2
a
1 + cos 2θ
a2
dθ =
2
2
a2
dθ +
2
cos 2θdθ =
a2 1
a2 θ a2
a2
θ+
sin 2θ + C =
+
2 sin θ cos θ + C =
2
2 2
2
4
2
a2
a2
a2
x
a2 x
x
θ+
sin θ cos θ + C =
arcsin +
1−
+C =
2
2
2
a
2 a
a
x
x 2
a2
arcsin +
a − x2 + C
2
a
2
17
Let us use this to compute some areas contained in circles of radius a:
First, the area under y =
a
2
x
x
a
arcsin +
a2 − x 2 dx =
a2 − x 2 =
2
a
2
−a
2
a a 2
a
−a
−a
arcsin +
arcsin
+
a − a2 −
a2 − (−a)2 =
a
2
2
a
2
a2
a2 π
−π
π a2
arcsin(1) + 0 −
arcsin(−1) + 0 =
−
=
,
2
2 2
2
2
a −a
a2
2
a2
2
√
a2 − x 2 from −a to a is that of a semicircle of radius a:
as expected.
Second, the area under y =
a
0
a2
2
a2
2
√
a2 − x 2 from 0 to a is that of a quarter circle of radius a:
a
x
x 2
a2
2
arcsin +
−
=
a −x =
2
a
2
0
2
a a 2
0
0
a
arcsin +
arcsin +
a − a2 −
a2 − 02 =
a
2
2
a 2
a2
a2 π
π a2
arcsin(1) + 0 −
arcsin 0 + 0 =
=
as expected.
2
2 2
4
a2
x 2 dx
√
√
a
a
3
to x =
is:
Third, the area under y = a2 − x 2 from x =
2
2
√
a2 3
2 x
x
a
2 − x 2 dx =
2 − x2 arcsin
+
a
a
=
a
a
2
a
2
2
2

 

