7

INTEGRATION AND ITS APPLICATIONS:
7.10 Integration is the process or act or an
instance of making up a whole or is a process
of uniting parts together. Integration is
considered also to be reverse of differentiation.
7.11. Integration as reverse of differentiation
in polynomial functions
In the process of Carrying differentiation
the following steps are done.
1. Multiple the old power by old coefficient
to get new coefficient of derivative
2. Reduce the old power by 1 (one) to get
new power of derivative i.e.
n n1
ax
nax
d
dx is differentiation
dx
coefficient
3. If a is old coefficient, n is old power, then
na is new coefficient of derivative while
n-1 is new power of derivative.
na
n1
x
is a derivative of
a
n
x
In the process of Carrying integration the
following steps are done.
1. Increase the old power by 1 (one) to get
new power of integral
2. Divide old coefficient by new power to
get new coefficient on integral
n a n1
ax dx x C dx is differential
n 1
coefficient
3. If a is old coefficientand n is old power
a
then n 1
is new coefficientof integral
while n + 1 is new power of integral.
a n1
n
x C
is the integral of ax
n 1
7.12. Indefinite integral
3 4 2
If we differentiate x x 3x
8with respect
to x, we obtain
3x
2 8x
3
. On integrating
3x 2 + 8x + 3 we cannot recover the constant
term (8). To recover this constant we normally
add an arbitrary constant, say K, to the integral.
The arbitrary constant added to the integral is
called constant of integration (which can be
any letter a part from x for this case).
244
2
3 2
I.e. 3
x 8x
3dx
x 4x
3x
K in this
case our K is equal to 8. Hence we cover back
our original function.
3 5x
3
4
Example: Find integral of x
x
with
respect to x.
Solution:
3 8
x 5x
dx
4
x
3
8
x dx 5xdx
dx
4
x
3
4
x dx
5xdx
8x
dx
1 31
5 11
8 41
x x x C
3 1
11
4 1
1 4 5 2 8 3
x x x C
4 2 4 1
1 4 5 2 8
x x C
4 2
3
3x
Example 2: integrate ax
to x
Solution:
ax
2
2
ax dx bxdx
cdx
a 3 b 2
x x cx A
3 2
bx c dx
Example 3: Integrate 2 3 2
Solution
4t
2t
3
2
2
dt
4t2dt
12tdt
9dt
4 3 2
t 6t
9t
C
3
12t
9 dt
Example 4
dy
Solve 3x
2 2x
1
dx
2
bx c with respect
t with respect to t.

Solution
dy
3x
dx
dy
2
dy 3x
dx 2x
dx 1dx
y C x
y x
3x
3
3
3
2
2
2x
1
x
x
2x
1
dx
2
2
2
x K
x K C
y x x x B
Since C and K are constants of integration but
(constant) – (other constant) gives another
constant. That is why K – C = B which is
another constant. For instance 5 – 3 = 2. And
for this reason we add a constant only on one
side at the end of integration.
NB: integration where there is no limits, you
have to add arbitrary constant call constant of
integration. This type of integration is called
indefinite integration and integral obtained is
indefinite integral like
3
2 x 2
x
xdx
2x
C
4
3
is indefinite
integral
7.13 Definite integral
When the limits are given the integral we
obtain after integration is called definite
integral. Therefore we do not add a constant at
the end of integration.
Example 1:
4
Evaluate x 3
4dx
2
Solution:
4 3
x
4dx
2
4
x
4
4x
k
4
2
4
4
64
16
k 4
8 k
80 k 12
k
68
4
2
k 4
4
4 4
2
2
K
NB: When limits are given, there is no need to
add the constant, K, of integration because it
will cancel each other.
5
Example 2: 2
Solution:
5 2
6x
dx
2
6
x
3
2
3
5
2
6x 2 dx
3 5 3 3
x
2 5
2
2
2125
8
266
Example 3: Evaluate
Solution:
2 3
4r
dr
0
4
r
3
1
r
4
16
2
0
4
2
0
16 0
31
4
2
0
2 4 0
dy 3 2
Example 4: If 8x
6x
4x
1, find y if
dx
from x = 1 to x = 2
r 3
dr
245

Solution
dy 3
8x
6x
dx
y
dy
8
y
4
y 2 16
y 21
3
dy 8x
2
8x
1
6x
2
8x
1
3
4 2 6 3 2 4 2 2
x
x
x
x
1
3
2
3
2
dx
6x
1 28
1 24
1 2
1
y 30 14
6 1
4x
1
2
6x
1
1
4x
1
dx
2
2
4x
1dx
2
dx
1
2
4xdx
1
Example 5 Given that 2y
3
the value a.
Solution
2
2 2
y
3ya
0
2
4
6
a
3a
2 a
a
a
2
2y
3 dy 0
3a
2 0
a
1a
2
a 1
Or a 2
a 1
2
3a
0
0
not 2
0
2
a
2
1
2
1dx
1
dy 0 . Find
EXERCISE 7 (a)
1. Find the integrals for the following
2
(i) t
t dt
2 3
(ii) 3
k 4k
dk
(iii) 2
da
a 2
(iv) x
12 xdx
(v)
2
y
3
2
y dy
1
2. Evaluate the following definite integral
3
(i) t dt 1
(ii)
2 2
t
1 dt
1
(iii)
3
1
x 4 dx
(iv)
3
1
dx
x
1 3
(v)
2
1
x dx
1
x
3. Find the following
(i) 4
x 3dx
(ii) 4
3y 2 dy
1
(iii)
dt
t 3
(iv) z dz
4. Find
x dt in each of the following
(i) x t
3 2 t 2
4
4t
3t
(ii) x
2
2t
6t
3
(iii) x
2 t
2
t 2
(iv)
t
5. Find the indefinite integrals of the
following
(i) m 4
dm
(ii) m 2
dm
1
(iii) dm
4
m
1
(iv) m 2 dm
6. Evaluate the following.
4
(i) e de 0
3
5
(ii) 4x 3dx
0
246

9
(iii)
4
7. Evaluate
1
f 2
df
x
5 3x
2 5x
1
(i) 1
2
(ii)
3 1 3 4
y
2
y y
2
2 16
Z Z
(iii) 9
3
t 3t
45
(iv)
5
t
1
Z
8. If
3t
2 dt 27 , find Z and give reason
O
t 2
9. If y
2y
6dy
0 find the value of t
O
and give reason.
2
a
x 1
10. Given that
30
1
dx
, find the
x
value of a.
7.20 Area Under Curve
Draw a rough graph (curve) within wanted
region by locating at least 3 points on the
curve. Indicate element by shading it as
indicated in the figure 129 below.
Figure 129
x
x
Shaded part is called an element. The area of
element = x( y y)
y x x
y
If both x and y are very small then
x y 0
x 0 and y 0 .
area of element = y x .
The area bounded by ABQP as x 0
b
of area of elements = yx
The required area =
is sum
a
Example 1: Find area under y 3x
2 6x
1
and bounded by x 1 and x 2,
and the x axis
Solution
Figure 130
Area of the element = y δx (Area of a shaded
part in figure 130)
A
2
1
3
= x 3x
2 x
= (-8 + 12 + 2) – (-1 + 3 + 1)
= 6 – 3 = 3
Example 2: Find the area enclosed between the
2
two curves y 9 x and y 2x
2 6x
Solution
2
xa
b
3x
6x
1
dx
2 1
2
2
When 9 x 2x
6x
give points of
intersection.
2
3x
6x
9 0
x
2
2x
3 0
x
3x
1
x 3
or
-1
0
ydx
247

Figure 131
The area of the element = y x (Shaded part in
figure 131)
Where y = y 1 – y 2 (as y 1 is above y 2 within the
interval between those intersection points)
The required area
3
3
2 2
ydx =
9 x
2
x 6x
dx
1
1
3
2
=
9 3x
6xdx
1
3 1
3 2
= 9x
x 3x
= (27 – 27 + 17) – (-9 + 1 + 3)
= 27 +5
=32
Example 3: Find the area enclosed by part of
y 9x
2 3 for which x is positive and y –
axis, and the lines y = 3 and y = 4.
y
x y
y = 4
The area of element is x y (Shaded part in
figure 132), the sum of areas of all element
between y = 3 and
y = 4 is
4
xy .
y 3
Hence the area enclosed
1
4
y 3
2
= dy
y
3
9
1
=
3
4 1
y
32
dy
y 3
1 2
=
3
3
y
3
3
2
2
3
= 32
9
y
3
2 3 3
= 4
32
3
32
9
2
3
2
3
= 1
2 0 2
3 9
2
9
but area enclosed by the part of y = 9x 2 + 3for
1
which x is positive and y axis is of the area
2
1
enclosed, hence the area required is sq. units.
9
Example 4: Find the whole area enclosed by
the curve y = sin x and the x – axis between
x = 0 and x = 2
Solution
A sketch of wanted area.
4
4
3
y = 3
Figure 132
x
Figure 133
248

