ba201 engineering mathematic unit 4 integration
DESCRIPTION
BA201 Engineering MathematicTRANSCRIPT
TOPIC 4
INTEGRATION 1
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
4.1 DEFINITION OF INTEGRATION
1) The process of obtaining
from y ( a function of x ) is known as differentiation
and the reverse process of obtaining y from
is called integration.
Differentiation process
Integration process
2) Integration of y respect to x , is denoted by ∫ ( ) .
3) Integration is divided into two parts which are indefinite integral and definite
integral and first we will approach to indefinite integral as the basic of integration.
4.2 INDEFINATE INTEGRALS
1) We shall use the following notation for the family of integration.
∫ ( )
2) If integration of y with respect to x, is denoted by ∫ ( )
𝑦
𝑦
TOPIC 4 : INTEGRATION
TOPIC 4
INTEGRATION 2
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
3) If 𝑦 ( ), then
( ) and ∫ ( ) 𝑦 is known as integration of y
respect to x where c is an arbitrary constant.
Formulas in integration :
i) a dx ax c ; where a is a constant.
ii) 1
1
nn ax
ax dx cn
; where n is an integer,
iii) ( ) ( ) ( ) ( )p x q x dx p x dx q x dx
Tips!
Note that a function is integrate respect to the function given and it is not compulsory to be
integrated respect to x only. (see example given)
Example 4.1
Find :
a) 3
5
x dx b) 3 r dr
c) 6 dk d) xdx
Solution:
a) 23 3
5 5 2
xx dx c
23
10x c
In indefinite integral, c must be stated. It is
known that as an arbitrary constant where
the value is yet to be determined.
TOPIC 4
INTEGRATION 3
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
Can only integrate terms by terms in
addition or subtraction. If the function
given is in multiplication, the
function must be expanded.
Tips! Operation
b) 4
3 4
rr dr c
c) 6 6dk k c
d) 1
2xdx x dx
32
32
32
2
3
xc
x c
4.3 Determining the integration of algebraic expressions involving operations.
Example 4.2
Find :
a) 2
6x dx b) 3 2 2 5x x dx
c)
4
4 5 2x xdx
x
d)
3
32
5x dx
x
Solution :
a) 2
26 12 36x dx x x dx
3 2
32
12 363 2
6 363
x xx c
xx x c
The given function is integrate respect to k,
therefore the integration will be in terms of k.
Make sure that x is always in index form.
TOPIC 4
INTEGRATION 4
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
b) 23 2 2 5 6 15 4 10x x dx x x x dx
2
3 2
3 2
6 11 10
6 11 103 2
112 10
2
x x dx
x xx c
x x x c
c) 2
4 4
4 5 2 20 8 5 2x x x x xdx dx
x x
2
4 4 4
4 3 2
3 2 1
3 2
20 3 2
20 3 2
20 3 23 2 1
20 3 2
3 2
x xdx
x x x
x x x dx
x x xc
cx x x
d) 3
32
5x dx
x
. (Let’s try this question in class)
Can only integrate terms by
terms in addition or
subtraction. If the function
given is in division, the
function must be simplified.
Tips! Operation
TOPIC 4
INTEGRATION 5
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
4.4 Definite Integrals of Algebraic Expressions
1. If x=a to x=b for ( )f x dx , then it is written as ( )
b
a
f x dx and it is called definite
integral.
( ) ( ) ( )
b
a
f x dx F b F a
2. For the definite integral ( )
b
a
f x dx , the constant a is known as the lower limit of
integration while b is known as the upper limit of the integration.
