6 In this chapter 6A Substi tut ion where the derivative is present in the integrand 6B Li near subst it ut ion 6C Anti deri vati ves in volvi ng trigonometric identities 6D Anti diffe ren tiati on using partial fractions 6E De finit e inte gral s 6F Ap pli cat ions of inte gration 6G V olume s of solids of revolution 6H Ap proxi mate eval uation of definite integrals and areas VCE coverage Area of study Units 3 & 4 • Calculus Integral calculus
derivative is present in the integrand
6B Linear substitution
6E Definite integrals
6H Approximate evaluation
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Integration techniques and applications You will have seen in your
Maths Methods course and elsewhere that some functions
can be antidifferentiated (integrated) using standard rules. These
common results are
shown in the table below where the
function f ( x ) has an antiderivative
F ( x ).
In this chapter you will learn how to find antiderivatives of more
complex functions
using various techniques.
Technique 1: Substitution where the derivative is present in the
integrand
Since , n
then it follows that:
Since ; f ( x ) ≠ 0
then it follows that .
The method relies on the derivative, or multiple of the derivative,
being present and
recognisable. Then, the appropriate substitutions may be made
according to the above
rules.
a --- c+
a --- c+
a --- c+
d x ---------------------------- n 1+
( )[ ] n=
f ′ x ( ) f x ( )[
]n d x ∫ f x ( )[ ]n 1+
n 1+ ------------------------- c n 1.–≠,+=
d loge f x ( )[ ] d x
----------------------------- f ′ x ( )
f x ( ) ------------=
f ′ x ( ) f x ( )
------------∫ d x loge f x ( )
c+=
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Find the antiderivative of the following expressions.
a ( x + 3)7 b 4 x(2 x2 + 1)4
c
THINK WRITE
x + 3 is 1. Let
u = x + 3.
a Let u = x + 3.
Find .
Make d x the subject. or
d x = du
Substitute for x + 3 and d x . So
∫ ( x + 3)7 d x
= ∫ u7 du
Antidifferentiate with respect to u. =
Replace u with x + 3 and state
answer
in terms of x.
2 x 2 + 1. Let
u = 2 x 2 + 1.
b Let u = 2 x 2 + 1.
Find .
Make d x the subject. or
Substitute u for 2 x 2 + 1 and for
d x . So
∫ 4 x (2 x 2 + 1)4 d x
= ∫ 4 x u4
out the 4 x .
Replace u with 2 x 2 + 1. =
+ c
c Recognise that 3 x 2 + 1 is the
derivative
of x 3 + x . Let
u = x 3 + x .
c Let u = x 3 + x.
Find .
5 -------------------------
1
1WORKEDExample
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THINK WRITE
Substitute
u for x 3 + x and
for
d x .
Antidifferentiate with respect to u. =
Replace u with x 3 + x .
=
Express in root notation. =
3 d x du
3 x 2 1+ ------------------=
∫ 3 x
2 ---
c+
Antidifferentiate the following functions with respect
to x.
a b
THINK WRITE
Recognise that x + 3 is half of the
derivative of x 2 + 6 x .
Let u = x 2 + 6 x . Let
u = x 2 + 6 x.
Find .
Make d x the subject. or
Substitute
u for x 2 + 6 x and for
d x . So =
Factorise 2 x + 6. =
form on the numerator. =
f x( ) x 3+ x2 6 x+( )3
--------------------------= f x( ) x2 1–( ) cos
3 x x3–( )=
1 ∫ x 3+
2
3
5 d x du
x 2 6 x +( )3
-------------------------d x ∫ x
3+
u3 ------------
du
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C h a p t e r 6 I n t e g r a l c a l c u l u s 215
THINK WRITE
Replace u with x 2 + 6 x .
=
Express the answer with a positive
index number. (Optional.)
Recognise that x 2 − 1 is a multiple of
the derivative of 3 x − x 3.
Let u = 3 x − x 3. Let
u = 3 x − x 3.
Find .
Substitute u for
3 x − x 3 and
for d x . So
=
Antidifferentiate with respect to u. =
Replace u with 3 x − x 3.
=
9 u 2–
4 ---------------------------– c+
11 1
4 x 2 x +( )2 -------------------------
c+–
1 ∫ x 2 1–( ) cos 3 x x 3–(
) d x
2
3
6 du
3 3 x 2– ----------------- ∫ x 2 1–(
) cos 3 x x 2–( ) d x
∫ x 2 1–( ) cos u du
3 3 x 2– -----------------×
3 1 x 2–( ) ----------------------×
3 -------------------------------- c+
Evaluate the following indefinite integrals.
a ∫ cos x sin4 x d x b c
d ∫ sin2 x cos 3 x d x
THINK WRITE
a Recognise that cos x is the derivative of
sin x . a
Let u = sin x . Let
u = sin x.
Tan 1– x
x ---------------- d x
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216 S p e c i a l i s t M a t h e m a t i c s
THINK WRITE
Make d x the subject. or
Substitute u for sin x and for d x .
So∫ cos x sin4 x d x
= ∫ Cancel out cos x . = ∫
Antidifferentiate with respect to u. =
u5 + c
Replace u with sin x . =
sin5 x + c
b Recognise that is half of the
derivative of Tan−1 .
Find .
for d x .
So d x
c Recognise that is the derivative of loge4 x .
c
Let u = loge4 x . Let
u = loge4 x.
Find .
Make d x the subject. or
d x = x du
3 du
d x ------
cos x ------------
2 --------------------------=
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C h a p t e r 6 I n t e g r a l c a l c u l u s 217
THINK WRITE
Substitute u for
loge4 x and x du for
d x
in the integral. So
Replace u by loge4 x . =
(loge4 x )2 + c
d Express cos3 x as
cos x cos2 x .
d ∫ sin2 x cos3 x d x
=
∫ sin2 x cos x cos2 x d x
Express cos x cos
2
x )(using the identity
sin2 x + cos2 x = 1). =
∫ sin 2
x cos x (1 − sin 2
x ) d x
Let u = sin x as its derivative is a
factor
of the new form of the function.
Let u = sin x.
Make d x the subject. or
=
Cancel out cos x . = ∫ u2(1 − u2)
du
Expand the integrand.
= ∫ (u2 − u4) du
Antidifferentiate with respect to u. = u3 −
u5 + c
Replace u by sin x . =
sin3 x −
sin5 x + c
5 ∫ loge4 x
x ----------------d x
and f (0) = 5,
find f ( x).
THINK WRITE
Express f ( x ) in integral notation.
f ( x )
= ∫ 4 xe x 2 d x
Recognise that 4 x is twice the
derivative of x 2.
Find .
1
2
3
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Substitution where the
integrand
1 Find the antiderivative for each of the following
expressions.
a 2 x ( x 2 + 3)4 b 2 x (6
− x 2)−3
c 3 x 2( x 3 − 2)5 d
2( x + 2)( x 2 + 4 x )−3
e f
g 3 x 2( x 3 − 5)2 h
i 4 x 3e x 4 j
(2 x + 3)
sin( x 2 + 3 x − 2)
k (3 x 2 + 5)
cos( x 3 + 5 x ) l
cos x sin3 x
THINK WRITE
Make d x the subject. or
Substitute u for x 2 and for d x .