Area of element of part 1 above x – axis
y 1 x(Shaded part in figure 133)
Area under curve above x – axis
sin xdx cos
x
0
0
= - (-1 – 1)
= 2.
Area of element of part 2 below x – axis
y 2
x (This area is equal to the area shaded in
part 1 above x-axis in figure 133)
Area of under curve below x – axis
2
2
sin xdx cos
x
cos 2
cos
11
2
Hence -2 indicates that the area is below
x – axis therefore the area is 2.
The whole area required
2
sin xdx
sin x dx
= 4 sq units.
NB: If there are some areas below and above
the x axis you work those areas separately but
not at once because you will add negative and
positive area which will give you less area than
what is required yet there is no area which is a
negative as you take their magnitude to get area
required.
Example 5: Find the area enclosed by a curve
whose parametric equations are
2
1
t 2t
x and y and the line y 1
1
t 1
t
Solution
x
1 2
Cartesian equation of the curve is x 1
Hence for curve to meet y = 1 we have
2
x 2x
1
1
x 1
x + 1 = x 2 – 2x + 1
x 2 – 3x = 0
x(x – 3) = 0 x = 0 and x = 3
.
y 1 = 1
y
Figure 134
Area of element y x = (y 1 – y 2 )dx (See figure
134)
Area required =
3
2
x 2x
1
1
dx
x 1
0
3
2
x 1
x 2x
1
dx
x 1
0
3
2
x 3x
dx
x 1
0
3
3
4
4 x dx
dx
x 1
0
0
3
x
2
x
4x
4ln
x
1 3 0
2 0
9
12 4ln
4 ln 1
2
15
4ln 1.9584
sq units.
2
Example 6: Find area bounded by the curve
represented parametrically by
1
2
y t and x t and the lines x 1 and x 4
t
Solution
From x = t 2 t = x
2
x 1
x 1
y =
x x
Hence Cartesian equation of the curve is
y
2
x 2x
1
y 2
x 1
x
249

x 1
y
x
1
x(1)
x
1
dy
2 x
dx
x
dy
for turning point 0
dx
1
0 x x
1
2 x
2x
x 1
2 x
0 = 2x – x – 1
0 = x – 1
x = 1
11
2 2
y(1) = or 2or
2
1 1 1
turning point are (1,2) and (1, -2)
X LHS 1 RHS
dy - 0 +
dx
Min
When x x
Or
X LHS 1 RHS
dy + 0 -
dx
When
x
x
Max
Hence (1,2) is minimum point while (1, -2) is
maximum point
Asymptotes vertical x 0
x = 0
Slanting
x
y =
x
x
1
x
1
x or y 2 = x
x 1
intercept when y = 0 0 =
x
x = -1 is outside the limit when x = 4
4 1
5
y = 2.5 or 2.5(1dp).
4 2
y
(0,0)
(1, 2)
Figure 135
Figure 135 is a sketch of the curve y = t + t
1
and x = t 2 .
(1, -2)
y
Figure 136
x =1
Area of element shaded = 2y x (See figure
136)
Area required =
4 1
4
x 1
2 dx
x
1
4 1 3
2
2
x dx x
1
1
1
2
y
y
2 ydx
dx
x =4
(4, 2.5)
x
(y = x)
(4, -2.5)
x
250

2
2
x
3
3
2
2x
1
2
4
1
3 3
1 1
4
2 2
2 2
4
1
44
1
3
4
8
1 42
1
3
40
= 3
However, since the area under curve is asked
there will be no need for asymptotes.
EXERCISE 7 (b)
x 1
1. Find area enclose by curve
f ( x)
x
from x = 1 to x = 9 and x = axis.
2. Show that the area enclosed by the curve y
x
= e and y – axis, and the line y = 2 is given
1
by 2ln 2
2
3. Find the area enclosed by the curve
y = x( 4 – x), the x- axis and the lines
x = 0 and x = 6.
4. Find the area bounded by the curve y = x 2 +
2, the x- axis and the line y = 3 and y = 5
5. Calculate the area enclosed by the curve y =
4-x 2 and y = -3x.
4
6. The line x+y = 5 and the curve y
x
intersect at A and B.
(i) What are the co-ordinates of A and B?
ii) Find the area enclosed the curve, line
and x-asis.
7. The line y = x meets the curve y = 5x-x 2 at
origin and point H.
(i) Find the co-ordinates of H.
(ii) Calculate the area of the region enclosed
between the curve and the line.
y
Figure137
If f(x) > 0 and area between y = f(x),
y = 0, x = a and x = b is rotated about x – axis
so that it forms the solid above. In case we
want to find the volume of the solid we divide
x – axis , from a to b into equal intervals each
of length x so that we form a series of very
thin cylinder or a disc of thickness x .
Volume of each of these discs
y 2 x.
Hence to get volume of solid range from x = a
to x = b will be the sum of volumes of those
discs.
x
b
xa
y 2 x
Required volume =
b y 2 dx
a
Volumes formed in this nature are always
called volumes of revolution.
Example 1: Find the volume of the solid
generated when region bounded by the curve y
= cos2x and x – axis for the value of x between
0 and is rotated about the x – axis.
4
Solution
7.21 Volumes of Revolution
Draw a rough graph (curve) within wanted
region and indicate element as in the figure 137
below.
Figure 138
Volume of each disc
138)
2
y x.
(See figure
251

Curve of volume of disc =
Required volume
4
x0
2
( y x)
(0 – 0)
Example 3: Find volume of area enclosed by a
8 8
line y = 8 and curve y x 2 4 when rotated
2
about;
8
(i) y = 8
Sum of volumes of each of those disc
2
y
The volume required
6
2 2
2
8 x 4
dx
x y
2
y 3
2
Required volume
2 2
4 x dx
6
x 2 dy
2
(ii) x – axis.
Example 2: Find the volume formed when area
4
enclosed by y x from y = 3 to y = 6 when it Solution
is rotated about x = 0.
(i)
y
Solution
y = x 4
(0,12)
(-2,8)
(2,8) y = 8
(0,4)
Figure 140
Figure 139
The points of intersection for the line and curve
2
x 4 8 x 2 4 x 2.
The volume of each circular disc x 2 y (See The volume of each cylindrical element
shaded part in figure 139)
y 2 x.
(See figure 140)
3
2
1 cos 4 x dx
x 1
sin 8
2
x
2 8
2
4
0
4
0
cos
2 2
xdx
4
0
252
6
3
y 2 1
2
y
3
2
6
3
3
2
dy
3
2
3
6
3
3
2
1
2
2
2
6 6 6 ( 3 3 3 )
3
3
2
( 9 . 500 )
3
6 . 33 19 . 9 ( 1 dp ).
1
2

(ii)
2
(16 8x
Figure 141
2
x
The required volume = the volume(1)generated
by rotation, about the x–axis, of the rectangle
ABDE minus the volume (2) generated by
rotation about the x–axis of the area y = x 2 + 4
From x = -2 to x = +2 ie ABCDE
4
) dx
2
5
2
8 3 x
16
x x
3 5
2
88
32
32
32
3 5
1 2
2 32 21 6
3 5
2 4
64
42
12
3 5
2 4
3 5
12 10
34
15
2
32
15
512
15
64
42 12
2 8 2
64
3
32
5
Volume (1) r
h 4 256
Element of (2) y 2 dx (See figure 141)
Volume (2)
2
2
2
2
2
( x 4) dx
2
2
5
x 8
3
x 16x
5 3
2
32 8
2 8
32
5 3
119.5
Required volume ( 256 119.5)
136.
5
2
EXERCISE 7 (c)
1. A shell is formed by rotating the portion of
the parabola y 2 4x
for which 0 x 1
through two right angled about its axis.
Find the volume of the solid form.
2. Find the volume of the solid by revolution
formed by rotating the area enclosed the
curve y 2x
2 1,
the y-axis and the line
y 2 and y 5 but lying in the first
quadrant.
3
3. An area is bounded by the curve y x
x
the x-axis and the ordinates x 1
and
x 3. calculate the volume of solid
obtained by rotating the area through 360 0
about the x-axis.
2
4. The curve y 4x
x . calculate the volume
generated when the area bounded by the
curve and the x-axis is rotated the four right
angle about x-axis.
5. The part of the curve y= x 3 from x=1 to x=2
is rotated completely round the
y-axis. Find volume of the solid generated.
6. Find the volume of the paraboloid
generated when the area in the first
quadrant and bounded by the parabola
y 2 = x and the lines x = 3, y = 0 and x = 0 is
rotated through four right angle about the x-
axis.
7. Find the volume generated when the area
between the curve y 2 = x and the line
2
( x 8x
16)
dx
253