√
√ √ 2
2
a 3
a 3
a
a
2
2
a 3  a
a 
a
2
=
arcsin 2 + 2
arcsin 2 + 2 a2 −
a −

−
2
a
2
2
2
a
2
2
√
a
3
2


√
√   2
2
a
a
a
1
a
3
3
1
3

−
arcsin
+
a
arcsin + a
=
2
2
4
4
2
2
4
4
√ √ π
1
π
π
π a2
1
3
3
2
2 π
a
+
−
+
=a
−
=
23
8
26
8
6
12
12
18
In general the area under y =
d
2
x
x
a
arcsin +
a2 − x 2 dx =
a2 − x 2 =
2
a
2
c
2
√
√
arcsin da + d2 a2 − d2 − a2 arcsin ac + 2c a2 − c 2 =
c
a2
2
a2
2
a2 − x 2 from x = c to x = d (c ≤ d) is:
d
√
√
√
arcsin da − arcsin ac + d2 a2 − d2 − 2c a2 − c 2
Thus the area of the circle x 2 + y 2 = a2 lying between the lines x = c and x = d is
2
c
d
+ d a2 − d 2 − c a2 − c 2
arcsin − arcsin
a
a
a
Setting c = 0, we get a formula for the area Ad between the y-axis and the line x = d:
Ad = a2 arcsin
d
+ d a2 − d2
a
Example: a2 + x 2 dx
We let x = a tan θ, so that a2 + x 2 = a2 + a2 tan2 θ = a2 (1 + tan2 θ) = a2 sec2 θ,
and dx = asec2 θ dθ, so we have
2
2
a + x dx =
a2 sec2 θ asec2 θdθ =
secθ tan θ + ln |secθ + tan θ|
+C
2
√
x 2 + a2
x
, so our integral in terms of x is:
Now tan θ = , and secθ =
a
a
√
√
x 2 +a2 x x 2 +a2 x
√
+
ln
+
x 2 + a2 + x 2
a
a
a
a
x
a
+C =
a2
ln +C =
x 2 + a2 +
2
2
a
2
2
a
sec3 θdθ = a2
a2 2
x 2
ln x + a2 + x + C
x + a2 +
2
2
19
We have used the previously derived formula
sec3 xdx =
secx tan x + ln |secx + tan x|
+C
2
Example: x 2 − a2 dx
Letting x = asecθ, we have
x 2 − a2 = a2 sec2 θ − a2 = a2 (sec2 θ − 1) = a2 tan2 θ
x
(so secθ =
and tan θ =
a
√
x 2 − a2
),
a
and dx = asecθ tan θ dθ, so
2
2
2
2
2
x − a dx =
secθ tan2 θ dθ =
a tan θasecθ tan θ dθ = a
secθ tan θ − ln |secθ + tan θ|
+C =
2
√
x √x 2 −a2 x 2 −a2
x
− ln a +
a
a
a
2
a
+C =
2
√
√
x+ x 2 −a2 2
2
2
x x − a − a ln a
+C =
2
a2
a2 x 2
ln x + x 2 − a2 + C
x − a2 −
2
2
We have used the previously derived formula
tan2 xsecxdx =
secx tan x − ln |secx + tan x|
+C
2
20
It is often necessary to complete squares before using a trig substitution:
dx
dx
dx
=
=
Example:
=
2
2
x + 4x + 5
x + 4x + 1 + 4
(x + 2)2 + 22
du
u2 + 2 2
where we have let u = x + 2.
Now we let u = 2 tan θ and have u2 + 22 = 22 sec2 θ and du = 2sec2 θ dθ,
so we have
du
1
1
u
1
2sec2 θ dθ
dθ = θ + C = arctan + C =
=
=
2
2
2
2
u +2
2 sec θ
2
2
2
2
1
x+2
arctan
+C
2
2
Example:#28, p.454 of Brown Stewart:
e2t − 9dt
2
Note that e2t − 9 = et − 32 , so if we let x = et , we get
2
e2t − 9 = et − 32 = x 2 = 32 and we can use the substitution x = 3secθ.
We must be very careful with the differentials: we have dx = et dt, so dt = e−t dx =
1
dx
dx =
t
e
x
dx
2t
Thus we have
e − 9dt =
x 2 − 32
x
Now we use x = 3secθ, dx = 3secθ tan θdθ, x 2 − 32 = 32 tan2 θ
√
x 2 − 32
x
and tan θ =
) to get
(so secθ =
3
3
3secθ tan θdθ
dx
=
= 3 tan2 θdθ = 3 (sec2 θ − 1)dθ = 3(tan θ −
x 2 − 32
32 tan2 θ
x√
3secθ
2
2
x
x −3
θ) + C = 3
− arcsec
+C =
3
3
x
et
+C
x 2 − 32 − 3arcsec + C = e2t − 9 − 3arcsec
3
3
21
Partial Fractions