So f ( x ) = ∫ 4 xe
u
Antidifferentiate with respect to u.
= 2eu + c
Replace u by x 2. f ( x )
= 2e x 2 + c
Substitute x = 0 and f (0) = 5.
f (0) = 2e0 + c = 5
Solve for c. 2 + c = 5
c = 3
State the function f ( x ).
Therefore f ( x )
= 2e x 2 + 3.
5 d x du
d f x ( )[ ]n 1+
d x ---------------------------- n 1+(
) f ′ x ( ) f x ( )[ ]n n
1–≠,=
f ′ x ( ) f x ( )[
]n d x ∫ f x ( )[ ]n 1+
n 1+ ------------------------- c n 1–≠,+=
d loge f x ( )
f ′ x ( ) f x ( )
------------∫ d x loge f x ( )
c+=
remember
6A
2 x 5+( ) x 2 5 x + 2 x
3–
x 2 3 x –( )4 ------------------------
3 x 2 4 x +
x 3 2 x 2+ ------------------------
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C h a p t e r 6 I n t e g r a l c a l c u l u s 219
2 Given that the derivative of
( x 2 + 5 x )4 is
4(2 x + 5)( x 2 + 5 x )3,
then the antiderivative of
8(2 x + 5)( x 2 + 5 x )3 is:
3
a The integral d x can be found by making the
substitution ‘u’ equal to:
b After the appropriate substitution the integral becomes:
c Hence the antiderivative of is:
4 Antidifferentiate each of the following expressions with respect
to x .
m −sin4 x cos x n
o sec2 x tan3 x p
A 2( x 2 + 5 x )4 + c B
( x 2 + 5 x )4 + c C
4( x 2 + 5 x )4 + c
D 2( x 2 + 5 x )2 + c E
( x 2 + 5 x )2 + c
A x 2 B x C D
x 2 + 3 E 2 x
A B C
D E
a 6 x 2( x 3 − 2)5 b x (4
− x 2)3
c x 2( x 3 − 1)7 d
( x + 3)( x 2 + 6 x − 2)4
e
( x + 1)( x 2 + 2 x + 3)−4
f
g h
i
(6 x − 3)e x 2 − x + 3
j x 2e x + 2
k ( x + 1)
sin( x 2 + 2 x − 3)
l ( x 2 − 2)
cos(6 x − x 3)
m sin 2 x cos42 x n cos
3 x sin23 x
o p
loge x
2 ---
2 ---
c+
2 x 5–
x 2 5 x – 2+( )6
---------------------------------- x 2 1–( ) 4
3 x – x 3+
loge3 x
2 x ----------------
x 2
x –-----------------------------------------------------
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5 Evaluate the following indefinite integrals.
6 Find the antiderivative for each of the following
expressions.
7 If and f (2) = 1
find f ( x ).
8 If and f (0) = 3
find f ( x ).
9 If g(1) = −2 and then find g( x ).
10 If and g′( x ) = 16 sin x cos
3
x then find g( x ).
a b
c d
e f
g h
i j
k l
m n
e f
g h
i j
k sin3 x cos2 x l
cos3 x sin4 x
m
2 --- d x ∫ x 1 x 2–
d x ∫
e x 3 2e x +( )4 d x ∫
sin x
cos3 x ------------- d x ∫
x 2 sin x 3 d x ∫
sin x e cos x d x ∫
cos x loge sin x ( )
sin x -------------------------------------------
d x ∫ e3 x 1
e3 x –( )2 d x ∫ 2– Cos 1–
x
3 ---
9 x 2– ---------------------------
d x ∫ 2 x 1+( ) x
x 2 3–+ d x ∫ x 2+( ) cos
x 2 4 x +( ) d x ∫
e x 1+
x 1+ ---------------- d x ∫
Sin 1– 4 x
1 x 2+
------------------- d x ∫
x
cos x
e2 x
sin x cos x +
------------------------------
x --------------------=
g
π
4---
0=
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Technique 2: Linear substitution
For antiderivatives of the form where g( x ) is a linear
function,
that is, of the type mx + c,
and f ( x ) is not the derivative of
g( x ), the substitution
u = g( x )
is often successful in finding the integral. Examples of this type
of integral are:
1. . In this example f ( x ) = 1 and
g( x ) = 4 x + 1 with
n = . By letting
u = 4 x + 1, and consequently
d x = du, the integral becomes which can
be
readily antidifferentiated.
2. . In this example f ( x )
= 4 x and g( x )
= x − 3 with n = 4. By
letting
u = x − 3, the function
f ( x ) can be written in terms of u, that is,
u = x − 3, thus
4 x
x = d
can be readily antidifferentiated.
The worked examples below illustrate how the use of the
substitution u = g( x )
simplifies integrals of the type .
f x ( ) g x ( )[ ]n d x n
0≠,∫
4 x 1+ d x ∫ 1
2 ---
1
4 ---
1
4 u
3+( ) u
d u
×∫ f x ( ) g x ( )[
]n d x ∫
i Using the appropriate substitution, express the following
integrals in terms of u only.
ii Evaluate the integrals as functions of x.
a b
THINK WRITE
a i Let u = x − 2. a i Let
u = x − 2
and x = u + 2.
Find .
Substitute u for x − 2, u + 2
for x and
du for d x .
Replace u with x − 2.
=
x x 2–( )52--- d x∫ x2
x 1+ ---------------- d x∫
2 --- d x ∫
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222 S p e c i a l i s t M a t h e m a t i c s
THINK WRITE
. =
=
Find . = 1
Make d x the subject. or
d x = du
Express x in terms of u.
x = u − 1
Hence express x 2 in terms of u.
x 2 = u2 − 2u + 1
Substitute u for x + 1,
u2 − 2u + 1
for x
So
Replace u with x + 1. =
Take out as a factor. =
Simplify the other factor. =
63 ------------------------------
c+
2 ---– d x ∫
2 ---– du∫
2 ---
2 --- x 2 2 x 1+ +( )
5 --------------------------------
2 x 1+( ) 1
2 --- 3 x 2 6 x 3 10 x – 10– 15+ +
+
15
----------------------------------------------------------------------
c+
15 ------------------------------ c+
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Note: Recall that the logarithm of a negative number cannot
be found.
a Find the antiderivative of . b State the domain of the
antiderivative.
THINK WRITE
a Since e2 x = (e x )2, it can
be
antidifferentiated the same as a linear
function by letting
u = e x + 1.
a Let u = e x + 1.
Find . = e x
Express e
x
= u − 1
=
Simplify the rational expression. =
Antidifferentiate with respect to u.
= u − logeu + c
Replace u with e x + 1.
= e x + 1 − loge(e
x + 1) + c
b e x + 1 > 0 for all values
of x as e x > 0
for all x . The function
loge f ( x ) exists
wherever f ( x ) > 0.
b For loge(e x + 1) to exist
e x + 1 > 0, which
it is for all x .