y = x is rotated completely about the x-
axis.
2
x
8. The portion of the curve y from
2
x = 0 to x = 2 is rotated about the x – axis
through 360 0 to from solid. Find volume of
solid generated.
9. Calculate the volume generated when the
region enclosed by the y = 1 + 2e -x and the
line x = 0, x = 1 and y = 1 is revolved
completely about the x – axis.
10. Find the volume of the solid generated by
rotating the area enclosed by the curve
y = 4-x 2 and x – axis about the y – axis.
11. Determine the volume of the solid
generated when the region bounded by
curve y = cos2x and the x – axis for values
of x between 0 and 4
is rotated about the x
– axis.
12. Find the volume of the solid of revolution
formed by rotating the area enclosed
13. Sketch the curve y 2 = x-1. The area
contained by curve, the y-axis and line
y 2 is completely rotated about the y-
axis. Find the volume of solid formed.
14. Find the volume generated when the finite.
Region enclosed by between the curve y =
ln x, the x- axis and the ordinate bat x = 2 is
revolved completely about the line x = -1.
15. The area enclosed by the curve y = x 2 +2
and line y = 2 for 2 x 0 is rotated
about y = 2. Find the volume generated.
16. Given that the curve y = 3x 2 – 4x + 1is
rotated through 2
about the x-axis, Find
the volume generated between x = 0 and x
= 2.
17. The area bounded between the curve y =
15x (2-x) and the x-axis is rotated through
2
radius about the x-axis. Find volume of
the solid generated.
18. Find volume of the solid formed by rotating
area enclosed between the curve y-2x = 0
and y + x 2 – 4x = 0.
19. Find the volume of the solid generated by
revolving about the line y = -4, the finite
region bounded by the parabola y = 1 – x 2
and the x-axis.
20. Area bounded by curve y = x 2 and y=2-x 2
calculate the volume of the solid generated
when area is rotated through radians
about the y-axis.
7.22 Non-uniform acceleration in a straight
Displacement, x, is distance covered in specific
direction that is why we can consider
movement in a straight line.
Velocity, v, is rate of change of displacement
dx
i.e. v .
dt
To obtain displacement we integrate velocity
with respect to time, t, while to obtain velocity
we integrate acceleration with respect to
time, t,
dv
(A) When consider a
dt
Example 1: A body of mass 2Kg initially at rest
at origin is acted upon by the force
2 i
t j
3t
k
~ ~ ~
t N.
Find (i) its acceleration at any time t.
(ii) its velocity after 3 seconds.
(iii) its position after 3 seconds.
(iv) the distance it has travelled after 3
seconds.
Solution:
(i)
(ii)
From F = ma
f 1
a
2t
i
t j
3t
k
m 2 ~ ~ ~
t 3
t i j t k
~ 2 ~ 2 ~
v a dt
2 2
t t 3 2
i
j
t C1
2 ~ 4 ~ 4
Since it started from rest then C 1 =0 i.e.
when t = 0 and v= 0.
2 2
t t 3 2
v i
j
t
2 ~ 4 ~ 4
254

(iii)
9 9 27
V (3) i
j
2 ~ 4 ~ 4 k
~
x v dt
3 3
t t 1
i
j
t
6 ~ 12 ~ 4
3
k
C
~
Since it started from origin C 2 = 0 i.e.
when t = 0 and x = 0 x
3 3
t t 1 3
i
j
t k
6 ~ 12 ~ 4 ~
9 9 27
V (3) i
j
k
2 ~ 4 ~ 4 ~
(iv) d(3) = x (3)
9
2
2
9
4
2
2
2
27
4
81 81 729
4 16 16
8.4187 m.
Example 2: A Particle Q moves in a straight line
AB so that its acceleration at time, t, is 3t 2 –16t +
20ms -2 . Given that its initial velocity is
-16tms -1 . Find (i) the time it comes to rest at B if
it is at rest at A after 2 seconds (ii) Distance
covered from A to B.
Find the (i) v a dt
t
8t
20t
c
3 2
=
1
3 2
t 0 ; 16
0
80
20(0) C
16
3 2
v t 8t
20t
16.
At rest v = 0 and t = 2 is one of solution.
t
3
8t
t2 3 2
t
2t
2
6t
20t
16
2
6t
20t
16
2
12t
8t
16
8t
16
0 0
But (t 2 – 6t + 8) = (t - 4) (t - 2) = 0
t=4 or 2
Hence at B, t = 4
1
(ii)
x
4
vdt
AB
2
3
t
8 3 2
x AB
t 10t
16t
4 3
2
64
8
64101616(4)
4 3
8
8
8 1041612
4
3
512
8 64
16 160
64 40 32
3
4 3
568
12
47.33 m.
Example 3: A and B are two adjacent schools
and C is a trading centre on a straight line in
between A and B. A teacher driving can only
stop at A and B has a velocity of
3 3 2
t t km/
min at t minutes past
8 8
8.00am, and it passes C at 8.00am.
Find (i) the time of departure from A and
arrival at B.
(ii) an expression for the distance of the car
from A in terms of t .
(iii) the average velocity between A and B in
nearest Km/hr.
(iv) the maximum velocity attained in Km/hr.
Solutions
(i) At A and B velocity is zero.
3 3 2
t t 0
8 8
2
3t
8t
3 0
t 8
100
6
3 or
1
3
minutes
Time of departure from A at 7:
59 : 40 am and time of arrival at B
at 8 : 03 am.
(ii) x v dt
2
t 1
t t
8 2 8
3 3
c.
4
255

1
at t and x 0
3
(iii)
(iv)
3 1 1 1 1 1
0 C
8 3 2 3 8 3
1 1 1
c
8 18 216
7
C
108.
2
3 t 1 3 7
x t t
8 2 8 108
9 9 27 7
x3
8 2 8 108
493
D Km
216
Total time
1
t 3 as t is time pass 8am at B
3
D
average velocity = T
dv
dt
formin
or max . v,
2
d v
2
dt
V max =
493 10
Km / min
216 3
493 3
60
216 10
41 Km / hr.
3
1
t
4
t
3
8
(B) from a =
4
3
4
3
dv
dt
3
8
25
Km
24
62.5Km
2
3
dx
500
2t
5
t 10seconds
.
(i) a = -cv 3
dv 3
cv
dt
3
v dv
c dt
2
v
ct
k
2
2
v 2ct
B
3
a cv
3
2 c10
1
c
4
500
0 t min
3
1 2
0
B
2
veindicates
max
10 500
1 2 1
t
2
v 500
2
4
2t
100
5
3
500
2 500
v
/ min
2t
5
1
(iii) v 500 2t
5 2
/ hrs.
10
1
dv
x 500 2
5
2
t dt
0
dt
10
dv
500 2t
5 1
2 c..
dt
dx
dx
dv dx
dx dt
dv
v
dx
dv
a v
Example 1: A particle moves in a straight
horizontal line and is subjected to retardation
whose magnitude is proportional toV 3 . the
initial speed of the particle is 10ms -1 and initial
retardation is 2ms -1 .
(i) show that its velocity time t later is V 2 =
(ii) Find the displacement of the particle after
When t = 0, v = 10 and a = 2
0
256