Properties Of Definite Integrals.
a) ( ) ( )
b b
a a
kf x dx k f x dx
b) ( ) ( ) ( ) ( )
b b b
a a a
f x g x dx f x dx g x dx
c) ( ) ( )
b a
a b
f x dx f x dx
d) ( ) ( ) ( )
b c c
a b a
f x dx f x dx f x dx
TOPIC 4
INTEGRATION 6
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
Evaluate definite integrals using the properties of definite integrals. Example 4.3 1) Evaluate each of the following definite integrals below.
a) 3
1
(3 5)x dx b) 0
1
( 1) 3x x dx
c) 2 2 3
1
4 5x x xdx
x
Solution
a)
33 2
1 1
3(3 5) 5
2
xx dx x
2 2
333 3 3 1
5(3) 5 12 2
= 22
b) 0 0
2
1 1
( 1) 3 3 3x x dx x x x dx
= 3
5
0
2
1
03 2
1
03
2
1
3
2
2 3
2 33 2
33
10 1 3 1
3
x x dx
x xx
xx x
TOPIC 4
INTEGRATION 7
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
c) 2 22 3
2
1 1
4 54 5
x x xdx x x dx
x
22 3
1
2 3 2 3
4 52 3
2 2 1 14(2) 5 4(1) 5
2 3 2 3
x xx
= 6
85
2) Given that 4
2
( ) 6g x dx and 8
4
( ) 10g x dx , find the values of
a) 4 4
2 8
( ) ( )g x dx g x dx b) 4
2
( ) 2g x x dx
c) 2 8
4 4
5 ( ) ( )g x dx g x dx d) k if 8
4
( ) 9kx g x dx
Solution :
a) 4 4
2 8
( ) ( ) 6 ( 10)g x dx g x dx
= 16
b) 4 4 4
2 2 2
( ) 2 ( ) 2g x x dx g x dx x dx
42
2
22
26
2
6 (4) 2
6 (16 4)
6 12
6
x
TOPIC 4
INTEGRATION 8
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
c) 2 8 4 8
4 4 2 4
5 ( ) ( ) 5 ( ) ( )g x dx g x dx g x dx g x dx
( 5)(6) 10
30 10
40
d) 8
4
( ) 9kx g x dx
8 8
4 4
( ) 9kx dx g x dx
82
4
10 92
kx
2 2
8 410 9
2 2
k k
32 8 1
24 1
1
24
k k
k
k
TOPIC 4
INTEGRATION 9
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
4.5 Integration for other functions
4.5.1 Integration of trigonometric functions
There are an overwhelming number of combinations of trigonometric functions which appear
in integrals. This section examines integration of trigonometric functions.
Formula integration of trigonometric functions
a) cos( )
sin( )
( )
ax bax b dx c
dax b
dx
sin( )ax b
ca
b) sin( )
cos( )
( )
ax bax b dx c
dax b
dx
sin( )ax b
ca
c) 2 tan( )sec ( )
( )
ax bax b dx c
dax b
dx
tan( )ax b
ca
TOPIC 4
INTEGRATION 10
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
Example 4.4
Find :
a) ∫ ( ) b) ∫ (
)
) ∫ (
)
Solution
a) ∫ ( ) ( )
) ∫ (
)
(
)
c)
cx
c
x
dxx
22tan
21
2tan
2
sec2
4.5.2 Integration of exponential functions
Formula of integration of exponential functions :
ca
e
c
axdx
d
edxe
ax
axax
TOPIC 4
INTEGRATION 11
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
Example 4.5
Find :
a) dxe x
35 b) dxeee xxx 32
c) dxe
eeex
xxx
35 24
Solution
a)
ce
dxex
x
3
3535
b) dxeee xxx 32 dxee xx 35
cee xx
35
35
c) dxe
eeex
xxx
35 24 dxee xx )24( 24
cee
x xx
24
44
cee
xxx
2
2
44
24
4.5.3 Integration of reciprocal functions
The integral of dxx
1 has an infinite discontinuity between x and 1 and has does not exist.
Thus, ln x is an antiderivative of x
1, therefore the integration of reciprocal functions is defined
by
ca
(ax)
c
axdx
d
axdx
ax
ln
)(
)ln(
1
TOPIC 4
INTEGRATION 12
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
Example 4.6
Find
a) dxx
5 b) dx
x
54
3
Solution
a) cxdxx
ln5 5
b) cx)(dxx
54ln3
54
3
4.6 Integration through substitution method
Integration of an expression in the form of nbax
Integration of expression in the form of nbax can be determined by using substitution
baxu where a and b are constants.