State the domain. Therefore the domain of the integral is R.
e2 x
e x e x
For antiderivatives of the form , make the substitution
u = g( x ) and so
[g( x )]n d x , n ≠ 0 becomes
g′( x ) un du, n ≠ 0. This technique can
be
used for the specific case where
g = mx + c since g′( x )
= m. The function f ( x ) needs
to be transformed in terms of the variable u as well.
f x ( ) g x ( )[ ]n d x n
0≠,∫ remember
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Linear substitution
i express the following integrals in terms of u
ii evaluate the integrals as functions of x .
2
a The integral can be found by letting u equal:
b The integral then becomes:
3
a b
c d
e f
g h
i j
k l
m n
o p
A B C x + 2 D 4 x E
2 x
A B C
x 3– ----------- d x ∫ 2
3 x 5+ --------------- d x ∫
4 x 1+ d x ∫ 3 2 x –
d x ∫ x x 1+(
)3 d x ∫ 4 x x 3–(
)4 d x ∫ 2 x 2 x 1+(
)4 d x
∫ 3 x 1 3 x –(
)5 d x
∫ 6 x 3 x 2–( ) 3
4 --- d x ∫ x 2 x 7+(
)
1
3 --- d x ∫
x x 3+ d x ∫ x
3 x 4– d x ∫ x 2+( )
x 4–( )
3
5
8 x – ---------------- d x ∫
u 5
x 2
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b The result of the integration is:
4 Find the antiderivative of each of the following
expressions.
5 a If and f (1) = −2,
find f ( x ).
b State the domain of f ( x ).
6 a If and f (0) = 1,
find f ( x ).
b State the domain of f ( x ).
7 a Given that and g(2) = 0, find g( x ).
b State the domain of g( x ).
8 a Given that g(0) = 2 – loge2 and , find
g( x ).
b State the domain of g( x ).
A B
C D
E
a x 2( x − 4)4 b x 2(5
− x )3 c
d e f
g h i
j k l
m n o
p q r
s t u
3 ---
4 ---
x 1+( )2 x 2– x 3–( )2
x 1+ e x
e x 1+
x 3
x 4+ ----------------
2 ---
( )
x 1–( )2 -------------------=
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Technique 3: Antiderivatives involving trigonometric
identities
Different trigonometric identities can be used to antidifferentiate
sinn x and cosn x ;
n ∈ J + depending on whether n is
even or odd. Functions involving tan2ax are
also discussed.
Even powers of sin x or cos x The double-angle
trigonometric identities can be used to antidifferentiate even
powers
of sin x or cos x . The first identity
is:
cos 2 x = 1 − 2 sin2 x
= 2 cos2 x − 1
or cos
x = (1 + cos 2 x )
The second identity is: sin 2 x = 2
sin x cos x
or sin x cos x = sin
2 x
These may be expressed in the following general forms:
sin2 ax = (1 – cos 2 ax) Identity 1
cos2 ax = (1 + cos 2 ax) Identity 2
sin ax cos ax = sin 2 ax Identity
3
1
2 ---
1
2---
1
2 ---
1
2 ---
1
2 ---
1
2 ---
a sin2 b 2cos2
Take the factor of to the front of the integral.
=
Antidifferentiate by rule. =
Simplify the answer. =
x
2 ---
x
4 ---
∫ 3
1
2 ---
1
4 1
5 x
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Odd powers of sin x or cos x For integrals involving
odd powers of sin x or cos x the
identity:
sin2 x + cos2 x = 1
can be used so that the ‘derivative method’ of substitution
then becomes applicable.
The following worked example illustrates the use of this identity
whenever there is an
odd-powered trigonometric function in the integrand.
THINK WRITE
Simplify the integral. =
Antidifferentiate by rule. =
x
a b
THINK WRITE
a Use identity 3 to change sin
x cos x .
Note: The integral could be
antidifferentiated using technique 1
a
=
Square the identity. =
Simplify the integral. =
Antidifferentiate by rule. =
Simplify the answer. =
∫ 4 sin2 x
2 --- cos2 x
2 --- d x
1
2 1
1 x
2 ---
2
4 sin2 x d x ∫ 5
1
6 1
2 --- x
7 x
8 WORKEDExample
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Find the antiderivative of the following expressions.
a cos3 x b cos x sin 2 x c
cos42 x sin32 x
THINK WRITE
Factorise cos3 x as
cos x cos2 x . =
Use the identity: (1 − sin2 x ) for
cos2 x . =
Let u = sin x so the derivative
method
can be applied.
=
+ c
Replace u with sin x .
= sin x −
sin3 x + c
b Express in integral notation. b
Use identity 3 in reverse to express
sin 2 x as 2
sin x cos x . =
Simplify the integrand. =
Let u = cos x so that the
derivativemethod can be applied.
Let u
= cos x.
Substitute u for cos x and for
d x .
So
=
1 cos3 x d x ∫ 2 cos
x cos2 x d x ∫ 3
cos x 1 sin2 x –(
) d x ∫ 4
5 du
d x ------
cos x 1 u2–( ) du
cos x ------------∫
1
3---
1 cos x sin
2 x d x ∫ 2
cos x 2 sin x cos x (
) d x ∫ 3 2sin
x cos2 x d x ∫
4
5 du
d x ------
sin x – --------------=
7 du
sin x – --------------
2 sin
x cos2 x d x ∫ 2
sin x u2( ) du
sin x – --------------∫
9 WORKEDExample
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C h a p t e r 6 I n t e g r a l c a l c u l u s 229
Using the identity sec2 x = 1 + tan2 x The
identity sec2ax = 1 + tan2ax is
used to antidifferentiate expressions involving
tan2ax + c where c is a constant since
the antiderivative of sec2 x is
tan x .
Otherwise, expressions of the form tann x
sec2 x can be antidifferentiated using the
‘derivative method’ of exercise 6A.
THINK WRITE
Replace u with cos x . = −
cos3 x + c
c Express in integral notation. c
Factorise sin32 x as sin
2 x sin22 x . =
Use the identity 1 − cos 2
2 x for sin 2
2 x . =
Find .
Substitute u for cos 2 x and
for d x .
Expand the integrand. =
Simplify the result. = u7 −
u5 + c
Replace u with cos 2 x . =
cos72 x −
cos52 x + c
8 2u2 du–∫ 9 2
3 ---
3 ---
1
cos42 x sin32 x d x ∫
2 cos42 x sin
2 x sin22 x d x ∫ 3
cos
4
2 x –( ) d x
6 d x du
2 sin 2 x – ----------------------=
7 du2 sin 2 x –---------------------- u4 sin
2 x 1 u2–( ) du 2sin 2 x –
----------------------∫ 8
1
9 1
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Find an antiderivative for each of the following expressions.
a b
THINK WRITE
=
c as one antiderivative only is required.
= x + tan x
b Let u = tan 3 x so that the derivative
method can
be applied.
Find .
