Download Unit IV, Some Topics On Integration

Unit IV
Some Topics on Integration
Unit IV
Some Topics on Integration
4.1 Integration by Trigonometric Substitution
4.1.1 Integration by Partial Fraction
In this section we set up a procedure how to integrate rational functions.
Some Algebraic Preliminaries
The following are some basic facts from algebra that are very important for our discussion.
1. A non-constant polynomial is called reducible if it can be written as the product of two polynomials of
smaller degree; otherwise it is called irreducible.
2. A polynomial of degree 2, ax2 + b x + c is irreducible only if b2 – 4ac < 0.
3. Any polynomial of degree at least 3 is reducible.
Preparing Rational Functions for Integration
We normally use the following procedures:
I. When applicable, divide the numerator of the rational function by its denominator, to insure that the
degree of the numerator of the resulting remainder is less than the degree of the denominator.
Example 1 Integrate
2 x3
 x 2  3 dx .
Solution Degree (2x3) = 3 and degree (x2 + 3) = 2. Hence we need to use long division.
Now
Thus
2 x3
x2  3
2 x3
= 2x −
6x
x2  3
.
x
 x 2  3 dx = 2  x dx − 6  x2  3 dx = x
Now let u = x2 + 3, then du = 2x dx and 6
Therefore,
2 x3
 x 2  3 dx
2
−6
x
x
 x2  3 dx
 x2  3 dx = 3 
du
= 3 ln u + c.
u
= x2 − 3 ln (x2 +3) + c.
II. Factor the numerator and the denominator of the proper rational expression, obtained in 1, into
expressions of the form:
Constants, (x − a)r , and (x2 + bx + c)s where r and s are in N. Factor quadratic factors whenever
possible. Finally, reduce the fraction if any common factors appear in both the numerator and the
denominator.
Example 2 Integrate
2x  4
 x 2  3x  2 dx .
Solution 2x + 4 = 2(x + 2) and x2 + 3x + 2 = (x + 1) (x + 2).
Hence
Therefore,
2x  4
x 2  3x  2
2x  4
=
2
and
x 1
 x2  3x  2 dx = 2 ln
2x  4
dx
 x 2  3x  2 dx = 2  x  1 = 2 ln
x 1 + c.
x 1 + c.
Lemma 4.1 The expression x − a is a factor of P(x) if and only if P(a) = 0.
Proof (  ) If x − a is the factor of P(x), then there is a polynomial R(x) such that
P(x) = (x − a) R(x).
Then P(a) = (a − a) R(a) = 0.
Therefore, P(a) = 0.
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Unit IV
Some Topics on Integration
() Let P(x) = c n x n + c n - 1 x n - 1 + … + c1 x + c0.
Assume that P(a) = 0.
n
Now P(x) = P(x) − P(a) =

j
j
c j (x  a ) =
j 0
but for any integer k  2, we know that
x k  a k = (x − a)
k 1

x
k  j 1
n
 c j (x j  a j)
j 1
j
a .
j 0
n 1

Hence P(x) = (x − a)
j k  j 1
c ja x
.
j 0
Therefore, (x − a) is the factor of P(x).
III. Rewrite the given rational function g(x) as:
n
g (x) = P(x) +
m
ci

i  1 (ax  b)
i
+

r jx  s j
j  1 (ax  b)
j
where ci , c j , r j , s j , a and b are constants and m, n  N.
Example 3 Integrate
2x2  1
 ( x 1)3
Solution Let g (x) =
2x2  1
dx .
.
( x 1)3
A
B
C
Then g(x) =
+
+
 A (x – 1)2 + B (x – 1) + C = 2x2 + 1, x  1,
2
3
x 1
(x 1)
(x 1)
where A, B and C are constants.
Now we need to determine values of the constants A, B and C.
The simples method is to replace x by the values of x for which the denominator is undefined.
If x = 1, then C = 3,
if x = 0, then A – B + C = 1, and hence B – A = 2,
i)
and if x = 2, then A + B + C = 9, consequently B + A = 6.
ii)
From i) and ii) we get:
B – A = 2 and B + A = 6  A = 2 and B = 4.
Thus,
2x2  1
 ( x 1)3
dx = 2 
dx
4
dx
dx
3
+ 4
+ 3
= 2 ln x 1 –
–
+ c,
x 1
x 1 2 (x 1) 2
(x 1)3
(x 1) 2
where c .
4
2x 2  1
3
dx = 2 ln x 1 –
Therefore,
–
+ c, where c .
3
x 1 2 (x 1) 2
(x 1)
2x2  2x  1
dx .
Example 4 Integrate
x3  x 2  2