STEPS
If baxu a
duudxbax n
n
adx
du c
na
u n
)1(
1
dxa
du c
na
bax n
)1(
)( 1
Integration of nbax can also be done by applying the following formula
cna
baxdxbax
nn
)1(
1
TOPIC 4
INTEGRATION 13
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
Example 4.7
Solve the integral below :
a) dxx3
24 b)
dxx
426
2
c) dxxx4252 d)
1
0
32 2 dxxx
Solution
a) By using substitution method,
xu 24 dxx3
24
2
3 duu
2dx
du; dx
du
2 duu 3
2
1
cu
42
1 4
cx 4
248
1
By using formula,
dxx3
24
cx
)2(4
244
cx
8
244
b) By using substitution method,
xu 26
dxx
426
2du
u
4
1
2dx
du; dxdu 2 duu
4
cu
3
3
TOPIC 4
INTEGRATION 14
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
cx
3263
1
By using formula,
cx
cx
dxxdxx
3
3
4
4
263
1
)3(2
262
26226
2
a) By using substitution method,
25 xu duudxxx 44252
xdx
du2 c
u
5
5
xdxdu 2
cx
5
552
b) By using substitution method,
22 xu
1
0
32 2 dxxx
2
3 duu
xdx
du2 ; xdx
du
2 duu 3
2
1
8
65
16818
1
2218
1
28
1
42
1
44
1
0
42
4
x
u
TOPIC 4
INTEGRATION 15
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
TIPS!
4.7 Using substitution method involving problems in integration
4.7.1 Trigonometric basic identities and Double-angle formulae
In certain cases, trigonometric integral needs to apply identity trigonometric in order to solve
the solution by using formula integration of trigonometric functions. You can see clearly in our
next examples.
Example 4.8
Find the integral below m:
a) dxx sin 2
b) dxx tan2
Solution
a) dxx sin2
By using double-angle formulae,
xx 2sin212cos
2
2cos1sin 2 x
x
dxx sin2
dxx
2
2cos1
dxx 2cos12
1
cx
x
2
2sin
2
1
b) dxx tan2
By using trigonometry identity;
xx 22 tan1sec
xx 22 tan1sec
dxx tan2 dxx 1sec2
cxx tan
It is recommended to use substitution method in
multiplication and division which cannot be
simplified to a single function.
TOPIC 4
INTEGRATION 16
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
4.7.2 Exponential function and Reciprocal functions
The best way to solve integration of exponential functions and reciprocal function in
multiplication or division operations is by using substitution method.
Example 4.9
Find :
a) dxee xx 242 3 b) dxxx
x
53
593
2
c) dxxe x )6sin(6cos
Solution
a) dxee xx 242 3
Assume 32 xeu
xedx
du 22
dxedu x2
2
dxee xx 242 3
2
4 duu
duu 4
2
1
ce
cu
x
10
3
52
1
2
5
a) dxxx
x
53
593
2
Assume xxu 53 3
59 2 xdx
du
TOPIC 4
INTEGRATION 17
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
cxx
cu
u
dudx
xx
x
53ln
)ln(
53
59
3
3
2
a) dxxe x )6sin(6cos
Assume )6cos( xu
)6sin(6 xdx
du
dxxdu
)6sin(6
dxxe x )6sin(6cos
6
dueu
ce
ce
due
x
u
u
)6cos(
6
1
6
1
6
1
TOPIC 4
INTEGRATION 18
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
Activity 4(a)
1. Integrate the functions below
a. 2
1 b. 2x
Feedback for Activity 4(a)
1. a.