Make d x the subject. or
=
Replace u with tan 3 x . =
tan33 x
2 tan2 x+( ) d x
∫ 3 tan23 x sec23 x d x
∫ 1 2 tan2 x +( ) d x ∫
1 sec2 x +( ) d x ∫ 2
1
3 d x du
3sec23 x
10 WORKEDExample
remem er1. Trigonometric identities can be used to
antidifferentiate odd and even powers
of sin x and cos x . These identities
are:
sin2ax = (1 − cos 2ax )
cos2ax = (1 + cos 2ax )
sin ax cos ax = sin 2ax
2. The identity sec2ax = 1
+ tan2ax is used to antidifferentiate
expressions
involving tan
1
2 ---
1
2 ---
1
2 ---
remember
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Antiderivatives involving trigonometric identities
1 Antidifferentiate each of the following expressions with respect
to x .
2 Evaluate the following indefinite integrals as functions
of x .
3 If a is a constant, then,
a is equal to:
b is equal to:
c is equal to:
a cos2 x b sin22 x c 2
cos24 x d 4 sin23 x
e cos25 x f sin26 x g cos2 h
sin2
i 3 cos2 j 2 sin2 k cos2 l sin2
a b
c d
e f
g h
i j
k l
A 2 x − sin 2ax + c B
x − 2asin 2ax + c C
D E
D E
A a cos
ax − 3a cos3ax + c B
a sin ax − 3 cos3ax + c
C D
Anti- differentiation
WORKED Example
8 2 sin x cos d x ∫ 4 sin
2 x cos 2 x d x ∫
3 x 3 x cossin
d x
∫ 2 sin 4 x cos
4 x d x –
∫ sin2 x cos2 x d x ∫
sin22 x cos22 x d x ∫
2
sin24 x cos24 x d x ∫
2
sin23 x cos23 x d x ∫
6 sin2 x
2 --- cos2 x
3 --- cos2 x
3 --- d x ∫
3 ------ cos24 x
3 ------ d x –∫
multiple choiceltiple choice
8 ---
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4 Find an antiderivative of each of the following
expressions.
5 Use the appropriate identities to antidifferentiate the following
expressions.
6 Antidifferentiate each of the following expressions with respect
to x .
7 Find the following integrals.
8 Find an antiderivative for each of the following
expressions:
9 Find the following integrals where n ∈ J+.
a sin3 x b cos32 x c 6
sin34 x d 4 cos3 3 x
e sin37 x f cos36 x g h
i j k l
a sin x cos 2 x b cos
2 x cos 4 x c sin
3 x cos 6 x
d cos 4 x cos 8 x e f
a sin x cos4 x b sin
2 x cos32 x c
d cos 3 x sin43 x e f
a b
c d
e f
g h
i j
k l
m n
a 1 + tan22 x b c
tan2 x sec2 x
d tan3 x sec2 x e 4
tan52 x sec22 x f
g tan2 x sec4 x h 6
tan22 x sec42 x i
j 3 tan33 x sec43 x k
l 12 tan56 x sec66 x
a b c
WORKED Example
9c
cos2 x sin3 x d x ∫
sin2 x cos
3 x d x ∫
cos22 x sin32 x d x ∫
sin23 x cos
33 x d x ∫ cos2 x
2 ---sin3 x
2 ------cos33 x
2 ------ d x ∫
4 cos2 x
3 --- sin3 x
3 --- d x
∫ 6– sin25 x
4 ------ cos35 x
4 ------ d x
∫ sin3 x cos4 d x ∫
cos32 x sin42 x d x ∫
2sin32 x cos52 x d x ∫
2
cos33 x sin63 x d x –∫
4 sin3 x
2 --- cos6 x
2 ------ sin73 x
2 ------ d x ∫
2 ---
5 ---
sin x cosn x d x ∫
cos x sinn x d x ∫
sec2 x tann x d x ∫
sin
3
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10 If f ′( x ) = 6
sin x cos2 x and ,
find f ( x ).
11 If f ′( x ) = 4
sin22 x cos22 x and ,
find f ( x ).
12 Find g( x ) if g′( x ) = sin3
cos4 and g(0) = .
The graph of a function and the graphs of its antiderivatives
Given f ( x ) = , what do the graphs of
its
antiderivatives look like?
fnInt(Y1,X,0,X).
(Remember that to insert the symbol Y1, press , select and
1:Function. Then select 1:Y1 and press (and similarly for any
Y variable).
As the given function is trigonometric, press
and select 7: ZTrig.
for every X-value on the screen, the antiderivative
graph can take some time to plot. You can speed it
up considerably by changing the value of Xres in
the WINDOW settings to 5.)
1 Which is the graph of f ( x ) = and
which is the graph of the antiderivative?
The antiderivative graph in the second screen
is the line that cuts 0 at x = 0, since the
integral from 0 to 0 of any function is 0. To see another
antiderivative graph, go to , press ,
select 9 and complete 9: fnInt(Y1,X,1,X) and
then press .
2 Generate another two antiderivative graphs on your calculator.
Sketch the
function and the four antiderivative graphs. Describe any
relationships you can
find.
3 Choose another function and investigate the relationship between
the graph of
the function and the graphs of its antiderivatives.
f π
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Technique 4: Antidifferentiation using partial fractions
Recall that rational expressions, in particular those with
denominators that can be
expressed with linear factors, can be transformed into partial
fractions. A summary of
two common transformations is shown in the table below. These
transformations are useful when the degree of the numerator is less
than the degree of the denominator;
otherwise long division is generally required before
antidifferentiation can be performed.
We have seen how this procedure simplifies the sketching of graphs
of rational func-
tions. Similarly, expressing rational functions as partial
fractions enables them to be
antidifferentiated quite easily. However, it is preferable to use a
substitution method, if
it is applicable, as the partial-fraction technique can be
tedious.
Rational expression Equivalent partial fraction
where f ( x ) is a linear function
where f ( x ) is a linear function
f x ( ) ax b+( ) cx d +( )
---------------------------------------- A
---------------------- A
ax b+( ) --------------------+
Find a, b and c if ax( x − 2)
+ bx( x + 1)
+ c( x + 1)( x − 2)
= 2 x − 4.
THINK WRITE
Let x = 0 so that c can be evaluated.
Let x = 0, −2c = −4Solve the equation for
c. c = 2
Let x = 2 so that b can be evaluated.
Let x = 2, 6b = 0
Solve the equation for b. b = 0
Let x = −1 so that a can be evaluated.
Let x = −1, 3a = −6
Solve the equation for a. a = −2
State the solution. Therefore a = −2, b = 0 and
c = 2.
1 2
3
4
5
6
7
11WORKEDExample
For each of the following rational expressions: i express as
partial fractions ii antidifferentiate the result.
a b
THINK WRITE
separate fractions with denominators
a i
original common denominator.
x 2+( ) x 3–( )
----------------------------------------------
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C h a p t e r 6 I n t e g r a l c a l c u l u s 235
THINK WRITE
so x + 7
= a( x − 3)
+ b( x + 2)
Let x = −2 so that a can be evaluated.
Let x = −2, and thus 5 = −5a
Solve for a. a = −1
Let x = 3 so that b can be evaluated.
Let x = 3, and thus 10 = 5b
Solve for b. b = 2
Rewrite the rational expression as
partial fractions.
fraction form.