This is a technique used to antidifferentiate proper rational functions: quotients of polynomials whose denominator has degree greater than the numerator. Using long division, any
rational function can be written as the sum of a polynomial and a proper rational function.
Theorem: Any polynomial with real coefficients can be written as the product of powers of
linear and quadratic factors.
We will usually only look at prefactored examples, or those where the factorization is easy.
Example:
1
du =
2
u −1
1
du
(u − 1)(u + 1)
The basic idea is that we wish to write
1
as the sum of the form
(u − 1)(u + 1)
A
B
+
,
u−1 u+1
where A and B are to be determined so that the equation
1
A
B
=
+
(u − 1)(u + 1)
u−1 u+1
is true for all values of u except −1 and 1.
The ordinary way of finding A and B is to multiply by u2 − 1 and to equate the coefficients of
the resulting polynomials:
1 = A(u + 1) + B(u − 1) = (A + B)u + (A − B) gives us two equations in two unknowns:
A + B = 0 and A − B = 1
which when solved give us A =
1
1
and B = −
2
2
22
A quicker way is to substitute the “illegal” values u = −1 and u = 1 into the equation
1 = A(u + 1) + B(u − 1)
u = −1: 1 = A(−1 + 1) + B(−1 − 1) = −2B so B = −
u = 1: 1 = A(1 + 1) + B(1 − 1) = 2A so A =
1
2
1
2
We then have
1
1 1
1 1
=
−
, so
(u − 1)(u + 1)
2u−1 2u+1
1
1
1
1
1
du =
du −
du =
(u − 1)(u + 1)
2 u−1
2 u+1
u − 1
1
1
+C
ln |u − 1| − ln |u + 1| + C = ln 2
2
u + 1
Definition: A partial fraction is an expression of the form
A
(cx + d)m
Bx + +C
.
(ax 2 + bx + c)n
or
At this point we (theoretically) have learned how to antidifferentiate partial fractions, and the
P (x)
next skill to learn is how to rewrite a proper rational function
as the sum of partial
Q(x)
fractions.
First Case: Q(x) = (c1 x + d1 )(c2 x + d2 ) · · · (ck x + dk ) is the product of distinct linear factors.
Then we can write
P (x)
P (x)
A1
A2
Ak
=
=
+
+ ··· +
Q(x)
(c1 x + d1 )(c2 x + d2 ) · · · (ak x + dk )
c1 x + d1 c2 x + d2
ck x + dk
Multiplying both sides of the equation by Q(x), we get:
P (x) = A1 (c2 x + d2 ) · · · (ck x + dk ) + A2 (c1 x + d1 )(c3 x + d3 ) · · · (ck x + dk ) + · · · +
Ak (c1 x + d1 ) · · · (ck−1 x + dk−1 )
which is easily solved by substituting the values xi = −
23
di
, i = 1, 2, 3, . . . , k.
ci
Example2, p456 of Brown Stewart:
x 2 + 2x − 1
dx
2x 3 + 3x 2 − 2x
x 2 + 2x − 1
A1
A2
A3
x 2 + 2x − 1
=
=
+
+
leads to
3
2
2x + 3x − 2x
x(x + 2)(2x − 1)
x
x + 2 2x − 1
x 2 + 2x − 1 = A1 (x + 2)(2x − 1) + A2 x(2x − 1) + A3 x(x + 2)
1
and we then substitute x = 0, −2, and :
2
x = 0: 02 + 2(0) − 1 = −1 =
A1 (0 + 2)(2(0) − 1) + A2 (0)(2(0) − 1) + A3 (0)((0) + 2) = −2A1 , so A1 =
1
2
x = −2: (−2)2 + 2(−2) − 1 = −1 =
A1 (−2 + 2)(2(−2) − 1) + A2 (−2)(2(−2) − 1) + A3 (−2)((−2) + 2) = 10A2 , so A2 = −
x = 12 : ( 12 )2 + 2( 12 ) − 1 =
A1 (
1
4
=
1
1
1
5
1
1
1
1
+ 2)(2( ) − 1) + A2 ( )(2( ) − 1) + A3 ( )(( ) + 2) = A3 , so A3 =
2
2
2
2
2
2
4
5
1
Thus
1
1
− 10
x 2 + 2x − 1
5
2
=
+
+
2x 3 + 3x 2 − 2x
x
x + 2 2x − 1
and therefore
2