Solution x3 + x2 – 2 = (x – 1) (x2 + 2x + 2 ).
Since 22 – 4  1 2 = – 4 and ( 2) 2  4  2  1 = – 4, both x2 + 2x + 2 and 2 x 2  2 x  1 are irreducible
and hence 2 x 2  2 x  1 and x2 + 2x + 2 are irreducible polynomials.
Now we need to determine A, B and C such that
2x2  2x  1
x3  x 2  2

=
A
Bx  C
+
2
x 1 x  2 x  2
2 x 2  2 x  1 = A (x2 + 2x + 2) + (B x + C) (x – 1)  x   \ 1.
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Unit IV
Some Topics on Integration
If x = 1, then A =1, if x = 0, then C = 1 and if x = – 1, then B = 1.
Thus
2x2  2x  1

x3  x 2  2
dx =
dx
+
x 1

= ln x 1 +

x 1
x2  2x  2
dx
x 1

dx .
x2  2x  2
Using substitution to integrate the right-hand side integral iii) we get:
iii)
Now let u = x 2  2 x  2 .
Then du = 2 (x + 1) dx and (x + 1) dx =
Thus,

Therefore,
x 1
x2  2x  2

dx =
2x2  2x  1
x3  x 2  2
Example 5 Integrate
1
du.
2
1 du 1
=
ℓn u + c, where c  .
2 u
2
dx = ℓn x 1 +
dx
 x(x 2  1)2


1
ℓn x 2  2 x  2 + c, where c  .
2
.
Solution Determine the constants such that:
1
x( x 2  1)2
=
A
Bx  C
Dx  E
+
+
2
x
( x 2  1)2
x 1
 A (x2 + 1) 2 + B x2 (x2 + 1) + C x (x2 + 1) + x (D x + C) = 1, x  0.
If x = 0, then A = 1. Now replace A by 1 and combine like powers.
Thus (1 + B) x4 + C x3 + (2 + B + D) x2 + (C + E) x = 0.
Hence B = – 1, C = 0, D = – 1 and E = 0.
Therefore,
dx
 x(x 2  1)2
=

dx
–
x
Now let u = x2 + 1, then x dx =
Thus
x
1
du
 x 2  1 dx = 2  u
Therefore,
dx
 x(x 2  1)2
and
x
x dx
 x 2  1 dx –  (x 2  1)2 .
1
du.
2
x dx
 (x 2  1)2
= ln x –
=
1
2
du
 u2 .
1
1
ln (x2 + 1) +
+ c, where c  .
2
2 ( x 2  1)
4.1.2 Integration by Trigonometric Substitution
To evaluate an integral
 f ( x) dx we can make a substitution of the form:
x = g(u) so that dx = g(u) du .
Hence
 f ( x) dx =  f ( g (u)) g  (u) du .
The corresponding definite integral formula is:
g (b )
b
 f ( x) dx =  f ( g (u)) g  (u) du .
a
g (a)
If g is a trigonometric function, then such substitution is called a trigonometric substitution. Trigonometric
substitutions are especially valuable when the integral contains square roots of the form a 2  x 2 ,
a 2  x 2 or x 2  a 2 , where a > 0.
Prepared by Tekleyohannes Negussie
80
Unit IV
Some Topics on Integration
I) Integrals containing a 2  x 2 .
a 2  x 2 with a > 0, then
If an integral contains
let x = a sin u where –


u
so that a cos u  0.
2
2
a 2  x 2 = a cos u.
Then dx = a cos u du and
x
a2  x2
dx