=
Antidifferentiate by rule. =
−loge( x + 2) + 2
loge( x − 3) + c
( x > 3)
Simplify using log laws. = loge + c
b i Factorise the denominator. b i =
Express the partial fractions with
denominators ( x − 4) and
( x + 1)
respectively.
original common denominator.
=
Equate the numerators. So 2 x − 3
= a( x + 1)
+ b( x − 4)
Let x = 4 to evaluate a.
Let x = 4, 5 = 5a
Solve for a. a = 1
Let x = −1 to evaluate b.
Let x = −1, −5 = −5b
Solve for b. b = 1
Rewrite the rational expression as
partial fractions.
fraction form.
=
Antidifferentiate by rule.
= loge( x − 4)
+ loge( x + 1) + c
( x > 4)
Simplify using log laws.
= loge[( x − 4)( x + 1)]
+ c ( x > 4)
or loge( x 2 − 3 x − 4)
+ c ( x > 4)
3
4
5
6
7
1–
2 x 3–
x 4–( ) x 1+( )
----------------------------------------------
9 2 x 3– x 2 3 x – 4–
-------------------------- 1
x 4– ----------- 1
1
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Find the following integrals.
Express into partial fractions with
denominators (1 − x ) and (1
+ x ).
=
original common denominator.
Equate the numerators. so 2 = a(1 + x )
+ b(1 − x )
Let x = 1 to find a.
Let x = 1, 2 = 2a
Solve for a. a = 1
Let x = −1 to find b.
Let x = −1, 2 = 2b
Solve for b. b = 1
Express the integrand in its partial
fraction form.
(−1 < x < 1)
same as the degree of the denominator
and hence the denominator should
divide the numerator using long division.
b
numerator.
x 2 + 5 x + 4
) x 2 + 6 x − 1
x 2 + 5 x + 4
x − 5
( x − 5).
x 4+( ) x 1+( ) -----------------------------------
d x∫
1 2
1 x 2–
----------------------------------------------
1 1 x – ----------- 1
1 x + ------------+
d x
x 4+( ) x 1+( )
----------------------------------
3
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THINK WRITE
Therefore
( x + 1).
original common denominator.
=
Equate the numerators. and thus x − 5
= a( x + 4)
+ b( x + 1)
Let x = −1 to find a.
Let x = −1, −6 = 3a
Solve for a. a = −2
Let x = −4 to find b.
Let x = −4, −9 = −3b
Solve for b. b = 3
Express the original integrand in its
partial fraction form. Therefore,
=
Antidifferentiate by rule.
= x − 2loge( x + 4)
+ 3loge( x + 1) + c,
( x > −1).
x 5–
x 4+( ) x 1+( )
-----------------------------------------------
1 2–
x 4+ ------------
Rational polynomials can be antidifferentiated by rewriting the
expressions as
partial fractions or by long division. If the numerator is of
degree less than the
denominator then use partial fractions; otherwise rewrite the
expression by long
division. Two common partial fraction transformations are shown
below.
Rational expression Equivalent partial fraction
where f ( x ) is a linear function
where f ( x ) is a linear function
f x ( ) ax b+( ) cx d +( )
---------------------------------------- A
---------------------- A
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Antidifferentiation using partial fractions
1 Find the values of a, b and c in the following
identities.
a ax + b( x − 1)
= 3 x − 2
b a( x + 2)
+ b( x − 3)
= x − 8
c a( x − 4)
+ b = 3 x − 2
d a(3 x + 1)
+ b( x − 2)
= 5 x + 4
e a(2 − 3 x )
+ b( x + 5)
= 9 x + 11
f a( x + 2)
+ bx = 2 x − 10
g a + b( x + 2)
+ c( x + 2)( x + 3)
= x 2 + 4 x − 2
h
a( x + 2)( x − 3)
+ bx ( x − 3)
+ cx ( x + 2)
= 3 x 2 − x + 6
2 Express each of the following rational expressions as partial
fractions.
3 Find the antiderivative of each rational expression in question
2.
4
5
a b c
d e f
g h i
j k l
m n
A a = 2, b = 3 B
a = −2, b = −3 C a = 3,
b = 2
D a = −2, b = 3 E
a = 1, b = −1
A 2loge( x + 6) − 3loge(4
− x ) + c B
−2loge( x + 6) − 3loge(4
− x ) + c
C 3loge( x + 6) + 2loge(4
− x ) + c D 3loge( x + 6)
− 2loge(4 − x ) + c
E
A 2loge( x + 3)
− loge( x − 2) + c B 2loge
C 2loge D loge( x + 3)
− 2loge( x − 2) + c
E loge( x + 1)
− 2loge( x − 6) + c
6D WORKED
Example 11
x 2–( ) x 2+(
)---------------------------------- 6 x
x 3+( ) x 1–(
)----------------------------------
9 x 11–
11 3 x –
a
---------------------------
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6 Antidifferentiate each of the following rational polynomials by
first expressing them
as partial fractions.
7 By first simplifying the rational expression using long division,
find the antiderivative
of each of the following expressions.
8 Evaluate the following integrals in terms of x .
9 a If and f (2) = 3loge2,
find f ( x ).
b State the domain of f ( x ).
10 a Find g( x ) if and g(4) = 4 − loge5.
b State the domain of g( x ).
a b c
d e f
g h i
j k l
6 x 1–
16 2 x –
x 4+
4
WORKED Example
x 2 3 x +-----------------
x 2 4 x – ---------------------------
x 2 4– --------------------------
x 2+( ) x 1+( )
----------------------------------
2 x 3 x 2 5–+
x 2 1– -----------------------------
2 x 2 9 x – 7+
x 2 6 x – 9+
------------------------------
4 x –
∫ 9 x 8+
∫ 5 x 1+( ) x 2
25–--------------------d x
∫ x 2 3+
x 2 9– --------------d x ∫
x 2 3 x 4–+
x 4–( ) x 2+( )
---------------------------------- d x ∫
x 2 4 x 1+ +
x 2 6 x 7–+ ---------------------------
d x ∫
x 3 x 2 4 x –+
x 2 4 x – 4+ -----------------------------
d x ∫ 4 x 2 6 x 4–+
2 x 2
x 2 4+ --------------d x ∫
4 x 2–
5 x 2
2 x 17+ + x 1–( ) x 2+( )
x 3–(
)---------------------------------------------------
d x
∫ x
2
18 x 5+ + x 1+( ) x 2–( )
x 3+(
)---------------------------------------------------d x
∫ x 2 8 x 9+ +
x 1–( ) x 2+( )2
------------------------------------ d x ∫
x 2 5 x 1+ +
x 2 1+( ) 2 x –( )
------------------------------------ d x ∫
f ′ x ( ) 6
x 2 1– --------------=
8/20/2019 Integral Calculus[1]
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Definite integrals
The quantity is called the ‘indefinite integral of the
function f ( x )’. However,
is called the ‘definite integral of the function
f ( x )’ and is evaluated using
the result that:
where F ( x ) is an antiderivative
of f ( x ).
The definite integral can be found only if the
integrand, f ( x ), exists for
all values of x in the interval [a, b]; that is,
a ≤ x ≤ b.