500 2t
5
500 5
62 m
5
1
2
5
EXERCISE 7 (d)
1. A body of 2 units moves under the action of
a force which depends on the time t given
2
24t
i 36t
16
j . Given that at t = 0
by
~
the body is located at 3i j and has a
velocity
~
~
~
~
6i
15j
. Find (i) velocity of the
body at t = 2
(iii) displacement of the body at t = 2.
2. The speed of a particle is given by
2
t 2t
4 and given that the particle moves
in a straight line.
Find (i) the value of t when the acceleration
is zero.
(ii) the distance covered by the particle
between the instants when t = 1 and t = 3
3. A particle starts from the point O with
initial speed u and moves in a straight line .
8
At and instant its acceleration is given by v
, where v is the speed of the particle . The
particle passes through the point p with
speed 24.
(i) Show that the particle moves from O to
3u
2
P in time and that the distance OP
16
7u 2
is .
24
4. A particle Q starts from rest at point O and
its acceleration at any time t seconds later is
given by (3 – 2t). Find the displacement of
particle Q from point O when it is next at
rest.
5. A body of 3Kg moves in a straight line is
acted upon bby the force (12t – 8)N. Given
that its initial velocity is – 8ms -1 , find (i)
the value of t at the point A and B when the
when the body is instantaneous by at rest.
(ii) the distance between A and B.
~
6. Given that a particle is moving in a straight
line Oy with retardation of 2t 2 ms -2 at any
time t seconds. The particle starts from a
point A with velocity of 5ms -1 . Describe its
motion after.
(i) 1seconds. (ii) 5 seconds
7. A particle is moving in a straight line with
an acceleration of 8t, where t is the time.
When t = 3 second the particle has a
displacement of 36 m from O and a velocity
of 37ms -1 ! Find
(i) Velocity and (ii) position of the
particle when t = 2 seconds.
8. The acceleration of particle moving in a
straight line from point O after leaving O is
(2 – 6t)ms -1 , when t is time after leaving O.
If its speed at O is 2ms -2 . Find the time at
which after leaving O.
(i) it first comes to rest.
(ii) the particle returns to O.
(iii)the particle is at a distance of 2 from
O.
9. A particle moves in a straight line and
passes affixed point A with acceleration (16
– 4t) a, where t is the time in second s after
passing point A. Given that velocity of the
particle is 38ms -1 when t = 3, find the
(i) Velocity in terms of t at any time after
point A.
(ii) displacement from point A after t = 4
10. A car starts from rest and its acceleration
t
after t seconds is p ms -1 . Until it
6
reaches a velocity of 60ms -1 at the end of 60
seconds. Find value of P and distance
traveled in this first minute if it traveled in
straight line.
11. A and B are two adjacent railway stations,
and C is a signal box on the straight line
between them. A train which stops at A and
2
3 t t
B has a velocity of
Km/min at t
8 2 2
minutes past 6.00pm, and the train passes C
at 6.00pm. Find
(i) an expression for the distance of the
train from A in terms of t
(ii) the average between A and B in Km/hr.
257

(iii)the maximum velocity attained in
Km/hr.
12. A particle moves along a straight line from
1
a fixed O with acceleration ms -2 .
2
t 1
find
velocity and (ii) displacement of particle
from O after time t if the particle starts from
rest and it is 2m away from O tat beginning.
13. A particle of mass 2 units moves under the
action of a force which depends on the time
2
t given by F 24t
i (36t
16)
j given
~ ~
~
that at t = 0, the particle is located at 3i
j
and has a velocity
. Find
(i) Velocity and kinetic energy of the particle
at t = 2sec.
(ii) displacement of the particle at t = 3sec.
14. A particle of mass 2kg initially at rest at (0,
2t
0, 0) is acted upon by the force t N
t
3
Find; (i) its acceleration at any time t.
(ii) its velocity after 3 seconds.
(iii) its position after 3 seconds.
(iv) the distance the particle has
travelled after 3 seconds.
7.14 Integration of trigonometrical function
1. Sine and cosine of angles
d
From (sin x)
cos x
dx
d sin x cos x dx
cos xdx
d
from cos
x
sin
x
dx
d(cos
x)
sin
xdx
sin xdx
d sin x
sin x C
~
cos
x C
d cos x
6i15
j
~
~
~
2. Odd powers of sine and cosine of angles
If x is any angle in radian then we must recall
these identities
2
2
sin x cos x 1
2
2
1
tan x sec x
2
2
1
cot x cos ec x
Example 1:
sin
3
Example 2:
but
sin
3
xdx
sin
xdx
sin
2
cos
1 2
sin xdx
xsin
xdx
x sin xdx
cos
2
xsin
xdx
Let u = cos x
2 du
cos
x u sin x
sin
x
2
cos x u du
1 3
cos x u C
3
5
4
cos xdx cos xcos
xdx
cos
2 2
x cos xdx sin
2
x
2 4
1
2sin x sin xcos
xdx
2
4
cos
x 2sin xcos
xdx
sin
2
cos xdx 2 sin xcos
xdx
cos xdx sin x K
1
2
2
2
sin
u
2
2
u 2
x cos x dx
cos x
dx
3
du
cos x
1
x cos
3
3
1 2
u sin x,
x cos x C
sin
cos xdx
xcos
xdx
4
du
cos x
xcos
xdx
dx
258

2
u
3
sin 4 xcos
x dx
If K
NB: When integrating sine or cosine of odd
powers. First reduce its power by one times
sine or cosine of order one respectively.
7
6
E.g. cos cos cos
so that a factor with
even powers can be expressed using identity
2
2
1 sin cos
7
2 3
cos 1sin
cos
3. Even powers of sine and cosine of angles
Suppose we have x to be any angle then we
can use double angle formula as the key factor
in the integration of sine and cosine of even
powers.
2
2
Recall cos 2x
cos x sin x
1
So that cos 2 x 1
cos 2x
2
1
sin 2 x 1
cos 2x
2
Example 1:
u sin x,
4
2
sin xdx sin x
1
cos
3
5
K
4 u
u du K 3
5
K
2
K
3
2
xdx sin x sin
3
2
1
sin
5
2
sin
3
5
5
1
2
4
x K
Cthen,
2
dx
1
cos 2x
du
cosx
2
dx
1 2
dx
1
2cos2x cos 2x dx
1 1
1
2cos 2xdx
cos
2 2xdx
4
4
3
3
x K
3
2
1
x sin
5
5
x C
Example 2:
x 2x
cos
2 dx
1 1
cos dx
3 2 3
1 1 2x
dx cos dx
2 2
3
x 3 2 x
sin K
2 4 3
x 1 1 1
sin 2x
1
cos 4xdx
4 4 4 2
x 1 x 1 1
sin 2 x sin 4 x C
4 4 8 8 4
Example 3:
1
1
dx
1
cos x 2 x
1
cos
sin
2
1
dx
2 x
1
2cos 1
2
1 x 1 2 x
sec
2 dx tan C
2 2 2 1 2
1
x
dx tan C
1 cos x 2
4. Product of sine and cosine of angles.
Make use of factor formulae
P Q P Q
Recall sin P sin Q 2sin cos
2 2
P Q P Q
sin P sin Q 2cos sin
2 2
P Q P Q
cos P cos Q 2cos cos
2 2
P Q P Q
cos P cos Q 2sin
sin
2 2
Example 1:
3 x 1 1
sin 4 xdx sin 2 x sin 4 x C
8 4 32
2sin 5xcos xdx
Comparing
P Q P Q
2sin
cos sin P cos Q
2 2
2
dx
x
2
259

P Q P Q
5x
and x
2
2
P Q 10 x ……………….(i)
P Q 2x
…………………(ii)
(i) + (ii) 2P 12 x;
P 6x
angle to a bigger angle)
(add smaller
Substituting for P in equation (ii) we have
6x Q 2x;
Q 4x
(Subtract smaller angle
from bigger angle) thereafter recall proper
factor-formula.
2sin 5 cos xdx
sin 6x
Example 2:
2
0
x sin 4x
dx
P = 2x + x = 3x
And Q = 2x – x = x
2 sin 2 x cos xdx
0
1
6
1 3 1
cos cos 0 cos
cos 0
6 2 2 2
1 3
6
2
3
Example 3:
2
3 cos 5
sin x xdx
0
P = 5x + 3x = 8x and Q = 5x – 3x = 2x
2
1 2
sin 3 x cos 5 x (sin 8 x sin 2 x ) dx
2
0
sin 2 x cos xdx
2
cos 3 x 0
cos x 0
1
cos
6
3
2
1 1
cos 6 x cos 4 x C
6 4
1
6
1
2
1
2
0
2
0
1
sin 3 x sin x dx
2
2
1
6
1
2
2
260
1 1
cos 8 x cos 2 x
16 4
0
1 1 1 1
cos 4
cos
cos 0 cos 0
16 4 16 4
1 1 1 1
16 4 16 4
1
1
1 1 1
16 4 16 4 2
Example 4:
3
sin 4
sin 2d
P = 4 + 2 = 6 and Q = 4 - 2 = 2
3
3sin 4
sin 2d
cos6
cos 2
d
2
1 3
sin 6 cos 2 K
4 4
EXERCISE 7 (e)
1. Evaluate
(i) 2
0
cos x sin 2xdx
and
ii) 2 cos2x
sin
0
2. Evaluate
3
tan x
i) 4 dx
0 3
sec x
ii) 1 cos
0
3. Evaluate
4 2
xdx
xdx
i) 2 cos 2x
sin 4xdx
0
ii) 6 cos
0
4. Evaluate
3
0
3
xdx
2
(1 cos3x
) dx
5. Find sin xdx
6. Evaluate
(i) 2
0
and
cos 2x
sin 4xdx