1
2
1
− 10
1
5

x + 2x − 1
 dx =
dx =  +
+
2x 3 + 3x 2 − 2x
x
x + 2 2x − 1
1 1
1
1
1
1
dx −
dx +
dx =
2 x
10 x + 2
5 2x − 1
1
1
1
ln |x| −
ln |x + 2| +
ln |2x − 1| + C
2
10
10
24
1
10
1
cos x
cos xdx
Example: secxdx =
dx =
dx =
=
2
cos x
cos x
1 − sin2 x
1
u−1
u+1
1
du = −
= − ln
+ C = ln
+C =
2
2
1−u
u −1
u+1
u−1
1
sin x + 1
ln
+C =
2 | sin x − 1|
sin x + 1
sin x + 1
sin x + 1
+ C = ln
+C =
ln
sin x − 1
sin x − 1
sin x + 1
sin x+1 x+1)2
(sin x+1)2
+C =
ln (sin
+
C
=
ln
+
C
=
ln
cos2 x
cos x
sin2 x−1
sin x
1 + C = ln |tan x + secx| + C
ln +
cos x
cos x Second Case: Q(x) = (c1 x + d1 )m1 (c2 x + d2 )m2 · · · (ck x + dx )mk
We then have to write
P (x)
=
Q(x)
A1,m1
A1,m1 −1
A1,1
+
+ ··· +
+
m
m
−1
(c1 x + d1 ) 1
(c1 x + d1 ) 1
c1 x + d1
A2,m2
A2,m2 −1
A2,1
+
+ ··· +
+
m
m
−1
2
2
(c2 x + d2 )
(c2 x + d2 )
c2 x + d2
..
.
Ak,mk
Ak,mk −1
Ak,1
+
+ ··· +
−1
m
m
(ck x + dk ) k
(ck x + dk ) k
ck x + dk
25
Example:
1
dx =
(1 − x 2 )2
We write
1
(x −
1)2 (x
1
(x −
1)2 (x
+
1)2
=
+ 1)2
dx
A1,2
A1,1
A2,1
A22
+
,
+
+
2
2
(x − 1)
x − 1 (x + 1)
x+1
and multiply by (x − 1)2 (x + 1)2 to get
1 = A1,2 (x + 1)2 + A1,1 (x − 1)(x + 1)2 + A2,2 (x − 1)2 + A2,1 (x − 1)2 (x + 1)
Substituting x = −1 and x = 1 gets us two constants quickly:
1 = A1,2 (−1+1)2 +A1,1 (−1−1)(−1+1)2 +A2,2 (−1−1)2 +A2,1 (−1−1)2 (−1+1) =
x = −1:
4A2,2
so A2,2 =
1
4
1 = A1,2 (1 + 1)2 + A1,1 (1 − 1)(1 + 1)2 + A2,2 (1 − 1)2 + A2,1 (1 − 1)2 (1 + 1) = 4A1,2
x = 1:
so A1,2 =
1
4
The quickest way to get the two remaining coefficients is to insert the coefficients just obtained
into the original equation:
1 = A1,2 (x + 1)2 + A1,1 (x − 1)(x + 1)2 + A2,2 (x − 1)2 + A2,1 (x − 1)2 (x + 1)
becomes
1
1
1 = (x + 1)2 + A1,1 (x − 1)(x + 1)2 + (x − 1)2 + A2,1 (x − 1)2 (x + 1)
4
4
or
2
1 = (x + 1)
1
2 1
+ A1,1 (x − 1) + (x − 1)
+ A2,1 (x + 1)
4
4
Differentiating, we get
1
1
0 = 2(x + 1)
+ A1,1 (x − 1) + (x + 1)2 A1,1 + 2(x − 1)
+ A2,1 (x + 1) + (x − 1)2 A2,1
4
4
26
Again we substitute the “illegal” values of x:
x = −1 :
1
1
2
0 = 2(−1+1)
+ A1,1 (−1 − 1) +(−1+1) A1,1 +2(−1−1)
+ A2,1 (−1 + 1) +(−1−1)2 A2,1
4
4
1
1
+ 4A2,1 = −1 + 4A2,1 , so A2,1 =
simplifies to 0 = −4
4
4
x=1:
1
1
+ A1,1 (1 − 1) + (1 + 1)2 A1,1 + 2(1 − 1)
+ A2,1 (1 + 1) + (1 − 1)2 A2,1
4
4
1
1
+ 4A1,1 = 1 + 4A1,1 , so A1,1 = −
simplifies to 0 = 4
4
4
0 = 2(1 + 1)
Thus we have
1
1
1
1
−4
1
4
4
+
=
+
+ 4 =
2
2
2
2
(x − 1) (x + 1)
(x − 1)
x − 1 (x + 1)
x+1
1
(x − 1)−2 − (x − 1)−1 + (x + 1)−2 + (x + 1)−1
4
Therefore:
1
1
−2
−1
−2
−1
(x
−
1)
dx =
dx
=
−
(x
−
1)
+
(x
+
1)
+
(x
+
1)
(1 − x 2 )2
4
1 (x − 1)−2+1
(x + 1)−2+1
− ln |x − 1| +
+ ln |x + 1| + C =
4
−2 + 1
−2 + 1
x + 1
1
1
1
x
x + 1
1 +C
−
−
+ ln +C =
+ ln
4
x−1 x+1
x − 1
2(1 − x 2 ) 4 x − 1 27
cos x
cos x
dx =
Example: I = sec xdx =
dx =
4
cos x
(cos2 x)2
cos x
1 1
u
u + 1 + C =
+
ln
dx
=
du
=
2
(1 − u2 )2
2(1 − u2 ) 4
u − 1
(1 − sin x)2
sin x
sin x + 1 sin x
1 sin x + 1 1 +C
+C =
+ ln + ln sin x − 1 2 cos2 x
4
sin x − 1 2(1 − sin2 x) 4
3
1
dx =
cos3 x
where we have made the substitution u = sin x.
By trigonometric manipulation, this can be worked into the somewhat useless formula we have
already derived using a complicated Integration by Parts calculation:
Example:
2 −6
− (1 − u )
sec3 xdx =
−11
(sin x)
u
−10
−10
(cos x)
du = −
secx tan x + ln |secx + tan x|
+C
2
dx =
(sin x)−12 (cos x)−10 sin xdx =
1
(u −
1)6 (u
+ 1)6 u10
du, (where u = sin x)
We write
1
=
(u − 1)6 (u + 1)6 u10
A1,5
A1,6
A1,4
A1,3
A1,2
A1,1
+
+
+
+
+
+
6
5
4
3
2
(u − 1)
(u − 1)
(u − 1)
(u − 1)
(u − 1)
u−1
A2,5
A2,4
A2,3
A2,2
A2,1
A2,6
+
+
+
+
+
+
6
5
4
3
2
(u + 1)
(u + 1)
(u + 1)
(u + 1)
(u + 1)
u+1
A3,10 A3,9 A3,8 A3,7 A3,6 A3,5 A3,4 A3,3 A3,2 A3,1
+ 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 +