When using substitution to evaluate definite integrals there is no
need to return to an
expression in terms of x providing the terminals
are expressed in terms of u. In fact it is
mathematically incorrect to show the integral in terms of
u but with terminals in terms
of x . Therefore when using a substitution, u
= f ( x ), the terminals should also
be
adjusted in terms of u.
f x ( ) d x ∫ f x (
) d x a
b
b
For each of the following integrals, state:
i the domain of the integrand ii whether the integral
exists.
a b
THINK WRITE
must be greater than 0.
a i The integrand exists if .
Solve the inequation for x .
x 2 < 9 –3 < x < 3
State the domain. The domain is (−3, 3).
ii The integral exists for all values of x
between the terminals −2 and 2.
ii The integral exists.
the denominator equal to zero.
b i x ≠ –3, 1
State the domain. Domain is R\{–3, 1}.
ii The integral does not exist for all values
of x between the terminals 0 and 4
(as 1 lies in the interval).
ii The integral does not exist.
∫ 2
–2
1
1 9 x 2– 9 x 2– 0>
2
3
1
2
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Use an appropriate substitution to express each of the following
definite integrals in terms
of u, with the terminals of the integral correctly adjusted.
a b
THINK WRITE
u = x 2 − 1 so the derivative method
can be
applied.
Find .
Adjust the terminals by finding
u when x = 2
and x = 3. When x = 2,
u = 22 − 1
= 3
= 8
linear substitution
u = x − 2.
b Let u = x − 2.
Find .
Express d x in terms of du. or
d x = du
Express x in terms of u.
x = u + 2
Adjust the terminals by finding
u when x
= 3
= 1
= 4
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242 S p e c i a l i s t M a t h e m a t i c s
Evaluate the following definite integrals.
a b
THINK WRITE
integrand.
Consider: =
=
original common denominator.
=
Equate the numerators. x − 2
= a( x + 4)
+ b( x + 1)
Let x = −1 to find a.
Let x = −1, −3 = 3a
a = −1
Let x = −4 to find b.
Let x = −4, −6 = −3b
b = 2
So
=
Antidifferentiate the integrand.
= [−loge( x + 1)
+ 2loge( x + 4)]2 0
Evaluate the integral. = [−loge3 + 2loge6]
− [−loge1 + 2loge4]
= −loge3 + 2loge6 − 2loge4
= loge2.25 − loge3
b Write the integral. b
Let u = 1 + sin x to
antidifferentiate. Let u = 1
+ sin x
Find .
∫
1 ∫ 2
2 x 2–
x 2–
x 2 5 x 4+ +
-----------------------------------------------
5
6
7
8
∫ 2
0
∫ 2
0
1–
π
2 ---
∫ 2
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THINK WRITE
x = 0 and x = .
= 1
π
2 ---
THINK WRITE
Find .
Make d x the subject. or
d x = cos θ dθ
Change the terminals by finding θ when
x = and x = 0.
When x = , = sin θ
θ = 0
1
2 ---
dθ ------ cos θ =
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To find the value of a definite integral, press and select
9:fnInt(. Then type in
the integrand, the function variable, the lower terminal and the
upper terminal. Press
to evaluate the integral.
Alternatively, if the function is already in Y1, press , select
9:fnInt(, complete
9: fnInt(Y1,X,0, 2) and press . (Remember that to insert the symbol
Y1, press
, select Y–VARS and 1:Function, then 1:Y1 (similarly for
any Y variable).
1 The screen shows both methods for (Worked example 16a).
2 To estimate cos2 d x , press , select
9:fnInt( and complete by entering
2(cos(X ÷2))2,X,0,π ) and pressing .
(1 + cos 2 θ
1
2 ---
0
π
6---
0
π
6 ---
0
π
6 ---
∫ 1
MATH
ENTER
MATH
ENTER
VARS
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A handy trick to use, if the answer is a simple fraction, is to
press , select
1: Frac and press — but it doesn’t work in this
case.
If the answer could possibly be a fractional multiple of π ,
first try dividing by π then
pressing , selecting 1: Frac and pressing . In this case, the
answer is
just π itself. (Don’t expect this trick to always
work!)
Definite integrals
1 For each of the following definite integrals , state i the
maximal domain
of the integrand f ( x ) and
ii whether the integral exists.
a b c
d e f
g h i
j k l
1. =
= F (b) − F (a), where F ( x )
is an antiderivative of f ( x ).
2. The definite integral can be found only if the
integrand, f ( x ), exists
for all values of x in the interval [a, b]; that
is, a ≤ x ≤ b.
f x ( ) d x a
b
b
b
2
0
1
3
1
∫ 3
2 -------
0
3
( ) 2 d x
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2 Evaluate the integrals in question 1 provided that the
integrand, f ( x ), exists for all
values within the domain of the integral.
3
The definite integral d x can be evaluated after
substituting u = x 3 + 1.
a The integral will then be equal to:
b The value of the integral is:
4
b The integral will then be equal to:
c When evaluated, the integral is equal to:
5 By choosing an appropriate substitution for u, express the
following integrals in terms
of u. (Do not forget to change the terminals.)
6 Evaluate each of the integrals in question 5.
A B C
A u = sin x B
u = cos x C
D u = cot x E u = 1
+ sin x
A B C D E
A 2 B C −2 D E
a b c
d e f
g h i
j k l
0
2
7
0
u 1 sin x +=
2
3 ---
2
∫ ∫ 21 4 x
x 2 3– 2 ---------------- d x x
x 2 1+ d x
0
1
2
4
1
2
∫ ∫ 3
0
x ------------- d x sin x e cos
x d x
π
3 ---
π
2 ---
1
0
π
2 ---
π 4 ---
π
2 ---
0
1
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7 Evaluate the following definite integrals.
8 By substituting x = sin θ , evaluate
.
9 By substituting x = 2sin θ , evaluate
.
10 By making the substitution x = tan θ ,
evaluate .
11 If , find the value of a.
12 If , find a.
13 If , find a.
14 If , find a.
0
∫ 4 x 7+( ) 2 x 2 7 x +
d x 0
1
3–
2–
π
3 ---
∫ ∫ 2
0
∫ 1
–1
1
1
∫ 1
–1
1–
4 x 1–( )2– ----------------------------
d x
sin3 x cos2 x d x
0
π
4 ---
π
2 ---
∫ ∫ 6
5
2 x 2
–1
1
π
2 ---
0
1
4 ---
π
3 ---
x 2 4– -----------------------------------------
d x
3
4
1
0
3
0
a
∫ π =
4
0
a
a
a–
a
∫ π
2 ---=
T 6.1
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You can check your answers by using the Mathcad file
‘Integrator’ found on the Maths
Quest CD-ROM.
Applications of integration In this section, we shall examine how
integration may be used to determine the area
under a curve and the area between curves.
Areas under curves You will already be aware that the area between
a curve
which is above the x -axis and the x -axis
itself is as shown in
the diagram at right.
Further, the area between a curve which is below the
x -axis, and the x -axis itself, is as shown
in the second
diagram.