Example 6 Integrate
a
.
2
x 16  x
Solution Let x = 4 sin u, then dx = 4 cos u du.
2
4
Thus

dx
x
2
=
16  x 2
16  x 2
= –
Therefore,
dx

16 x
16  x 2
+ c
16  x 2
+ c, where c  .
16 x
=–
2
x
1
1
csc 2 u du = –
cot u  c

16
16
16  x 2
2
x dx
Example 7 Integrate
.
16  x 2
Solution Let x = 4 sin u, then dx = 4 cos u du.
x

2
x dx

Thus
16  x 2
1
(1– cos 2u) and sin 2u = 2 sin u cos u,
2
since sin 2u =

4

= 16 sin 2 u du
1
1
sin 2 u du = u – sin 2u + c. Hence
2
Therefore,

4
2
x dx
16  x 2
 x
4
= 8 sin – 1   –

x
16  x 2
2
x dx
= 8u – 4 sin 2u + c
16  x 2
x
16  x 2 + c, where c  .
2
5
2
Example 8 Integrate

25  4 x 2 dx .
 5
2
Solution Let 2x = 5 sin u, then 2 dx = 5 cos u du.

5
2
Thus

25  4 x 2 dx =
25 2
cos 2 u du

2


 5
2
25
4

2
 (1  cos 2 u ) du =

2
25
1
(u  sin 2u )
4
2
5
2
Therefore,

 5
2
25  4 x 2 dx =
2x
2

=
5
25  4 x 2
2

2
=
25
.
4
25
.
4
Prepared by Tekleyohannes Negussie
81
Unit IV
Some Topics on Integration
II) Integrals containing a 2  x 2 .
a 2  x 2 with a > 0, then
If an integral contains
let x = a tan u where –


u
so that a sec u  0.
2
2
Then dx = a sec 2 u du and
Example 9 Integrate
x
2
x
a 2  x 2 = a sec u.
dx

a2  x2
a
.
1  x2
Solution Let x = tan u, then dx = sec 2 u du.
1 x 2
Thus

dx
x
2
sec u
 tan 2u
=
1  x2
1
= 
+c =–
sin u
Therefore ,

dx
x
2
16  x 2
Example 10 Integrate

du =
cos u
 sin 2u
x
du
1
1  x2
+c
x
1  x2
+ c, where c  .
x
=–
dx
.
4  16 x 2
Solution Let 2x = tan u, then 2 dx = sec 2 u du.
1 4 x 2
Thus

dx
=
4  16 x 2
1
1
sec u du =
ln sec u  tan u + c

4
4
2x
1
1
ln ( 1 4 x 2  2 x) + c
4
dx
1
=
ln ( 1 4 x 2  2 x) , where c  .
4
4  16 x 2
=
Therefore,

III) Integrals containing x 2  a 2 .
If an integral contains
x 2  a 2 with a > 0, then
let x = a sec u where 0  u < or   u <
3
so that
2
x
x2  a2
x 2  a 2 = a tan u and dx = a tan u sec u du.
3
Example 11 Integrate

a
x2  9
dx .
x
6
Solution Let x = 3 sec u, then dx = 3 sec u tan u du.
3
Thus

6
x2  9
dx = 3
x

= 3 (tan u  u )
3
Therefore ,

6
4
3

 (sec
2
u  1) du
x
x2  a2
2x
4
3
a
=  − 3 3.
x2  9
dx =  − 3 3 .
x
Prepared by Tekleyohannes Negussie
82
Unit IV
Some Topics on Integration
6
Example 12 Integrate
dx

4
3 2 x
.
x2  9
Solution Let x = 3 sec u, then dx = 3 sec u tan u du.