u10
u
u
u
u
u
u
u
u
u
28
Third Case: Q(x) = (a1 x 2 + b1 x + c1 )(a2 x 2 + b2 x + c2 ) · · · (ak x 2 + bk x + ck ) is the product
of distinct irreducible quadratic factors.
Then we can write
P (x)
P (x)
=
=
2
2
Q(x)
(a1 x + b1 x + c1 )(a2 x + b2 x + c2 ) · · · (ak x 2 + bk x + ck )
A1 x + B1
A2 x + B2
Ak x + Bk
+
+ ··· +
2
2
a1 x + b 1 x + c 1 a2 x + b 2 x + c 2
ak x 2 + bk x + ck
Example:
1
dx =
(x 2 + 1)(x 2 + 4)
A1 x + B 1
dx +
x2 + 1
A 2 x + B2
dx
x2 + 4
has to be solved for A1 , B1 , A2 , and B2 . Multiplying by the common denominator (x 2 +1)(x 2 +4),
we get
1 = (A1 x + B1 )(x 2 + 4) + (A2 x + B2 )(x 2 + 1)
There are two ways of solving this for the desired constants. One is to use the fact that the
coefficients of the two polynomials must be equal so as to get 4 equations in 4 unknowns:
1 = 0x 3 + 0x 2 + 0x + 1 = (A1 x + B1 )(x 2 + 4) + (A2 x + B2 )(x 2 + 1) =
A1 x 3 + B1 x 2 + 4A1 x + 4B1 + A2 x 3 + B2 x 2 + A2 x + B2 =
(A1 + A2 )x 3 + (B1 + B2 )x 2 + (4A1 + A2 )x + (4B1 + B2 ) gives us
A1 + A2 = 0, B1 + B2 = 0, 4A1 + A2 = 0, 4B1 + B2 = 1
Using B2 = −B1 , we get 3B1 = 1, so B1 =
1
1
and B2 = − .
3
3
Usin A2 = −A1 , we get 3A1 = 0, so A1 = A2 = 0. Thus we have
1
1
1
1
1
dx −
dx =
dx =
2
2
2
2
(x + 1)(x + 4)
3 x +1
3 x +4
1
11
x
1
1
x
arctan x −
arctan + C = arctan x − arctan + C
3
32
2
3
6
2
29
An alternative methof of solving
1 = (A1 x + B1 )(x 2 + 4) + (A2 x + B2 )(x 2 + 1) for the desired constants is to substitute artfully
selected values of x. In this case we shall use complex numbers:
x=i: Then we have
1 = (A1 i + B1 )(i2 + 4) + (A2 i + B2 )(i2 + 1) =
(A1 i + B1 )(−1 + 4) + (A2 i + B2 )(−1 + 1) = 3(A1 i + B1 )
or 1 + 0i = 3B1 + 3A1 i, which immediately gives us A1 = 0 and B1 =
1
3
x=2i: Then we have
1 = (A1 2i + B1 )((2i)2 + 4) + (A2 2i + B2 )((2i)2 + 1) =
(2A1 i + B1 )(−4 + 4) + (2A2 i + B2 )(−4 + 1) = −3(2A2 + B2 )
or 1 + 0i = −3B2 − 6A2 i, which immediately gives us A2 = 0 and B2 = −
1
3
Fourth Case:
Q(x) = (a1 x 2 + b1 x + c1 )n1 (a2 x 2 + b2 x + c2 )n2 · · · (ak x 2 + bk x + ck )nk is the product of not
necessarily distinct irreducible quadratic factors.
Then we can write
(a1
x2
+ b1 x + c 1
)n1 (a
2
x2
P (x)
=
+ b2 x + c2 )n2 · · · (ak x 2 + bk x + ck )nk
A1,n1 x + B1,n1
A1,n1 −1 x + B1,n1 −1
A1,2 x + B1,2
A1,1 x + B1,1
+
+···+
+
+
2
n
2
n
−1
2
2
1
1
(a1 x + b1 x + c1 )
(a1 x + b1 x + c1 )
(a1 x + b1 x + c1 )
a1 x 2 + b1 x + c1
A2,n1 x + B2,n2
A2,n2 −1 x + B2,n2 −1
A2,2 x + B2,2
A2,1 x + B2,1
+
+
·
·
·
+
+
+
(a2 x 2 + b2 x + c2 )2 a2 x 2 + b2 x + c2
(a2 x 2 + b2 x + c2 )n1 (a2 x 2 + b2 x + c2 )n2 −1
.
+ .. +
Ak,nk x + Bk,nk
Ak,nk −1 x + Bk,nk −1
Ak,2 x + Bk,2
Ak,1 x + Bk,1
+
+ ··· +
+
2
n
2
n
−1
2
2
(ak x + bk x + ck ) k
(ak x + bk x + ck ) k
(ak x + bk x + ck )
ak x 2 + bk x + ck
30
Example:
We write
1
(x 2
+
1)2 (x 2
+ 4)2
dx
1
=
(x 2 + 1)2 (x 2 + 4)2
A1,2 x + B1,2 A1,1 x + B1,1 A2,2 x + B2,2 A2,1 x + B2,1
+
+
+
(x 2 + 1)2
x2 + 1
(x 2 + 4)2
x2 + 4
and multiply by the common denominator to get
1=
(A1,2 x + B1,2 )(x 2 + 4)2 + (A1,1 x + B1,1 )(x 2 + 1)(x 2 + 4)2 +
(A2,2 x + B2,2 )(x 2 + 1)2 + (A2,1 x + B2,1 )(x 2 + 1)2 (x 2 + 4)
which will result in 8 equations in 8 unknown constants. Most reasonable people would have
this computation done by a CAS: a Computer Algebra System.
31
Improper Integrals
So far, definite integrals have been used to compute areas of finite regions. It is possible to
extend the notion of area to regions lying under the graph of a function over infinite intervals,
and to functions which have vertical asymptotes.
Definition: Improper Integrals of Type I are defined to be those of the form:
∞
−∞
f (x)dx = lim
T →∞
a
∞
T
f (x)dx =
Example:
1
=
1 − lim
T →∞ T