Area = –
=
The modulus is required here since, for a curve segment that lies
below the x -axis,
the integral associated with that curve segment is a negative
number. Area is a positive
number and in this case the integral is negative.
hc a d
b
b
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Similarly, the area between a curve and the y-axis can
be
found if the rule for the curve is expressed as a function
of
y, that
is, x = f ( y).
Area = (integral measures to the right of the
y-axis are positive) or
y-axis are negative)
Area =
If the graph crosses the x -axis, then the areas of
the
regions above and below the x -axis have to be
calculated
separately. In this case the x -intercepts must be
determined.
In the figure at right a single intercept, c, is shown.
Area =
=
Similarly the shaded region in the figure at right has an
area given by:
b
a
b
b
∫ y
b
c
b
b
b
∫ y( ) d y–
If y = , find:
a the x-intercepts b the area bounded by the curve,
the x-axis and the line x = 3.
THINK WRITE
a For x -intercepts, y = 0, when
2loge x = 0. a x -intercepts
occur when 2loge x = 0.
Solve for x . That is, x = 1.
b Sketch a graph showing the region
required. (A graphics calculator may
be used.)
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A graphics calculator should be used here to verify the
result.
To find the area under a curve between two x -values,
first graph the curve by entering
its equation as Y1 in the Y= menu.
Consider y = in worked example 18. Press
Y= and type in (2ln(X))÷X at Y1.
Then press .
To find the area bounded by this curve and the
x -axis between x = 1 and
x = 3, press [CALC]
and select 7: ∫ f(x) dx. Type in 1 for the lower value
(press ) and 3 for the upper value (press
). Compare this result to that obtained in
worked example 18.
Express the area as a definite integral. Area =
Antidifferentiate by letting
u = loge x to apply the derivative
method.
Let u = loge x.
Make d x the subject. or
d x = x du
Express the terminals in terms of u.
When x = 1, u = loge1
= 0
Area =
1.207 square units.
2loge x
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The shaded area shown in the figure in worked example 19
could also have been calculated relative to the x -axis
by sub-
tracting the area between the curve and the x -axis from
the
area of the rectangle as shown in the figure at right. That
is:
Area =
Using symmetry properties In some problems involving area
calculations, use of symmetry properties can simplify
the procedure.
a Express the rule as a function of y.
b Find the area of the shaded section.
THINK WRITE a Write down the rule. a
y =
Square both sides of the equation.
y2 = x − 1
Add 1 to both sides to make x the subject.
or x = y2 + 1
b Express the area between the curve and
the y-axis in integral notation.
b Area =
y
2
5
∫ –
Find the area inside the ellipse in the figure at right.
THINK WRITE
symmetrical about the x -axis and y-axis
and so finding the shaded area in the figure allows for the total
enclosed area
to be determined.)
for the top half of the ellipse.
= 1 − x 2
of the ellipse.
y
x
y2
— 4
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Areas between curves When finding the areas between two curves that
intersect, it
is necessary to determine where the point of intersection
occurs. In the figure at right, two functions,
f and g, inter-
sect at the point P with x -ordinate c.
The area contained within the envelope of the two func-
tions bounded
by x = a and x = b is
given by:
THINK WRITE
area).
Express the total area as four times this integral.
Total area of ellipse =
To antidifferentiate, let x = sin θ .
Let x = sin θ .
Find . = cos θ
Make d x the subject. or
d x = cos θ dθ
Express the terminals in terms of θ .
When x = 0, sin θ = 0
θ = 0
θ =
Rewrite the integral in terms of θ . Area = cos
θ dθ
Simplify the integrand using identities. =
=
=
State the area. The exact area is 2π square units.
3 2 1 x 2– d x 0
1
0
1
0
1
∫ 5
π
2 ---
π
2 ---
2 --- sin 0+[ ]–
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C h a p t e r 6 I n t e g r a l c a l c u l u s 253
Area = +
tive to the y-axis.
Area =
Note that on the interval [a, b], g( y)
≥ f ( y) and hence the
integrand is g( y) − f ( y) and
not f ( y) − g( y).
When an area between a curve and the x-axis (or between
curves) gives an
integrand which cannot be antidifferentiated, it may be possible to
express the
area relative to the y-axis, creating an integrand which can
be antidifferentiated.
g x ( ) f x ( )–[ ] d x a
c
b
∫ y
g y( ) f y( )–[ ] d ya
b
∫
Find the area bounded by the
curves y = x2 − 2
and y = 2 x + 1.
THINK WRITE
the curves intersect. If they do, solve
x 2 − 2 = 2 x + 1 to
find the x -ordinate of
the point or points of intersection for
the two curves.
( x − 3)( x + 1)
= 0
x = 3 and x = −1
The curves intersect at x = 3
and x = −1.
Express the area as an integral.
(Use , as without a graph we cannot
always be sure which function is above
the other. Here is a valuable use for the
graphics calculator.)
= 10
State the solution. The area bounded by the two curves is
10 square units.
2 x 2 2– 2 x 1+( )–[ ] d x
1–
3
3
1–
3
3 ---– 1– 3+( )]–
9– 1 2
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254 S p e c i a l i s t M a t h e m a t i c s
Consider using the TI calculator for worked example 21.
1. To graph the two curves with
equations y = x 2 – 2
and y = 2 x + 1 enter
Y1= X2 – 2 and Y2= 2X + 1. Then
press . Use TRACE to locate the points of
intersection. Adjust the WINDOW settings if
necessary.
2. To show the area bounded by the two curves, press
, position the cursor to the left of the Y1 symbol
and press successively to obtain the ‘shade
below’ style. Repeat for Y2 to obtain the ‘shade
above’ style. Press . The required area is
shown unshaded.
3. To determine the value of the area bounded by the
curves on the required interval (in this case, between
x = –1 and x = 3), press
, select 9 and
complete 9: fnInt(Y2–Y1,X,–1,3) and press .
Remember, to insert Y1, press and select
Y–VARS, 1:Function and 1:Y1 (or 2:Y2 to enter
Y2).
Note that in this case we are subtracting Y1 from Y2
(seen by viewing the graph). However, if it is entered the opposite
way, it only
produces the negative of the required answer.
Graphics CalculatorGraphics Calculator tip!tip! Showing and
finding the area bounded by two curves
GRAPH
remem er
1. The area between a curve f ( x ),
the x -axis and
lines x = a and x = b is
given by:
Area = where F ( x ) is the antiderivative
of f ( x ).
2. Area measures can also be evaluated by integration along
the y-axis. The area
between a curve f ( y), the y-axis and
lines y = a and y = b is
given by:
Area = where F ( y) is the antiderivative
of f ( y).
3. If an area measure is to be evaluated over the interval [a, b]
and the curve
crosses the x -axis
at x = c between a and b, then the
integral has to be decomposed into two portions.