x + 6 and 3 +
x or any
2
1x + C
b. x2 and 6 + x2 or any x2 + C
TOPIC 4
INTEGRATION 19
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
Activity 4(b)
1. Evaluate
a. kdk5
b. dxx27.0
2. Evaluate
a. dxx 54
b. dzz
4
5 2
Feedback for Activity 4(b)
1. a. ck
2
5 2
b. cx 223.0
2. a. cx
4
1
b. cz
12
5 3
TOPIC 4
INTEGRATION 20
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
Activity 4(c)
Evaluate :
1. dx3
2. dxx )14(
3. dxxx )123( 2
4. dtt
tt )9
2(2
5.
dx
x
x2
1
Feedback for Activity 4(c)
1. 3x + c
2. 2x2- x + c
3. x3 + x2 –x + c
4. ct
tt
9
23
4 2
2
3
5. cxxx 2
1
2
3
2
5
23
1
5
2
TOPIC 4
INTEGRATION 21
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
Activity 4(d)
Integrate x.
1. a. b.
2. a. b. c.
3. a b.
Feedback for Activity 4(d)
1. a.
b.
2. a.
b.
cxdxkosxdxkosx sin333
cxdxxsekdxxsek
tan4
1
4
1
4
22
dxx 2
1dx
x 3
dxkosx3
dxxsek
4
2
dxx sin3
1
dxe x
2
dxe x
4
cxdxx
dxx
ln2
11
2
1
2
1
cxdxx
dxx
ln31
33
TOPIC 4
INTEGRATION 22
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
c. ckosx
ckosxxdxxdx
33
1sin
3
1sin
3
1
3. a. b.
ce
dxex
x 4
44 c
edxe
xx
2
22
TOPIC 4
INTEGRATION 23
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
Activity 4(e)
Evaluate the given integral
1. a.
b.
2. a.
b.
c.
3. a
b.
dtt 3
1
dzz
13
dxkosx3
dxxsek
2
2
dkk sin2
1
dxex
2
1
dxe x
4
TOPIC 4
INTEGRATION 24
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
Feedback for Activity 4(e)
1. a.
b. cz ln13
2. a. - 3sin x + c
b. cx tan2
1
c. - ckosk 2
1
3. a. ce x
4
4
b. ce
x
2
1
2
ct ln3
1
TOPIC 4
INTEGRATION 25
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
Activity 4(f)
1. Evaluate the following integrals.
a. dxxx ]4[ 23
b. dtt
t ]1
3[3
3
c. dxx
]32
[2
2. Evaluate
a. dkkk ]44[ 2
b. dzz 2)32(
c. dxx
x
2
542
Feedback for Activity 4(f)
1. a. cxx
34
3
4
4
b. ct
t 2
4
2
1
4
3
c. cxx
32
2. a. ckkk
423
23
b. czzz 963
4 23 c. cxx
42
TOPIC 4
INTEGRATION 26
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
Activity 4(g)
1. Using simple substitution, evaluate.
a. dxx 3)24( b. dzz 6)83(
Feedback for Activity 4(g)
1. a. cx
16
)24( 4
b. cz
56
)83( 7
TOPIC 4
INTEGRATION 27
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
Activity 4(h)
1. Evaluate:
a. dxx 4)32( b. dzz 3)63(
c. dtt 5)75( d. dxx 3)84(6
e. dxx 3)27( f. dt
t 2)31(
g. dxx 3)54(
1 h. dx
x
4)53(2
3
2. Evaluate
a. dkkk 732 )1(
b. dzzzz )33()3( 233
c. dppp
p
3 3
2
3
1
TOPIC 4
INTEGRATION 28
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
Feedback for Activity 4(h)
1. a. cx
10
)32( 5
b. cz
12
)63( 4
c. ct
42
)75( 6
d. cx
8
)84(3 4
e. cx
2)27(14
1 f. c
t
)31(3
g. cx
2)54(8
1 h. c
x
3)53(6
1
2. a. ck 83124
1
b. css 323
3
1 c. cpp 3
23 3
2
1
TOPIC 4
INTEGRATION 29
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
Activity 4(i)
Evaluate the given indefinite integrals
a.. dxx4sin b. dxxx 2cos
c. dxx7tan2 d. dxx
x
2sin
cos2
Feedback for Activity 4(i)
a. - Cx 4cos4
1 b. Cx 2sin
2
1
c. Cxx 7tan7
1 d. -cot x – csc x + C
TOPIC 4
INTEGRATION 30
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
Activity 4(j)
Evaluate the given integrals
1. x
dx
3 2.