a
−∞
∞
f (x)dx +
a
f (x)dx
∞
a
x −2 dx = lim
−∞
f (x)dx = lim
T →−∞
a
T
f (x)dx
f (x)dx, if both limits exist.
T
T →∞
1
a
1
T
−1 T
x
−1 −1 −1
−2
= lim
−
=
x dx = lim
= Tlim
T →∞ −1 T →∞ x
→∞
T
1
1
1
1−0=1
Example:
Example:
∞
1
x
−1
−∞
−1
x
dx = lim
T
T →∞
−1
dx = lim
1
x −1 dx = lim ln |x||T1 = lim ln T − ln 1 = ∞
T →∞
−1
T →−∞
−∞
T →∞
x −1 dx = lim ln |x||−1
T = lim ln | − 1| − ln |T | = −∞
T →−∞
T →−∞
If the limit defining an improper integral exists, we say that the integral converges or is convergent, otherwise we say it diverges or is divergent.
Example:
∞
0
If s is considered to be a constant,
e−sx dx = lim
T →∞
T
0
e−sx dx = lim
T →∞
T
1 −sx = lim 1 e−s(T ) − 1 e−s(0) = 1 − lim 1 e−sT = 1
e
T →∞ −s
−s
−s
s T →∞ −s
s
0
32
Definition: If f is defined on the interval 0, ∞), and treating s as a constant in integrating, we
can define a function of F (s) of s by letting
F (s) =
∞
0
e
−sx
f (x)dx = lim
T
T →∞
0
f (x)e−sx dx
This function is called the Laplace Transform of f . In the previous example we showed that
the Laplace transform of the constant function 1 is 1s . Most of the integration techniques
learned in this course will be used far more often in practical applications of Calculus via the
Laplace transform than in the computation of areas.
Example:(closely related to Example 2, p.489 of brown Stewart)
∞
0
xe
−sx
dx = lim
T →∞
We integrate
xe
−sx
dx =
x
1
− e−sx +
s
s
so
T
T
0
xe−sx dx
xe−sx dx by parts with u = x, dv = e−sx dx, so that du = dx and v = − 1s e−sx :
udv = uv −
1 −sx
1
vdu = x − e
− − e−sx dx =
s
s
x
1
e−sx dx = − e−sx − 2 e−sx
s
s
T
x −sx
1 −sx =
lim
xe
dx = lim − e
− 2e
T →∞ 0
T →∞
s
s
0
T −sT
0 −sx
1
1 −sT
1 −s(0)
lim − e
− − e
= 2
− 2e
− 2e
T →∞
s
s
s
s
s
−sx
33
Note: Examples 3 & 4 of brown Stewart 7.9 are very important:
0
T
∞
dx
dx
dx
= lim
+ lim
=
2
2
T →−∞ T 1 + x
T →∞ 0 1 + x 2
−∞ 1 + x
lim arctan x|0T + lim arctan x|T0 =
T →−∞
T →∞
lim (arctan 0 − arctan T ) + lim (arctan T − arctan 0) =
T →−∞
T →∞
0 − lim arctan T + lim arctan T − 0 =
T →−∞
T →∞
π
π
− −
=π
2
2
∞
1
dx
= lim
T →∞
xp
T
1
−p+1 T
x
=
x −p dx = lim
T →∞ −p + 1 1
1−p+1
T −p+1
−
=
T →∞ −p + 1
−p + 1
lim
T 1−p
1
+ lim
=
p − 1 T →∞ 1 − p
1
if and only if p > 1
p−1
The other type of improper integral occurs when the integrand f (x) has a vertical asymptote
at say x = c.
Then, assuming a < c < b, we define
c
T
f (x)dx = lim−
f (x)dx
a
T →c
b
a
c
f (x)dx = lim+
T →c
and
b
c
b
f (x)dx =
f (x)dx +
f (x)dx (if both integrals exist).
a
a
c
34
b
T
f (x)dx
Comparison Theorems: If we have f (x) ≤ g(x) on the (possibly infinite) interval (a, b),
b
b
then we know that
f (x)dx ≤
g(x)dx. Thus we can conclude for proper integrals that
if
b
a
a
f (x)dx diverges, then so must
b
so must a f (x)dx.
a
b
a
g(x)dx, and conversely, if
b
a
g(x)dx converges, then
1
Example: The region R = (x, y)|x ≥ 1, 0 ≤ y ≤
is revolved about the x-axis to form
x
a solid object, known as Gabriel’s Horn. It volume is
T
∞
T
−1 T
π
x
−1 −2
= π lim −1 − −1 = π
= π lim
V =
dx = π lim
x dx = π lim
2
T →∞ 1
T →∞ −1
T →∞ x
T →∞ T
1
1 x
1
1
However, as we shall soon see, the formula for its surface area is
∞
∞
1
1
1
2π
1 + 4 dx ≥
2π = ∞ !!!
S=
x
x
x
1
1
Thus we now have a pattern for a paint bucket whose volume is finite, but which cannot be
painted because its surface area is infinite!
35
Approximate Integration
We have already seen how Riemann sums are approximations to definite integrals. Recall that
the regular partition P of [a, b] with n intervals is formed by letting
b−a
, xj = a + j∆x, so that x0 = a,,x1 = a + ∆x, …, xn−1 = b − ∆x, xn = b. We have
n
||P|| = ∆x.
∆x =
The formulas for the special Riemann sums are:
(1) The Left-hand sum of f :
n
Ln (f ) =