(ii) 4 2sin 3x
cos 2xdx
6
7.15 Integration of exponential function
Example 1:
Let t 5x
10e dx
5x
dt
dx
5
t dt t
10e
2e
dt
5
2 e t K
Example 2:
xe
( x 2 1)
dx
2
Let m 2x
1
dm
dx
4x
m
m
xe
dm
4x
4
1 e m c
e
4
Example 3:
sin
x 2 dx
x
e
Example 4:
1
2
4xe x 1 0
Let
2dx
d
dx
x 2
10 e
xe
e
2
x
dx 2 e
5 x 5 x
2xe
2
x x 2
2
x
e
2
2xe dx dx
2
x 1 x
e xdx
e
2 1
e dx 2
x e
2x
dm
2
x 1 1 2
x 1
dx e
4
K
C
x
x
sin x 2 e dx cos x 2 e K
x 2
Or
Let t = x 2
dt
dx
2x
4
4
2
2e
xe
t
2
x
1dx
t dt
xe 1
2x
t 1
e dt dt
2 t
t cC
4x
e
2x
EXERCISE 7 (f)
1. Find (i)
e x 1
dx ,
2
e
x
(ii)
e
dx ,
2x
1
e
2x
(iii)
e
2 x
dx , (iv) 2x
x e dx
1
e
(v)
x
2
x
1
0
2 1
1
0
2
x0
2e
1 2
e 0
e x 2e 1
2 0
1 2
4xe x 1
dx
2e
1
4.44(2dps)
0
xe x2 dx (vi) xe ax2 dx
7.16 Integration by inspection or
Recognition
Whenever we recognise that there exist
function and its derivative (function and its
derivative exist) we change the function to
different variable and carry out the integration.
2 3 5
Example 1: x 4x
5 dx
2
Since x is a factor of derivative of 4x
3 5
Let u 4x
3 5
du 2 du
12x dx
2
dx
12x
2 3 5
2 5 du
x 4x
5
dx x u
2
12x
1
u 5 du
12
1 6
u K
6 12
2 3 5 1 3 6
x 4 x 5 dx 4 x 5 C
72
K
where C
72
2
261

Example 2:
cos 2
2xsin
2
xdx
We can consider sin 2x
as a factor of
derivative of cos 2x.
Let t cos 2x
dt
dx
2sin 2x
2
2 dt
cos 2xsin
x dx t sin 2x
2sin 2x
1
t 2 dt
2
3
t
C
6
2
1 3
cos 2 x sin 2 xdx cos 2 x C
6
2
Example 3: x
3
x
1dx
1
3
m
Let x
3
3x
2
1
m
dx dm
2
since x is a factor of derivative
dx
1
2
x
2
x
dm
2
3x
3
x 1 dx
dm
3 2 3
x 1 dx x 1
2
x
Example 4: dx
6
1
x
3
Let u x
du
2
3x
dx
2
x du
2 2
1
u 3x
1 du
3
2
1
u
2
2
9
m
1
2
3
2
C
9
x
2
m
1
2
3
2
dm
3 x
3
of x
2
C
1
x
1
sin
3
1
dx sin
3
Example 5:
x 2
dx
2
x 4x
4
let u = x 2 – 4x + 4
du
dx
2 x 4
x 2 du
u 2x
4
1 1
du
2
u
1
ln u C
2
1 ln x
2 4 x 4 C
2
1 2
EXERCISE 7 (g)
cos x
1. Evaluate
2 dx
0 2
1
sin x
3
2
2. Find
x
x e dx
3. evaluate
3x
4x
1
dx
3 2
x 2x
x
x
4. Show that
2
1 cos x
2
2
x dx
5. Evaluate 0 3
( 1
x )
x 2 C ln x C
ln
2
2
2
x
2
x 1 3
6
1
u C
dx
C
1
2
1
2
6. Find ( 1 3cos x)
2
sin 2xdx
e
7. Evaluate 2 dx
e xln
x
8. Evaluate
2
x
tan
2
C
262

(i)
(ii)
0
2
x ( 1
x
2 ) 3 dx
2
x 2
dx
x 4x
1 2
9. Evaluate 2 2
sin x cos
0
5 x
1
10. Find 3 2 dx
x 2x
xdx
cos x
11. Evaluate
2
0 2 1
sin
dx
x
4
2
12. Evaluate x x
13. Evaluate
(i)
3
2 2
x
0
0
sin( x
9 dx
3
) dx
7.17 Integration using trigonometric inverse
function
Recall
y sin 1
x
2
sin y x but cosy = 1 sin y
cos ydy dx
2
as sin
2
y cos y 1
dy 1 2
2
cos y 1
sin y
dx cos y
1
2
1
sin y
1
2
1 x
d 1
1
sin
x
dx
2
1
x
1
When y cos 1
cos y x
x
1 1
x
2
dx sin
2
sin ydy dx but sin y 1
cos y
2
2
dy 1 as sin y cos y 1
2
2
dx sin
y
sin y 1
cos y.
x
K
dy
dx
When y tan 1
tan y x
dy
sec 2 y 1
dx
dy 1
but sec
2
dx sec y
dy
dx
1
2
1
tan y
Example 1: Find
Example 2:
x
dy 1
2
dx 1
x
d 1
1
tan x
dx 1
x
1
dx tan
2
1 x
2
Let sin u x
cos udu dx
1
2
1 sin u
2
but 1
sin u cosu
1
ucu
cosu cos
du u k but u sin
1
1
x
dx sin
1 1
2
1
x
2
d
dx
1
2
1
1
cos
y 1
tan
x
-1
1
1
cos udu
1
x
2 2
a b x
C
2
1
x
2
x 2
x
k
dx
x
1
dx cos
dx
2
1
x
y
2
1
x
K
dy
dx
1
1
2
1
cos y 2
263

264
Example 3:
Example 4:
0.2272
2
3
=
72
3
4
3
6
3
dx
x
a
b
a
2
2
2
1
1
1
u
u
dx
udu
b
a
a
b
dx
du
u
x
a
b
u
2
2
cos
sin
but 1
cos
cos
Let sin
udu
b
a
u
a
cos
sin
1
1
1
2
du
u
u
b
cos
cos
1
dx
x
dx
x 2
2
9
4
1
1
3
1
4
9
1
udu
dx
x
u
cos
2
3
3
2
Let
udu
u
cos
2
3
sin
1
1
3
1
2
u
udu
cos
cos
2
3
3
1
x
u
3
2
sin
but
1
dx
du
u
dx
udu
x
u
2
2
tan
1
2
3
sec
2
3
3
2
Let tan
2
3
2
3
1
3
2
tan
6
3
x
1
tan
3
tan
6
3 1
1
=
6
3 u
6 3
du
. 0.3128
. 0.2272
2
2
2
tan
1
2
3
tan
1
1
3
1
3
4
1
1
3
1
u du
u
dx
x
dx
x
dx
x 2
2
3
2
3
2
3
4
1
1
3
1
4
3
1
3
2
sin
2
1
4
9
1 1
2
C
x
dx
x
C
u
du
2
1
2
1
C
x
a
b
b
x
b
a
1
2
2
2
sin
1
1
x
a
b
u
C
u
b
du
b
1
sin
but
1
1
0.3128
2
3
u
x
0.3128
0.2272
0.3128
0.2272