6
3
1
dx
cos 3 u du
Thus
=
4 x2  9
81 
3 2 x

=
x
2

5
4
3


3
3
4
4
x 9
1
1
cos u du −
sin 2 u cos u du

81 
81 


3
3
1 3 3
5 2
1
1
3

.

=
−
=
sin u
sin u
 8

81
12
3

81
81




4
4
6
dx
1 3 3
5 2

.

Therefore, 
=
 8

4
2
81
12


x 9
3 2 x
IV) Integrals containing b x 2  c x  d .
b x 2  c x  d can be expressed as a 2  x 2 ,
By completing the square in x,
x 2  a 2 or
x2  a2
for suitable a > 0 and then use trigonometric substitution.
Example 13 Integrate
dx

9x2  6x  2
.
Solution Since 9 x 2  6 x  2 = (3x + 1)2 + 1,
let 3x + 1 = tan u, then 3 dx = (1 + tan2 u) du.

Thus
dx
9 x2  6 x  2
=

1
ln sec u  tan u + c
3
3x + 1
1
(3x  1) 2  1 and tan u = 3x + 1.
But sec u =
Therefore,
9 x2  6 x  2
= 3 sec u du
dx

9 x2  6 x  2
Example 14 Integrate

=
1
ln 3x 1 (3x 1)2 1 + c.
3
x 2  6 x  5 dx .
Solution x2 + 6x + 5 = (x + 3)2 − 4.
x3
Now let x + 3 = 2 sec u, then dx = 2 sec u tan u du.
Thus

5
x 2  6 x  5 dx = 4  tan 2 u sec u du

2
x  6x  5

= 4 sec 3 u du − 4 sec u du
2
= 2 sec u tan u − 2 ln sec u  tan u + c.
1
x3
= ( x  3) x 2  6 x 5  2 ln

2
2
Therefore,

x2  6x  5
+ c.
2
1
x3
x 2  6 x  5 dx = ( x  3) x 2  6 x 5  2 ln

2
2
Prepared by Tekleyohannes Negussie
x2  6x  5
+ c.
2
83
Unit IV
Some Topics on Integration
4.2 Improper Integrals
b
We have defined
 f ( x) dx only for a continuous function f on [a, b]. From the maximum-minimum
a
theorem such a function f is bounded on [a, b]. i.e. there is a real number M such that f (x)  M,
 x  [a, b].
A function that is not bounded on a given interval I inside its domain is said to be unbounded on I.
f(x) =
1
is an example of a function that is not bounded on + .
x
Definite integrals with either the integrand or the interval of integration is unbounded are called
Improper integral.
4.2.1 Integrals with unbounded integrands.
We say that f is unbounded near c if f is unbounded either on every open interval of the form (c, x) or (x, c).
f(x) =
1
1
and g(x) =
are examples of functions that are unbounded near 0.
x
x
I) Consider a function f that is continuous on (a, b] and unbounded near a.
b
Now f is continuous on [c, b] for any c  (a, b) and hence f ( x) dx is defined for such c. If the one sided
c

limit
b
lim
 f ( x) dx
c  a c
Exists, then we define
b
b
a
a
 f ( x) dx to be that limit and say that the integral  f ( x) dx converges. Otherwise we
b
say that

b
 f ( x) dx = , where f is non-negative.
f ( x) dx diverges and we write
a
a
2
Example 15 Show that
dx

1
1 x (ln x) 2
1
Solution Let f(x) =
x (ln
1
x) 2
converges and compute its value.
. Now f is continuous on (1, 2] and is unbounded near 1.
Let c  (1, 2). We need to compute:
2
dx

1
c x (ln x) 2
1
Let u = ln x  2 , then du =
2
Thus

dx
1
c x (ln x) 2
dx
1
x (ln x) 2
= 2 (ln
1
x) 2
.
.
2
= 2 (ln
1
2) 2 −
2
1
(ln c) 2
c
1
1


2
and
=
2 (ln 2)  2 (ln c) 2 
1

c 1 
c 1 
c x (ln x) 2


lim
2
dx
lim
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84
Unit IV
Some Topics on Integration
2
Therefore,
1
dx

