integral calculus[1]

derivative is present in the integrand
6B Linear substitution
6E Definite integrals
6H Approximate evaluation
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212 S p e c i a l i s t M a t h e m a t i c s
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+=
8/20/2019 Integral Calculus[1]
C h a p t e r 6 I n t e g r a l c a l c u l u s 213
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
THINK WRITE
Substitute u for x 3 + x and for
d x .
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 .
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
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
=
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
THINK WRITE
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 --------------------------=
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.
=
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
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
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+ ------------------------
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 –-----------------------------------------------------
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=
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 ∫
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+
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
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
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 ).
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 ).
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 -------------------=
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
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. =
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. =
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
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
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
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 .
=
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
Antiderivatives involving trigonometric identities
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 ---
4 Find an antiderivative of each of the following expressions.
5 Use the appropriate identities to antidifferentiate the following expressions.
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
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
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
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–( ) ----------------------------------------------
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.
=
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
Find the following integrals.
Express into partial fractions with
denominators (1 − x ) and (1 + x ).
=
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
THINK WRITE
Therefore
( x + 1).
=
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
f x ( ) ax b+( ) cx d +( )
---------------------------------------- A
---------------------- A
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
---------------------------
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 ).
10 a Find g( x ) if and g(4) = 4 − loge5.
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– --------------=
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
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
Evaluate the following definite integrals.
a b
THINK WRITE
integrand.
Consider: =
=
=
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
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 =
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
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
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
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
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
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.)
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
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
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+[ ]–
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
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
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
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
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
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
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
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

integral calculus[1]

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