In general
1 1 dx
dx
2 2 2 2
a b x a 2
b
2
1 x
2
a
1
2
a
a 2
sec udu
b
2
1
tan u
2
1 sec udu
ab
2
sec u
1
du
ab
1 u K
ab
1 1
tan
ab
b
Let tan u
a
a
sec
b
2
u tan
udu dx
1
but 1
tan
x
b
x
a
2
u sec
Example 1:
1
dx
2
25 x
3
1 1
dx
25
2
x 3
1
5
x 3
Let tan u
5
2
5 sec udu dx
2
1 5sec u
2
2
du but sec u 1 tan u
25
2
1 tan u
2
1 sec u 1
du u k
5
2
sec u 5
1 1 1 x 3
dx tan K
2
25 x 3 5 5
Example 2:
1
dx
2
4 x 1
b
x K
a
2
u
1
2
1
2
1 x 1
dx 2sin u
2
2
x 1
1
dx 2cos udu
2
1
2cos udu
2
1
sin u
Hence
du u k
EXERCISE 7 (h)
1
1. Find (a) dx
4 3x 2
dx
2. Find (i)
2
9 25x
dx
(iii)
2
64 9x
Let
Example 3:
1
dx
x 18
1
dx
2 2
2 12x
21 2x
6x
9
21
dx 1 dx
2 3
2 2
x
3 3 1
3
2 x
3
2
tan u x
3
3
dx
3
secudu
2
2
but 1
tan u sec
3 2
sec udu
1 2
=
3 2
1
tan u
1
3
2
1 1
tan
6
4
1 1
dx sin
2
2
3
2
x 1
u
sec udu
2
sec u
2 x 3
1
u K
6
K
x 1
2
K
3
(b) dx
81144 x 2
2dx
(ii)
2
9 16
x
dx
(iv)
16 x
2
265

5dx
3. Find (i)
2
9 4x
dx
4. Find (i) 2
x 6x
13
dx
5. Find (i) 2
x 2x
10
6dx
(ii) 16
x
2
5dx
(ii) 2
2x
5x
6
5dx
(ii)
2
7 36 x
7.18 Integration of partial fraction
1. Partial fractions
What to be done when we want to express
fraction into partial fractions.
Step: check the highest power of x in
numerator and denominator after expansion to
ensure the fraction is proper. If;
a) Highest power of x in numerator is less
than that of denominators then the fraction
is proper.
b) Highest power of x in numerator is equal or
greater than that of denominator then the
fraction is not proper or is improper.
Improper fractions must be expressed as
mixed fraction before as expressing it into
partial fraction by carrying long division
3
x
Example 1: express as mixed fraction
2
x 1
Solution
3
x
2
x 1
x
2
x 1
3 2
x 0x
0x
0
3
x 0x
x
x
as 3 2
3
x x x
x x
2
2
2
x 1
x 1
x 1
3
x 3
Example 2: Express
2
x 2 x 1
fraction
Solution:
3
x 3
2
x
2x
1
3
x 3
2
x 1
i.
e 3 3
as a mixed
x
3
2x
2
3
x 3
x
1
x 2 x
3
3
0x
2x
2x
2
2
2
0x
3
x 2
x 1
2
2x
x 1
1
2
2
x
2x
1 x
2x
1
4
x 2 x1
Example 3: express
2 2
x
32
x
fraction.
Solution:
4
x 2x1
2 2
x
32
x
4
x 2x
1
4 2
x x 6
1
i.
e.4
as a mixed
4
4 3 2
4 3 3
x 0x
x 0x
6 x 0x
0x
2x
1
4
x 2x
1
4 3 2
x 0x
x 0x
6
2
x 2x
5
2
x 2x
5
1
2 2 2
2
x
32
x x
3
2
x
2
x
2x
5
1
2
2
x
3 2
x
Step 2: Check whether the denominator can be
factorised and if so factorise it completely
(i) Find out there is difference on the two
squares at denominator
2x
x 14
Example 1: Express
2
4
x 1 x 3
simpler fraction
Solution:
2
2x
x 4
2x
12
x 1
2
4
x 1 x 3 2x
12x
1x
3
2
2
2x
x 4
into
4x
2 1is different of the two squares i.e
2
4x
1
266

Example 2: Express
fraction
x
x
4
2
1
in to simpler
Hence
2x
3 2x
3
2x
3
4 3 2 2
x x 2x
x
x
2 x
1 xxx
2x
1
Solution:
4
x 1
2
x
4
x 1
2 2
2
x
1x
1 x
1x
1x
1
2
x
2
x
1 x
1 x
1
3 2
3x
x 2x
Example 3: Express
2 2
4x
19
x
Solution:
4x 2 1
1
2x1
2x
2
9
x
3
x3
x
3 2
3x
x 2x
3 2
3x
x 2x
2 2
1
4x
9
x 1
2x1
2x3
x3
x
2. Find out if there is common factor at
denominator.
2
x 1
Example 1: Express
3 2 into
x 4x
3x
simpler fraction.
Solution:x is common factor and pull it out.
2
2
x 1
x 1
3 2
2
x 4x
3x
x x 4x
3
2
but x 4x 3 can be factorised as
2
x 4x
3 x
3x
1
2
2
x 1
x 1
3 2
x 4x
3x
x x 3 x 1
Then
2x
3
Example 2: Express 4 3 2
x x 2x
Solution:
Common factor is x 2 .
4 3 2 2 2
x x 2x
x x x 2
2
x
x
2x
1
Step 3: Check the nature of factors at the
denominator.
1. Check whether the factor at denominator is
linear in order to choose the constants for
partial fractions. If linear factor (constant)
put capital letter on top of it because
derivative of the linear factor is a constant.
Example 1: Express
fraction form.
Solution:
2
x 1
A
x
x
2
x 1
x
1x
3
xx
1x
3 x
1 x
3
Example 2: Express
x
12x
1
fraction form
Solution:
3x
1
x 1
2x
1
B
3x
1
x 1
2x
1
A
B
C
in to partial
in partial
Step 4: Check whether the factors at
denominator are repeated so that you start with
power 1 and increase until you reach the power
given of repeated factor as you put constant
(capital letter) on top of them
6x
Example 1: Express
2 in to
x 2 x 4
partial fraction form.
Solution:
6x
A
x
2x
4 4 x 2 x 4 x
4 2
Example 2: Express
3
partial fraction form.
Solution
B
2
x 2x
8
x 1
x 3
C
into
267

2
x 2x
8
A
x
1x
4 2 x 1
x 3 x
3 2 x
3 3
B
C
D
then compare the co-efficiencies of x of
different powers.
Step 5: check whether there is quadratic factor
at denominator. Before you conclude that the
factor is quadratic factor make sure it cannot be
factorised to become linear factors. Like x
2 4
is not quadratic factor because
2
x 4 x 2 x 2 , but if factorised into
surds i.e. x 2 – 2 = (x – 2)
( x 2)
is
considered not factorisable, hence x
2 2 2(1)
A(1
1)
B(1
1)
2 2A
A 1
remains a quadratic factor . If it is a quadratic When x = -1
factor then express its numerators as Ax B 2(
1)
A(
11)
B(
11)
2
2B
B
where A and B are constants 2(
1)
because A(
1when 1) Byou
( 11)
2
2B
B 1
2
differentiate ax bx c you obtain 2 ax b
since constants 2 a a a A as one constant. Example 2
This is the reason why all constants are capital
11x
1
Express
in the form
letters.
2
x 1x
1
2
x x 3
A Bx C
Example 1: Express
x
2
2x
2 into partial
9
x 1 2
x 1
fraction form
Solution:
Solution
x
2
2
x 3 Ax B Cx
D
11x
1
Ax
1 ( Bx C)(
x 1)
x
2
2x
2
9 x
2
2 x
2
9
When x = 1
12 A(2)
A 6
Example 2: Express
form
Solution:
x
2
5
x
4
1
x
2
5
2
x 5
x
4
1
into partial fraction
x
2
1 x
2
1 x
1x
1x
2
1
x
2
5 A B Cx
D
x
4
1
x 1
x 1
x
2
1
x
2
5
Step 4: How to find the constants
NB:
1. Start by eliminating linear factors.
2. If all the linear factors have been eliminated,
put x = 0
3. In case the two above have been done and
there are still some constants left expand
both sides and bring them under same LCM
Example 1.
2x
A B
Find A and B
x 1
x 1
x 1
x 1
denominator with only linear factors
Solution
2x
A x 1
B x 1
When x = 1
2
2
11x
1
Ax A Bx Bx Cx C
Comparing coefficients of x on both sides for
2
x A B 0 6 B 0
B 6
for x 11 C B substituting for B
11 C ( 6)
C 5
11x
1
6 6x
5
2
1 2
x 1
x 1
x x 1
6 6x
5
2
2
x 1
x 1
x 1
Example 3
1
If =
2
( x 2)( x 1)
A B D
then find A, B and C.
x 2 x 1
2
( x 1)
1
268