= 2 (ln 2) 2 .
1
1 x (ln x) 2
2
Example 16 Show that

0
ln x
dx diverges.
x
ln x
Solution Let f(x) =
. Then f is continuous on (0, 2] and is unbounded near 0.
x
2
2
lim 1
ln x
2
Now let c  (0, 2).
dx 
(ln x)
c  0  x
c  0 2
c
c
1
lim
1
(ln c) 2 = − 
= (ln 2) 2 −
 2
c0
2
lim
2
Therefore,
ln x
dx diverges.
x

0
b
II) If f is continuous on (a, b) and is unbounded near both a and b, we say that
 f ( x) dx converges if for
a
some point d in (a, b) both integrals
d
b
b
a
d
a
 f ( x) dx and  f ( x) dx converge. Otherwise, we say that  f ( x) dx
diverges.
1

Example 17 Determine whether
3x 2  1
3 x3
dx converges.
x
0
Solution The integrand is unbounded near both 0 and 1 and is continuous on (0, 1).
Let d =
1
. Now we need to analyze the convergence of
3
1
3

0
3x 2  1
3 x3
x
1
dx and 
3x 2  1
3 3
1 x x
dx .
3
For 0 < c < 1 we’ve:
3
1
3

c
2
3x  1
3 x3
1
3
lim
x
dx


1


2
2
3 3
3 3
2
3
3
x x
c c3
=
=
−
2
c
3x 2  1
2
3
and
dx = −
 
c  0 c 3 x3  x
3
2
2 c
c
3 3
3x 2  1
1
x x3 1 =
dx =
For < c < 1, 
3
3 x3  x
2
1
3


2
3

 lim
3

c c

 c 0


2
3 = 2

3


2
2
3 3
c c3 −
3
2
3
and
c 1 

3  lim
3
c c
dx
=

3 3

c 1
2
1 x x

c
lim
3x 2  1
  − 23 = − 23 .
2
3

3
Therefore,
1
3x 2  1
0
3 x3

x
dx = 0 and hence converges.
b
III) If f is continuous on [a, b] except at a point d in (a, b) near which f is unbounded. We say
 f ( x) dx
a
converges if both integrals
d
b
b
a
d
a
 f ( x) dx and  f ( x) dx converge. Otherwise, we say that  f ( x) dx
diverges.
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Unit IV
Some Topics on Integration
2 1
Example 18 Show that x 3 dx converges.
1
1

Solution f(x) = x 3 is continuous on [- 1, 0)  (0, 2] and f is unbounded near 0.
For – 1 < c < 0.

lim
c 0 
c 1
x 3 dx =
1

2
3 3
lim
x

c 0 2
c
=−
1
3
.
2
For 0 < c < 2.
2 1
2 2
3 3
3 3
lim
x 3 dx = lim
=
x
4 .
 c

c0
c0 2
2
 

c


2 1
3
Therefore,
x 3 dx =
1  3 4 and hence converges.
2
1
2
1 
1
Example 19 Show that
 
 dx diverges.
x x 2 

1


Solution f(x) =
1
1
is continuous on [- 1, 0)  (0, 2] and f is unbounded near 0.

x x2
Let – 1 < c < 0.
lim
c 0 
Then
c
1 
1
1

 
 dx = lim   ln c   = − .
2
c 0 
c
x x 
1 

2
1 
1
 
 dx diverges.
x x 2 

1

Therefore,
4.2.2 Integrals over Unbounded Intervals
Integrals of the form

a
a

 f ( x) dx and  f ( x) dx are also called improper integrals.
If f is continuous on [a, ), then

we say
b
b
a
a
 f ( x) dx is a proper integral for any b  a. If  f ( x) dx exists, then
 f ( x) dx converges, and
a

b
lim
 f ( x) dx = b    f ( x) dx .
a
a
Otherwise the integral diverges.
a
The improper integral
f ( x) dx is handled in an analogous way.