f (xi−1 )∆x = [f (x0 ) + f (x1 ) + · · · + f (xn−1 ] ∆x =
i=1

b
−
a
 f a + (i − 1)
b−a
n
n
i=1
n
(2) The Right-hand sum of f :
n
Rn (f ) =


f (xi )∆x = [f (x1 ) + f (x2 ) + · · · + f (xn )] ∆x =
i=1
n

f a+i
i=1
b−a b−a
n
n
(3) The Midpoint sum :
xi−1 + f (xi )
Mn (f ) =
∆x =
f
2
i=1
x0 + x1
x 1 + x2
xn−1 + xn
f
+f
+ ··· + f
∆x =
2
2
2


n
b
−
a
1
 f a+ i−
b−a
2
n
n
i=1
n
36
(4) The Inscribed sum of f : In (f ) = [m1 + m2 + · · · + mn ]∆x
(5) The Exscribed sum of f : En (f ) = [M1 + M2 + · · · + Mn ]∆x
The Trapezoidal Estimate or Trapezoidal Rule Tn (f ) =
2f (xn − 1) + f (xn ))∆x =

1
(f (x0 ) + 2f (x1 ) + 2f (x2 ) + · · · +
2

b
−
a
b−a
f (a) + 2
f a+i
+ f (b)
n
n
i=1
n−1
which is the sum of the areas of the trapezoids passing (xi−1 , 0)(xi−1 , f (xi−1 ))(xi , f (xi ))(xi , 0)
turns out to be the average of the left and right hand sums:
n
Ln (f ) + Rn (f ) =
n
f (xi−1 )∆x +
i=1