Area =
4. The area bounded by two curves f ( x ) and
g( x ) where f ( x )
≥ g( x ) and the lines
x = a and x = b is
given by:
Area =
5. Where possible use a graphics calculator to draw the function or
functions to
determine whether the integrals have to be decomposed into portions
and to
check and verify the correct use of the modulus function.
f x ( ) d x a
b
b
c
b
f x ( ) g x ( )–[ ] d x a
b
remember
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C h a p t e r 6 I n t e g r a l c a l c u l u s 255
Applications of integration
For the following problems, give exact answers wherever possible;
otherwise give answers
to an appropriate number of decimal places. (Use a graphics
calculator to assist with, or
verify, any graphing required.)
i the x -intercepts
ii the area between the curve, the x -axis and the given
lines.
2 For each of the graphs below:
i express the relationship as a function of y (that is,
make x the subject of the rule)
ii find the magnitude of the shaded area between the curve and the
y-axis.
a y = , x = 0
and x = 9 b
y = x −
, x = 1 and x = 2
c y = , x = 2
and x = 5 d y =
, x = 3 and x = 4
e y = , x = 1
and x = f
y = cos2 x , x = 0
and
g y = 2 x cos x 2,
and x = 0 h y =
, x = 0 and x = 1
a b c
d e f
d
x 2 4– ---------------
π
2 ---=
– 3 π
– 2 π
– 4 π
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256 S p e c i a l i s t M a t h e m a t i c s
3 Find the magnitude of the shaded areas on each graph below.
4
a The definite integral that correctly gives the area bounded by
the curve
y = 4 x − x 2
and the x -axis is:
b The area, in square units, is equal to:
5 a Which of the graphs below correctly shows the area bounded by
the curve
y2 = x + 1 and
the y-axis?
b The definite integral which gives the area bounded by
y2 = x + 1 and
the y-axis is:
a b c
d e f
A B C
A 10 B 2 C 5 D 8 E −5
A B C
y
2
1
0
0
4
2
0
∫ 2
3 ---
1
3 ---
1
3 ---
1
3 ---
1
0
0
∫ y
2
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C h a p t e r 6 I n t e g r a l c a l c u l u s 257
c The value of the area, in square units, is equal to
6 Find the area bounded by the graph with equation
y = ( x − 2)2( x + 1)
and the x -axis.
7 Find the area bounded by the graph with equation
y2 = x + 4 and
the y-axis.
8 a Show that the graphs of f ( x )
= x 2 − 4 and g( x ) = 4
− x 2 intersect at x = −2
and x = 2.
b Find the area bounded by the graphs
of f ( x ) and g( x ).
9 a On the same axis sketch the graphs
of f ( x ) = sin x and
g( x ) = cos x over [0,
π ].
b Show algebraically that the graphs intersect at .
c Find the area bounded by the curves and the y-axis.
10 a On the same axis sketch the graphs of y =
and y = x + 3.b Find the value
of x where the graphs intersect.
c Hence find the area between the curves
from x = −1
to x = 2.
11 Find the area bounded by the curves
y = x 2 and y = 3 x + 4.
12 Find the area enclosed by the curves
y = x 2 and .
13 Find the area bounded
by y = e x and y = e− x and
the line y = e.
14 Examine the figure at right.
a Find the area enclosed by f ( x ),
g( x ) and the y-axis.
b Find the shaded area.
15 Find the area of the ellipse with equation .
Hints:1. Use symmetry properties.
2. Antidifferentiate by using the
substitution x = asin θ .
16 Find the area between the circle
x 2 + y2 = 9 and ellipse
.
Hint : Make use of symmetry properties.
17 a Sketch the curve y = e
x + 2
.b Find the equation of the tangent at
x = −2.
c Find the area between the curve, the tangent and the
y-axis.
18 a Sketch the graph of .
b Find the area bounded by this curve and the x -
and y-axes.
19 a Show algebraically that the
line y = x does not meet the curve
.
b Find the area enclosed by the curve, the lines
y = x and , and
the y-axis.
A B 2 C 1 D 5 E 2 2
3 ---
2
3 ---
1
3 ---
1
3 ---
d
2
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258 S p e c i a l i s t M a t h e m a t i c s
Volumes of solids of revolution If part of a curve is rotated about
the x -axis, or y-axis, a figure called a solid of
revol-
ution is formed. For example, a solid of revolution is
obtained if the shaded region in
figure 1 is rotated about the x -axis.
The solid generated (figure 2) is symmetrical about
the x -axis and any vertical cross-
section is circular, with a radius equal to the value of
y at that point. For example, the
radius at x = a is f (a). Any thin vertical
slice may be considered to be cylindrical, with radius
y and height
δ x (figure 3).
The volume of the solid of revolution generated
between x = a and x = b is
found by
allowing the height of each cylinder, δ x , to be
as small as possible and adding the vol-
umes of all of the cylinders formed between
x = a and x = b.
That is, the volume of a
typical strip is equal to
π y2 δ x .
Therefore the volume of the solid contained from
x = a to x = b is
the sum of all the
infinitesimal volumes:
=
The value of y must be expressed in terms of
x so that the integral can be evaluated.
From the figure
above y = f ( x ) and thus the
volume of revolution of a curve f ( x )
from
x = a to x = b is .
Similarly if a curve is rotated about the y-axis, the
solid
of revolution shown in the figure at right is produced.
The volume of the solid of revolution is likewise
For regions between two curves that are rotated about the
x -axis:
π y2 δ x
b
b
y = f ( x )
y = g ( x ) V π f x (
)[ ]2 g x ( )[ ]– 2 d x
a
b
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C h a p t e r 6 I n t e g r a l c a l c u l u s 259
Consider the volume of the solid of revolution formed in Worked
example 22. 1. The line with
equation y = 2 x is rotated about
the x -axis
to form a cone. To graph the line, enter Y1 = 2X and
press
. Press to locate particular coordinates. 2. To determine the
volume of the cone, press , then
select 9: fnInt( and insert Y1 2, X, 0, 2) and press
. To insert Y1, press and selectY–VARS, 1:Function and
1:Y1 (similarly any Y variable).
(Note that Y1 2 provides the integrand, X is the
variable,
0 and 2 are the terminals of the integral.)
Can you verify the formula V = for this cone? What is the
radius for this cone?
You can try to convert your volume answer to a fraction of . Press
and [ ],
then . Select 1: Frac and press . This is also shown in the
screen above.
a Sketch the graph of y = 2 x and show the
region bounded by the graph, the
x-axis and the line x = 2.
b Find the volume of the solid of revolution when the region is
rotated about the
x-axis. THINK WRITE
Shade the region required.
b State the integral that gives the
volume. (The volume generated is bounded
by x = 0
and x = 2.)
b V =
State the volume. The exact volume generated is cubic units.
1 y
1 π 2 x ( )2 d x 0
2
0
2
22 WORKEDExample
Graphics CalculatorGraphics Calculator tip!tip! Finding the
volume of a solid of revolution
GRAPH TRACE
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260 S p e c i a l i s t M a t h e m a t i c s
a Sketch the region bounded by the curve y =
loge x, the x-axis, the y-axis and
the line y = 2.
b Calculate the volume of the solid generated if the region is
rotated about the
y-axis. THINK WRITE
calculator if necessary.)
Shade the region required.
b Write the rule y = loge x . b
y = loge x
Take the exponent of both sides to get y
as a function of