Example 20 Find the area of the region bounded by the graph of y = (x + 1) – 3 and the non-negative x-axis.

dx
Solution A =
. For b  0 we get:
3
(x

1
)
0
b
b 1
dx
1
2
2
= ( x 1)
=
(1 − (b 1)
).
3
2
2
(
x

1
)
0
0
b
2
1
dx
lim
lim
(b  1) = 1 .
Hence
=
−
3
b 
b 
2
2
0 ( x 1)



Therefore, A =
1
square units.
2
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Unit IV
Some Topics on Integration

dx
1 
Example 21 Show that
diverges.
x
1
Solution For b  1 we’ve:
b
b
dx
=
2
((1
+
)
−
ln
(1
+
))
= 2 (1 + b ) − 2 ln (1 + b ) − 4 + 2 ln 2.
x
x

1
11  x
= 2 (− 1 + b ) − 2 ln (
b
lim
Now
b 

Therefore,
x
1

We say that
1
dx
1 
dx
1 
lim ( (− 1 + b ) − ln ( 1  b )) = .
b 
2
=2
x
1 b
).
2
diverges.
d


d
 f ( x) dx converges if both  f ( x) dx and  f ( x) dx converge, for some d.

In this case


d


 f ( x) dx =  f ( x) dx +  f ( x) dx .

d
 xe
Example 22 Determine whether
 x2
dx converges.

Solution The integrand is continuous on (− ∞, ∞).
0
Now consider the integrals
 xe
 x2

dx and

0
 xe


and  x e
 x2
 x2
0
Therefore,
dx =
dx =

 xe
 x2
 xe
 x2
dx .
0
0
1  x2 0
1
 x2
lim
lim
x
e
dx
=
=−
e

b
b 2
2
b
b
b
1  x2 b 1
 x2
lim
lim
x
e
dx
=
= .

e
b 
b  2
2
0
0
dx = 0 and hence converges.

4.3 More on Applications
4.3.1 Volume.
I) Cross Section Method.
y
Consider a solid region D. Suppose x  [a, b] the
plane Perpendicular to the x-axis at x intersects D
A(x)
in a plane having a cross sectional area A(x).
If A(x) = A0 for a  x  b, then v = A0 (b − a)
a
and if A(x) =  r2 for a  x  b, then
x
b
x
v =  r2 (b − a) = volume of right circular cylinder
of radius r and height b − a.
Suppose the cross sectional area A of a three dimensional solid D is a continuous function on [a, b].
Let Ρ = x0, x1, x2, x3, … , xn be a partition of [a, b]. For each k  ℕ between 1 and n, let tk be an arbitrary
number in the sub-interval [xk – 1, xk].
As Δxk = xk − xk - 1 tends to zero Δvk = The volume of the part of D between xk and xk – 1
 A(tk)Δxk.
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Unit IV
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Since v 
n

k 1
vk 
V=
n
 A(tk )xk . Which is the Riemann Sum for A on [a, b].
k 1
n
 A(tk ) .  xk
lim
P k 1
b
=
 A( x) dx
a
Therefore if a solid region D has a cross sectional area A(x) for a  x  b and if A is continuous on [a, b],
then we can define the volume of D by the formula
b
V=
a A( x) dx .
Example 23 Find the volume v of the solid D whose cross section at x is semicircular with radius
sin x for 0  x 

.
2
Solution The cross sectional area at x is given by:

A(x) =
1
 sin 2 x and v =
2

2
1
 2  sin
2 x dx
0

2


2
1
2
Hence v =
=
( x  sin 2 x)
(1  cos 2 x) dx =
4 0
4
2
8
0
2
Therefore, v =
cubic units.
8
Note that: We can reverse the role of x and y and integrate with respect to y by the corresponding formula:
d
v=
 A( y) dy
c
Example 24 Find the volume v of the pyramid with square base 3 units on a side with height 4 units.
Solution Let the altitude of the pyramid coincides with the positive y-axis and the base with the x-axis.
Then the cross sectional area A(y) for any y  [0, 4] is:
A(y) =
9
4
4 y
(4 − y)2 , since
=
from similarity theorem for triangles
3
16
s( y)
where s(y)  = side length of the square at y.
=
Hence v =
9
(16 − 8y + y2).
16
4
4
9
2 ) dy = 9 (16 y  4 y 2  1 y 3 ) = 12
(
16