f (a) + 2
n−1
f a+i
i=1
f (xi )∆x =
i=1

b−a
b−a
+ f (b)
= Tn (f )
n
n
We would like to know how close these sums are to the actual value of
b
b
approximation to
f (x)dx, we define E(A) = f (x)dx − A)
a
a
b
a
f (x)dx. If A is an
So far, we only know the following theorem:
Theorem
If f has continuous derivative f on [a, b], and |f (x)| < M for all x
in [a, b], then En (f ) − In (f ) ≤ M
(b−a)2
,
n
so
lim En (f ) − In (f ) = 0
n→∞
This was used to prove part of the Fundamental Theorem of Calculus, but is not particulary
useful because of the difficulty of finding the Incribed and Exscribed Sums. We
 can make
 some
n
n
b−a
use of it by noting that for any Riemann sum Rn (f ) =
f (ti )∆x =
f (ti ) over a
n
i=1
i=1
regular partition we have In (f ) ≤≤ En (f ), so the theorem tells us that E(Rn (f )) ≤ M
(b − a)2
.
n
In terms of the examples just preceding the Substitution Method, we have, letting y = f (x) =
b
1
x 2 , a = 0, b = 1, (so that a f (x)dx = 3 ),
37
Ln (f ) =
1
1
1
1
1
−
+
−
, so E(Ln (f )) =
2
3 2n 6n
2n 6n2
Rn (f ) =
1
1
1
1
1
+
+
+
, so E(Rn (f )) =
3 2n 6n2
2n 6n2
Tn (f ) =
Ln (f ) + Rn (f )
1
1
1
,
so
E(T
(f
))
=
= +
n
2
3 6n2
6n2
Mn (f ) =
1
1
1
−
, so E(Mn (f )) =
2
3 12n
12n2
We state without
proof some &useful facts: If f exists and is continuous on [a, b], and if
% M2 = max |f (x)| : a ≤ x ≤ b , then
E(Tn (f )) ≤ M2
(b − a)3
(b − a)3
and
E(M
(f
))
≤
M
n
2
12n2
24n2
In our example, we have f (x) = 2, so we take M2 = 2 and get
E(Tn (f )) ≤ 2
(1 − 0)3
1
(1 − 0)3
1
(b − a)3
=
and
E(M
(f
))
≤
M
=
2
=
n
2
2
2
2
2
12n
6n
12n
24n
12n2
Example:
3
Use the Trapezoidal and Midpoint Rules to estimate
42 − x 2 dx to within
0
0.001 of the true value.
Solution:
12
We have f (x) = 42 − x 2 ,
− 12
1
x
16 − x 2
(−2x) = −
1 ,
2
(16 − x 2 ) 2
1
1
1
2 2
2 −2
16
−
x
(1)
−
(x)
(−2x)
16
−
x
16 − x 2 + x 2
16
2
f (x) = −
=
−
=−
3
3
2
16 − x
(16 − x 2 ) 2
(16 − x 2 ) 2
'
(
'
(
16
1
16
so M2 = max −
= 16 max
= √
3 : 0 ≤ x ≤ 3
3 : 0 ≤ x ≤ 3
(16 − x 2 ) 2 7 7
(16 − x 2 ) 2
f (x) =
16 (3 − 0)3
36
36000
√
= √ 2 < 0.001 if n2 >
or n >
Thus E(Tn (f )) ≤ √
2
7 7 12n
7 7n
7 7
36000
√
44.1. We
7 7
10.797845 + 10.707562
take n = 45 and get L45 (f ) = 10.797845, R45 (f ) = 10.707562, so T45 (f ) =
=
2
10.752704.
For the midpoint rule, we take n = 23 and get M23 (f ) = 10.753927. The actual value (to 15
decimal places) is 10.753123598448735.
38
Simpson’s Rule
A more sophisticated approach is make n be even and to approximate the graph of y =
f (x) on the intervals [xi , xi+2 ] with the parabolas passing throught the points (xi , f (xi )),
(xi+1 , f (xi+1 ), and (xi+2 , f (xi+2 ). The result is the approximation
Sn (f ) = [f (x0 ) + 4f (x1 ) + 2f (x2 ) + 4f (x3 ) + · · · + 2f (xn−2 ) + 4f (xn−1 ) + f (xn )]
%
&
If f exists and is continuous on [a, b], and if M4 = max |f (x)| : a ≤ x ≤ b ,
E(Sn (f )) ≤ M4
b−a
3n
(b − a)5
180n4
Simpson’s Rule can be expressed in terms of the Trapezoidal and Midpoint Rules:
Sn (f ) =
Tn (f ) + 2Mn (f )
3
In our first example, we have f (x) = 0, so we take M4 = 0 and get
E(Sn (f )) ≤ M4
Using Sn (f ) =
(b − a)5
=0
180n4
Tn (f ) + 2Mn (f )
, we get Sn (f ) =
3
1
3
+
1
6n2
+ 2( 13 −
3
1
)
12n2
=
1
.
3
3
In our second example, we have, using f (x) = −16(16 − x 2 )− 2 ,
5
3
x
(16 − x 2 )− 2 (−2x) = −48
(x) = −16 −
5
2
(16 − x 2 ) 2
f
f
5
(x) = −48
5
3
(16 − x 2 ) 2 (1) − x 2 (16 − x 2 ) 2 (−2x)
= −48
16 − x 2 + 5x 2
= −192
4 + x2
7
7
(16 − x 2 )5
(16 − x 2 ) 2
(16 − x 2 ) 2
'
(
'
(
4 + x2 4 + x2 :0≤x≤3 ≤
Thus M4 = max −192
= 192 max 7 : 0 ≤ x ≤ 3
(16 − x 2 ) 72 (16 − x 2 ) 2 192
4 + 32
7
7
2
=
2496
√ 2.75
343 7
Thus E(Sn (f )) ≤ M4
9
1.237
(3 − 0)5
9
= M4
< 2.75
=
< 0.001 if n4 > 1237 or n > 5.9.
4
4
4
180n
20n
20n
n4
39
11.06815 + 10.39103
=
2
10.72959 + 2(10.764584)
T6 (f ) + 2S6 (f )
10.72959. Also, M6 (f ) = 10.764584, so S6 (f ) =
=
=
3
3
32.258758
= 10.752919
3
We take n = 6 and get: L6 (f ) = 11.06815, R6 (f ) = 10.39103, so T6 (f ) =
Is it worth any extra effort to get a better estimate of M4 ? The answer is that it depends on
difficult the expression for f (x) is. We obtained a good guess for the maximum value of
|f (x) by substituting x = 3 into it. It is now easy to sketch the graphs of functions with
computers:
y
0
-1 0
-2
-3
x
1
2
3
The graph confirms the suspicion that the maximum value of |f (x)| occurs at x = 3.
1
Example:
Evaluate
Solution:
The error must be less than
0
2
e−x dx to 5 decimal place accuracy.
1
1
2
0.00001 =
. With f (x) = e−x , we have
2
20, 000
2
f (x) = −2xe−x , and
2
2
f (x) = −2e−x − 2x(−2x)e−x = −2e−x
2
1 − 2x 2 ,
2
2
2
f (x) = −2 (−2x)e−x 1 − 2x 2 + e−x (−4x) = 4e−x (3x − 2x 3 ),
2
f (x) = 16x 2 e−x (x 2 − 3)
y
2
0
-2 0
-4
-6
-8
-10
-12
-14
x
1
2
3
40
We have M4 ≤ 13, so E(Sn (f )) ≤ M4
n4 >
(1 − 0)5
13
1
if
=≤
<
4
4
180n
180n
20, 000
260, 000
26, 000
13, 000
=
=
1444.4 or n > 6.16, so we take n = 8.
180
18
9
Then L8 (f ) 0.78537315, R8 (f ) 0.7063581,
so T8 (f ) =
L8 (f ) + R8 (f )
= 0.74586564.
2
Also M8 (f ) = 0.7473036,and S8 (f ) = 0.74682426.
As a matter of fact,E(S8 (f )) ≤
of S8 (f ).
13
0.000007, so we expect six decimal place accuracy
180(10)4
Using n = 100, we have
L100 (f ) = 0.7499786,
R100 (f ) = 0.7436574,
T100 (f ) = 0.746818,
M100 (f ) = 0.7468272,
S100 (f ) = 0.7468242574357303.
We have E(S100 (f )) ≤
13
0.0000000007, so our first 9 decimal places are accurate,
180(100)4
i.e.,
1
2
e−x dx = 0.746824257 is a true statement.
0
41