Solution
2x
Example 1: Express
as partial
1 A B D
x 1 x 1
2
2
( x 2)( x 1)
x 2 x 1
( x 1)
2x
fraction. Hence find
dx
2
1 A(
x 1)
B(
x 1)(
x 2) D(
x 2)
x
1 x 1
When x = 1
2x
A B
1
denominator with
1 3D
D
x 1
x 1
x 1
x 1
3
only linear factors
When x = -2
2x
Ax
1 Bx
1
2 1
1 A(
3)
A
When x = 1
9
2(1)
A(1
1)
B(1
1)
2 2A
A 1
2
2
1 Ax 2Ax
A Bx Bx 2B
Dx 2 When x = -1
3
comparing coefficients of x on both sides. 2(
1)
A(
11)
B(
11)
2
2B
B 1
2(
1
1) A(
11)
B(
11)
2
2B
B 1
0 A
B B
9
2x
1 1
dx dx dx
1
hence
x
1 x 1
x 1
x 1
2
( x 2)( x 1)
ln( x 1 ) ln( x 1 ) K
1 1 1
9( x 2) 9( x 2) 2
3( x 1)
Example 4
2
x 2x
7 A Bx D
If
, find A, B
2
2
( x 3)( x 1)
( x 3) ( x 1)
and D
Solution
2
2
x 2x
7 A(
x 1)
( Bx D)(
x 3)
When x = 3
9 6 7 A(10)
A 1
2
2
2
x 2x
7 Ax A Bx 3Bx
Dx 3D
Comparing coefficients and constants on both
sides;
1 A B ……………………(i)
2 D 3B ………………..(ii)
7 A 3D ………………….(iii)
Hence
4 3
x 2x
4x
4 1
x 1
2
( x 3)( x 1)
x 3 x
Integration using partial fraction
2
1
2
2 x
dx ln( x 1 )( x 1 ) K
x 1 x 1
Example 2: Express
A Bx C
x 1 2
x 1
hence find
11x
2
11x
1
A x 1
( Bx C)(
x 1
When x = 1
12 A(2)
A 6
in the form
2
x 1x
1
11x
1
2
x
1x
1 dx
2
2
11x
1
Ax A Bx Bx Cx C
Comparing coefficients of x on both sides for
2
x A B 0 6 B 0
B 6
for x 11 C B substituting for B
11 C ( 6)
C 5
11x
1
6 6x
5
2
1 1 1
2
x x x x 1
6 6x
5
2
2
x 1
x 1
x 1
11x
1
dx
2
x 1
x 1
ln( x
2 1 ) K
1
)
269

6 6x
5
dx dx
1
2
x 1
2
x
dx
x 1
1
6x
6ln( x 1)
tan ( x)
dx
2
x 1
Let u x 2 1
du dx
2x
6x
6x
6x
du
2 2
x 1
x 1
u 2x
du
= 3
u
3 ln u K
3 ln( x 1 ) K
11 x 1
dx
2
( x 1 )( x 1 )
6 ln( x 1 ) 5 (tan
1
( x ) 3 ln( x 1 ) K
1
Example 3: If =
2
( x 2)( x 1)
A B D
then find A, B and C.
x 2 x 1
2
( x 1)
1
Hence find
2
x
2x
1 dx
1 A B D
2
2
( x 2)( x 1)
x 2 x 1
( x 1)
1 A(
x 1)
When x = 1
2
B(
x 1)(
x 2) D(
x 2)
1
1 3D
D
3
When x = -2
2 1
1 A(
3)
A
9
2
2
1 Ax 2Ax
A Bx Bx 2B
Dx 2
3
comparing coefficients of x on both sides.
1
0 A
B B
9
1
hence
2
( x 2)( x 1)
1 1 1
9( x 2) 9( x 2) 2
3( x 1)
2
1
dx
2
( x 2)( x 1)
1
( x 2 )( x 1 )
Example 4: Express f(x)
4 3 2
x 2x
x 4x
4
as a partial fraction.
2
x 1
x 3
Hence find
Solution:
4 3 2
x 2x
x 4x
4
2
( x 3)( x 1)
f
dx
x
2
x 2x
7
x 1
2
( x 3)( x 1)
because of carrying
long divisions.
We can express
2
x 2x
7
2
( x 3)( x 1)
in partial
fraction.
2
x 2x
7 A Bx D
2
2
( x 3)( x 1)
( x 3) ( x 1)
2
2
x 2x
7 A(
x 1)
( Bx D)(
x 3)
When x = 3
9 6 7 A(10)
A 1
1
dx
9( x 2)
1
1
dx
9( x 1)
3
1 1
1
ln( x 2 ) ln( x 1 ) K
9 9 3 ( x 1 )
1 x 2 1
ln K
9 x 1 3 ( x 1 )
2
dx
2
2
2
x 2x
7 Ax A Bx 3Bx
Dx 3D
Comparing coefficients and constants on both
sides;
1 A B ……………………(i)
2 D 3B ………………..(ii)
7 A 3D ………………….(iii)
Hence
4 3
x 2x
4x
4 1 2
x 1
2
2
( x 3)( x 1)
x 3 x 1
2
( x 1)
dx
270

x
4
2x
3
x
2
4x
4
dx
( x 3)( x
2
1)
1 2
( x 1)
dx dx dx
( x 3) x
2
1
Example 5: Find
Solution:
EXERCISE 7 (i)
1. Use partial fractions to evaluate
6 dx
(i) 4 2
x 2x 3
3 2
6 2x
2x
2
ii) Find 4 2 2
(1 x)
(1 x )
2x
x 14
2. Express f ( x)
as a partial
2
2
(4x
1)(
x 3)
fractions and hence evaluate 3
3. Find
( x 3)
(i)
( x 1)(3x
1)
2
2
2
dx
5x
16
(ii)
dx
4
x 16
2x
x
4. Find (i)
2 2 3
dx
( x 2)(3x
1)
ii)
2
( x
2
5. Express
2
x
dx
3
1)(3x
1)
2x
2
x 4
( x 3)(4x
2
x 2
dx
( x 1)(
x 1)
x 2 3
dx 1
( x 1)(
x 1)
( x 1)(
x 1)
3
3
1dx
dx dx
2( x 1)
2( x 1)
2
x
2
x ln( x 3 ) 2 tan
2
x 2
dx x
( x 1 )( x 1 )
2
1)
1
3
2
( x ) K
x
In
x
and
1
dx
1
C
1
f ( x)
dx
6. Express
f ( x)
dx
7. Express
f
x
x 4
x)
( x 3)(1 x
(
3
2
hence evaluate
8. Find
(i)
ii)
3
2
hence find
)
1
into partial fractions,
x 1
3
2
x
x
dx
2
x x 20
3
x
dx
2
( x 1)( x 4)
2
2
1
dx .
x 1
12x
x
9. Express as a partial fractions.
2
( x 1)(2x
1)
7x
2
10. Express f ( x)
as a partial
2 2
( x 1)
(4 x )
fraction hence, find
f ( x)
dx
x(
x 1)
11. Find the integral of
dx
2
( x 2)
11x
12
12. Express
in partial
3 2
2x
x 15x
18
fractions hence evaluate 6
ydx
5
3
2
3x
2x
3x
1
13. Express
into partial
2
x(
x 1)
fractions
2
3x
1
x
14. Express f ( x)
into partial
3
(1 x)(
x 2)
fraction. Hence evaluate 4
2
3
2
f ( x)
dx
2x
14x
3
15. Express f ( x)
into partial
2
2x
9x
4
fractions hence find
f ( x)
dx
4
x 2x
16. Express dx as partial fractions.
2
( x 1)(
x 1)
2
( x 2x)
Hence find
dx
2
( x 1)(
x 1)
271