8
y

y
16
3
16 0
0
Therefore, v = 12 cubic units.
II) The Disc Method
When a non-negative continuous function f on [a, b] is revolved about the x-axis it generates a solid
region having cross sections that are circular discs of radius f(x) for each x  [a, b].
Thus A(x) =  (f(x) ) 2 .
b
Therefore, v =
   f ( x) 
2 dx .
a
Example 25 Find the volume of the sphere of radius r.
Solution Let f(x) =
r 2  x 2 for − r  x  r.
Revolve f about the x-axis.
A(x) =  ( r 2  x 2 ) for − r  x  r.
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Unit IV
Some Topics on Integration
1 3
2
2
2
 r  x dx =  (r x  3 x )
r
Hence v = 
r
=
r
Therefore, v =
r
4
 r3
3
4
 r 3 cubic units.
3
Example 25 Find the volume of the frustum of the cone on [0, 1] whose lateral side passes through (0, b)
and (h, a).
 a b 
 x  b for − r  x  r.
 h 
Solution. Let f(x) = 
Now revolve f about the x-axis.
h
h
h
ab

3
a  b 2
=
b

x
dx
(
b

x
)





3 (a  b)
h
 h  
0 
0
4
h
3
3
(a  b )  r 3
=
3
3 ( a  b)
1
Therefore, v =  h (a 2  ab  b 2 ) cubic units.
3
Then v = 
III) The Washer Method.
If f and g are two non-negative continuous functions on [a, b] such that 0  g(x)  f(x) for a  x  b, then
the volume v of the region, bounded by f and g on [a, b], generated by revolving the graphs of f and g
about the x-axis is:
v= 
   f ( x) 
b

2  g ( x)  2 dx .
a
If g(x)  0 for a  x  b, then the method of finding the volume v of the solid is called the Washer Method.
Example 27 Find the volume v of the solid generated by revolving the region between the graphs of

].
4


Solution Observe that tan x  1  x  [0, ] and sin x, cos x  0  x  [0, ].
4
4

Hence cos x − sin x  0  x  [0, ].
4
y = cos x + sin x and y = cos x − sin x about the x-axis on [0,

 (cos x  sin x )
4
Thus v = 
0

4
= 2

2  (cos x  sin x ) 2 dx

 sin 2 x dx
=   cos 2 x
0
Therefore, v =  cubic units.
4 =
0
Note that: The Washer method holds true if and only if both f and g are of the same parity on [a, b].
IV) The Shell Method.
Suppose a rectangle is bounded by the x-axis, the line y = c and the lines x = a and x = b where b  a  0 and
c  0. If we revolve this rectangle about the y-axis, then the volume v of the cylindrical shell is the difference
between the volumes of the outer and inner cylinders.
Thus v =  c (b2 − a2).
Now let f be a non-negative continuous function on [a, b] with a  0.
We wish to define the volume v of the solid region obtained by revolving about the y-axis, the region
between the graph of f and the x axis on [a, b].
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Unit IV
Some Topics on Integration
Let P = x0, x1, x2, . . . ,xn be any partition on [a, b]. For k between 1 and n let tk be the mid-point of
[xk – 1, xk].
 f (tk ) ( xk
lim
Now vk 
2
2
 x k 1 )
y
xk  0
lim  f (t ) ( x  x
x
x
k
k
k 1) ( k  k 1)

xk  0
lim 2  t f (t )  x
k
k
k

xk  0
Thus v =
lim
xk  0
2  t k f (t k )  xk =
x
b
 2 x f ( x) dx
a
Which is the Riemann Sum for 2 x f ( x) on [a, b].
Therefore, v = 2 
b

x f ( x) dx .
a
This method of finding the volume v of a solid is called the Shell Method.
Example 28 Let R be the region between the graphs of y = x + x3 and y = x2 + 4x − 4 on [1, 2]. Find the
volume v generated by revolving R about the y-axis.
Solution Since x + x3  x2 + 4x − 4 x  [1, 2].
2
2
3
2
v = 2 x( x  x  3 x  4) dx = 2 ( x 4  x3  3 x 2  4 x) dx
1
1