dxx
x
12 3. dx
x
xln
4. dxe x10 5. dxex x322 6. dt
e
et
t
1
Feedback for Activity 4(j)
1. Cx ln3
1 2. Cx )1(ln
2
1 2
3. Cx 2)(ln2
1 4. Ce x 10
10
1
5. - Ce x 32
6
1 6. -e-t + t + C
TOPIC 4
INTEGRATION 31
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
Activity 4(k)
1. Evaluate a.
b.
c.
2. Integrate
a.
b.
c.
3. Integrate
a.
b.
c.
dxx3
1
dt
t 32
1
dk
k
k
33
dte t 23
dxe
eex
xx
2
3
dxeee xxx )( 32
dxx
kos3
dxx
3sin 2
xkosxdxsin2
TOPIC 4
INTEGRATION 32
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
Feedback for Activity 4(k)
1. a.
b.
c.
2. a.
b.
c.
3. a.
b.
c.
cx ln3
1
ct )32ln(2
1
ckk
)3ln(3
1 3
ce t 23
3
1
cee xx 3
3
1
cee xx
43
43
cx
3sin3
cx
x 3
2sin
4
3
2
1
cxkos
2
2
TOPIC 4
INTEGRATION 33
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
Activity 4(l)
Evaluate each of the following:
1 a. 3
2
2 )5( dxxx
b. dxx )33
2(
3
0
2. a. 4
2 )21)(31( dttt
b. 1
1 )32( dkk
3. Find dxx
xx
51
2 3
4
4. An arrow was following a straight line with velocity, v = 6t 4 + 3t2 at time t. What is the
distance of the arrow that moves from t = 1till t =10 ?
TOPIC 4
INTEGRATION 34
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
Feedback for Activity 4(l)
1 a. b.
2 a. b.
3
4. 120 997.8 meter
2
2
25
3
2
2
35
3
3
2
5
3
)5(
2323
3
2
23
3
2
2
xx
dxxx
6
)0(36
)0(2)3(3
6
)3(2
323
2
)33
2(
22
3
0
2
3
0
xx
dxx
115
)2(22
22)4(2
2
44
22
)61(
)21)(31(
32
32
4
2
32
4
2
2
4
2
tt
t
dttt
dttt
6
)1(3)1()1(31
3
)32(
22
1
1
2
1
1
kk
dkk
1
)2(
5
2
)2(
)1(
5
2
)1(
5
2
)5(
5
22
1
2
2
1
2
2
1
2 3
4
x
x
dxxx
dxx
xx
TOPIC 4
INTEGRATION 35
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
Activity 4(m)
1. If 2
7 )(
1
2 dxxf and
2
3 )(
2
1 dxxf , evaluate :
a. 2
2 )( dxxf
b.
2
1
1
2 )(2 )( dxxfdxxf
c.
1
2
1
2 )(2 )( dxxfdxxf
2. Evaluate each of the following if 1 )(3
2 dxxf and 4 )(
3
1 dxxg
a. dxxf )1)(3(3
2
b. } )( )({23
2
3
1 dxxfdxxg
TOPIC 4
INTEGRATION 36
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
Feedback for Activity 4(m)
1. a.
52
3
2
7
)( )(
)(
2
1
1
2
2
2
dxxfdxxf
dxxf
b.
2
16
2
13
2
32
2
7
)(22
7
)(2 )(
2
1
2
1
1
2
dxxf
dxxfdxxf
c.