2
 x5 x 4
29
3
2 


=2
=

 x  2x
 5

4

 1 10
Therefore, v =
29
 cubic units.
10
Example 29 Find the volume v of the solid generated by revolving the region between the
graphs of y = x  2 and y =
Solution y = x  2 and y =
Since  x  [1, 3],
1
1
( x  2 )2 +
about the y-axis.
2
2
1
1
( x  2 )2 +
 y2 −2y +1 = 0  y = 1.Hence x = 1 or x = 3.
2
2
1
1
( x  2 )2 +
 x2 ,
2
2
3
2
3
1
1

V = 2   ( x  2) 2   x  2  dx =   ( x3  2 x 2  x) dx +   ( x3  6 x 2  9 x) dx
2
2

1
1
2
=
Therefore, v =
7
3
4
   = 
12
4
3
4
 cubic units.
3
4.3.2 Length of Curves
Let f have continuous derivatives on [a, b].
If f is linear, then the length L of the graph of f on [a, b] is given by:
L=
b  a   f (b) 
f (a)  2
If f is not linear, then let P = x0, x1, x2, . . . , xnbe any partition of [a, b] and approximate the length of the
graph of f by a polygonal line whose vertices are (x0, f (x0)), (x1, f(x1)), . . . , (xn, f(xn)).
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Unit IV
Some Topics on Integration
Let Lk be the length of the portion of the graph of f on [x k – 1, xk].
If xk = x k – 1 − xk tends to zero, then
 x  x
   f ( x )  f ( x
)  2
k
k

1
k
k

1

 

Lk 
But by the mean value theorem we get:
f ( xk )  f ( xk 1) = ( xk  xk 1 ) f ' ( xk ) for some tk  ( xk  xk 1 ).


n
 1   f ' (tk )  2
1  f ' (tk ) 2 ( xk  xk 1 ).
Hence Lk 
Thus L 
k 1
n
lim
P  0 k
1
Therefore,
xk , which is the Riemann Sum for

b

1  f ' (tk ) 2 xk =


1  f ' (tk ) 2 on [a, b].
1   f ' ( x)  2 dx .

a
Definition 4.1 Let f have a continuous derivative on [a, b], then the length L of the
graph of f on [a, b] is defined by:
b
L=

1   f ' ( x)  2 dx
a
Example 30 Find the length L of the graph of f on [1, 2] where f(x) = ln x −
Solution f ΄ (x) =
1 1
− x.
x
4
2
2
1 1 2
 1   x  4 x  dx = 
1
1
2

1 1 
=    x  dx =  ln x 

x 4 

1
Thus L =
Therefore, L =
1 1 2
  x  dx
x 4 
x2
8
 2
3

= + ln 2

8
 1
3
+ ln 2 units.
8
Example 31 Find the length L of the graph of f on [0, 1] where f(x) =




1
1 3
1
x + x − tan x .
4
3

2

2
 4 x2  1  1 
and (f ΄ (x)) = 
 .
 4 x2  1 
4 x2  1


 
Solution f ΄ (x) = x 2  1 −
Thus
1 2
x .
8
1

2


2

2
1
 4 x2  1  1 
1 
 x2  1 
.

2
2
4 x 1
 4 x 1 



1
Hence L =
0




1
x 2  1 dx 
1
 

dx
4  x 2  1
0
1 1

1 3 1
1

1
tan x
= x x 
+
=
+
3  0
4

0 4 16

1
Therefore, L =
+
units.
4 16
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