2
16
2
13
2
32
2
7
)(22
7
)(2 )(
2
1
1
2
1
2
dxxf
dxxfdxxf
2. a. b.
2
)23(1
1
1)(3
)1)(3(
3
2
3
2
3
2
3
2
x
dxdxxf
dxxf
10
28
)1(2)4(2
)(2 )(2
) )( )((2
3
2
3
1
3
2
3
1
dxxfdxxg
dxxfdxxg
TOPIC 4
INTEGRATION 37
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
Activity 4(n)
1. Evaluate :
a.
1
0
3)12(2 dxx where u = 2x + 1
b.
dz1z2
z43
2
22
where u = 2z2 +1
2. Evaluate these definite integrals by using a suitable substitution.
a.
2
1
43 dt)1t5(t
b. dk1k
k3
02
2. Management cost for a building increased from time to time. If the rate of the
management cost increased and given by 20t
1500
dt
dx
where x is cost in thousand ringgit
and t is time in years, evaluate the management cost for 5 years.
TOPIC 4
INTEGRATION 38
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
Feedback for Activity 4(n)
1. a. 20
b. 171
10
2. a. 1638
1
b. 1
3. RM 15 835.92 million
TOPIC 4
INTEGRATION 39
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
SELF ASSESSMENT 4 (a)
1. Evaluate the following integral
a. dz7
b. dtt32
c. dxx 4
10
2. Evaluate
a. dxxxx 96 2
b. dzx2
52
3. Rewrite each expression and using derivative of a sum, to solve the following
integrals
a. (3x - 2)2
b. 5
2 )1(
x
xx
c. 2
)1)(1(
k
kk
4. Evaluate
a. dss334
b. dzx 2)76(
c.
dkk
k)6(
TOPIC 4
INTEGRATION 40
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
SOLUTION : SELF ASSESSMENT 4 (a)
Have you tried ?
1. a. 7z +c
b. ct
5
22 5
c. cx
33
10
2. a. cxxx 323
3
2
2
92
b. cxxx
25103
4 23
3. a. cxxx 463 23
b. cxx
1
2
12
c. ck
k 1
4. a. css 4
4
34
b.
c. √ -
TOPIC 4
INTEGRATION 41
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
SELF ASSESSMENT 4 (b)
Evaluate
1. a.
b.
2. An epidemic struck Klang. The affected population is rising at the rate of
per day where t is the number of days after the outbreak of the epidemic. Find
the epidemic function S(t) for new cases each day.
73 43 ttdt
dS
,)31( 4 dxx
,52 dtt
TOPIC 4
INTEGRATION 42
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
SOLUTION: SELF ASSESSMENT 4 (b)
1. a b.
2.
Let u = (t4 + 7) 2
1
32
14 47
2
1tt
dt
du
2
1
4
3
)7(
2
t
t
= duut
2
2
3 = cu
t3
2
1 = ct
tct
t 2
343
2
14 7
2
1]7[
2
1
dxx 4)31(
cx
cx
15
)31(
)14(3
)31(
5
14
c
t
ct
dtt
dtt
3
52
)12
1(2
52
52
52
2
3
12
1
2
1
dtttS 73 43
3
2
1
443
2
)7(73
t
dutttS
TOPIC 4
INTEGRATION 43
PREPARED BY SITI NORSHAFINAZ BINTI MD NOH
SELF ASSESSMENT 4 (c)
1. Evaluate :
a.
3
1
2 dx)5x4x( b.
200
100
2 dt)120t2t03.0(
2. The rate of sale for a shop is given as J (t) = -3t2 + 300t where t is the number of days
after the sale. Determine the total of sale J(t) for the first week after the campaign.
3. The rate of toll collection at a highway is given as this model
23 t240t20dt
dk
Where t is time in hours and k is the sum of money that has been collected. This model is
used from time 1200 (t = 0) till time 1000. What is the total of collection that has been made
in this period?