4x
17. Express 2 3x
2
dx in partial fractions.
( x 1)(2x
3)
x 3x
2
Find
dx
( x 1)(2x
3)
4 2
3
5x
2x
5x
18. Express f ( x)
in partial
4
x 1
fractions hence evaluate 2
f ( x)
dx . Correct
to 4 decimal places.
7.19 Integration by Substitution method
1
Example 1. Find dx by using cos x
1 cos x
2
2 x
cos x 2cos 1
2
1
dx
2 x
(1 2cos 1)
2
1 1 2 x 1 x
sec tan C
2 x 2 2 2 2
2 cos
2
Example 2. Find x dx
x (3 1 )
2
( u u
2 u 3x
1
1) 2
u du
2
3 3 3x
u 1
2u
dx du
3
2 2 2
(
u 1)
u du
9
2 4 2
u
u du
9
2 5 2 3
u u K
45 27
Example 3 Find
Use u
2 3
u
( 3 u
2 5 ) K
135
2
3
3 x 1 2 3 3 x 1 5 K
135
2
1
( x 2)( x 1)
(3x
1)
4
dx
Let u x 1
x u 1
dx du
4
( u 3) u du
u du
u 5 4
( 3 )
1 5
( x 1)
(5x
13)
k
30
x(
x 4)
Example 4 dx
2
( x 2)
( x 2 )( x 1 )
( u 2)( u 2)
du
2
u
2
6
u
6
u
3
u
5
5
( u 2)2u
2
du (2u
u
2 3
u 3
K
6 5
4
5
K
dx ( x 1 )
u 4 u K
3
Let u x 2
x u 2
dx du
x 1 3
K
6 5
u 4 4 4
du 1 du u K
2
2
u u u
x ( x 4 )
4
dx ( x 2 ) K
2
( x 2 )
( x 2 )
1 ( x 2 ) 4 2 K
x 2
x
Example 5 dx
( x 2)
Let
u x 2
2
u 2 x
dx 2udu
2
4) du
5
272

Example 6 Use the substitution
evaluate.
Let
x sin
dx cos
d
1
2 dx
1 2
2 x 1 x
2
4
6
4
6
cot
( 1) (1.7321)
0.7321(4dps)
Example 7: Evaluate
Solution:
Let x
x
dx
( x 2 )
cos 2
dx 2sin 2d
1
0
2
(sin )
1 x
1 x
0
2
cos
( 1
cos ec d
4
6
4
4
sin
1
2
dx
x 2 )
dx
x
1
2
1
2
x
0
1
2
d
)
1
0 1
4
6
4
1
x
dx
x
x sin to
4
0
1 cos 2
( 2 sin 2 ) d
1 cos 2
1 cos 2
2 sin 2 d
0 1 cos 2
4
2
(
3
3
2
( x 2 ) K
Example 8: Evaluate
Solution:
Let
2 1
t tan
2
1 2
dt sec d
2 2
2dt
d
2
sec
2
2dt
2
1
tan
2
2dt
2
1
t
3
0
=
2
sin
2 sin 2 d
cos
0
4 2 sin cos
2 d
cos
2
4 sin d
0
0
4 1
4 ( 1 cos 2 ) d
2
0
2 sin 2 0
2 3
0 1 sin
d
2 d
2 tan
1 2
2
1 tan
2
2
2 sin
sin 2 d
2 cos
0
2
4
4
2
4
273

2
1 tan
4
= 3
2 d
0 2
1 tan tan
2 2
2 2dt
(1 t )
1
2
3
1
t
0
2
1
t 2t
1 dt
3
2 0
2
(1 t)
6
1
t
Example 9: Find
EXERCISE 7 (j)
1. Use substitution t tan to find
2
d
i) 1
cos
d
ii) 1
sin cos
2. Use the substitution x = sec to evaluate
(i)
ii)
5
2
1 4 2
5
x
1
Let
0
x
2 2
x u
2
x u
x
x
dx
6
2
dx
(x 1)
3
dx
( x 1)
dx 2udu
2u
du
2
( u 1)
u
2
1
du 2 du
2
u u
2
( 1) 1 u
dx
( x 1 )
2 tan
2 tan
x
1
1
(
x
( u ) K
x ) K
3. Use the substation x 2 = to find
2
x
dx
2
1 cosx
4. use the substitution u = x 3 + 2x 2 + x to
2
3
3x
4x
evaluate
dx
1 2 2
x 2x
x
5. By using substitution x = tan to evaluate
12
1
5
dx
4
3
3
2 2
)
x(1
x
1
6. By using x or x = sec 2 to evaluate
u
2 dx
x
x
1 2
1
2 2 x
7. Use x = 2cos2 to evaluate dx
0
2 x
8. Using the substitution u 2 = 2x – 1, evaluate
5 3x
dx
1
2x
1
b) Find 3x
dx
x 3
9. Using the substitution x = e y , find
3 ln x 3 ln x
(i) dx (ii) dx
2 2
x
x
10. Using the substitution u = 2-x show that
1 x
dx 3 4ln 2
0 2 x
11. Evaluate the integral
1
1
2 2
1
i) x 1 x dx ii)
0 dx
0 2
(1 x
2 )
12. a) Use u tan to evaluate
2
5
2
d
0 3sin 4 cos
x 3
b) Find
x e dx
2
13. a) Use the substitution u=x to evaluate
3 1
dx
1
x x
cos x
2
b) Evaluate dx , correct to 3
0 2
1
sin x
decimal places.
274

1
14. a) Find dx
3
2
(4 x
2 )
1
b) Evaluate the integral
2
dx
0
1
cos x
2 1
15. Use x = 2cos to evaluate
dx
1 2 2
x 4 x
16. (i) Use the substitution x = sin to evaluate
1
2
1
x
dx
0
1
x
(ii) Use the substitution x = cos 2 to
evaluate
17. Integrate
1
2
0 1
1
x
dx
x
1
2
2
(1 9x
)(1 9x
)
x
1
sin x 1 8
18. Show that
2
dx
ln
0
5 cos x 3 3
19. Show that if x = cos2,
1
2
with respect to
1 1
x
dx
1
0
1
x 2
3 x
20. Express f ( x)
as partial
2
(1 x)(1
x )
fractions find
f ( x)
dx
12
5
dx
21. Evaluate 4
by using
3
3
2
x( 1
x
2 )
substitution x = tan .
d
22. Evaluate
4
0 2
2
3sin cos
1
23. Show that 5 1
25x
2 2
2
dx
0
40
24. Use the substitution u = e x to show that
x
ln 3 e
0 2x
1
e 12
3
8x
25. Find the integral
dx
1 x
8
26. Evaluate 0
sin( 2x
)cos 5x
cos 2xdx
27. Find (i) sec 2 xsin xdx
1
ii)
dx
2
1 sin x
x
iii)
dx
16 x
2
1 1
28. Show that
2
x
dx
( 2 2)
0
1
x 2 2
29. Show that
x 1 2 1 4x
1
dx ln( 2x
x 1)
tan
2
2x
x 1
4
2 7 7
x
Find dx
5
2x
1
1
31. Integrate
with respect
3sin 2 4cos2
32. Evaluate
sin x
3
dx
2sin x cos x
2
0 sec
33. Evaluate (i)
d
4
cos
sin
(ii) 4 2
tan d
0
34. Using the substitution t = tan x
1
(i)
4
0 2
1
sin x
dx
(ii)
4
0 2
2
4cos x 9sin x
2 2
35. Evaluate (i) x ( 1
x ) dx
1
0
1
2
(ii)
4 2
(sec 1)
dx
0
36. Use the substitution t = sin x to evaluate
4cos x
2
2 to two decimal places
3 sin x
6
x
Find (i) dx
1 x
2
(ii)
x tan 1
(
3
x dx
5
2 x
37. (a) Evaluate dx
0 4
(16 x )
(b) Using the substitution x = 4sin 2 u to
2
evaluate x ( 4 x)
dx
0
3
2
C
275

38. Use the substitution x = 2cos to
dv
x
2 1
dx
dx
1 2 2
x 4 x )
2
x
v
2
7.110 Integration by parts
1
u sin ( x)
d du dv
Recall ( uv)
v u
du 1
dx dx dx
d du dv
dx 2
1
x
( uv)
dx v dx v dx
dx dx dx
2
2
x 1
x 1
du dv
uv v dx u dx
d ( )
sin ( x)
dx
2
dx
2 2
dx
1
x
dv du
2
uv u dx v dx
x
dx dx
From
1
dx
2
2
dv
du
1
x
u dx uv v dx
dx
Let
dx
Example 1: Evaluate
sin u x
2 2
2
x sin u
x cos xdx
0
cos udu dx
2
dv
sin
u
1
Let u x cos x
cos udu
dx
2
2
1
sin u
du
1
and v sin x
1 2
dx
sin udu
2
1 1
1
1
1
cos 2u du
2 2
cos 2u
du
2
4
4
x cos xdx
0
u 1
sin 2 u C
4 8
2 2
x sin x sin x 1 dx
0 0
1 1
sin
1 ( x ) sin u cos u C
4 4
2
2
x sin x cos x
0
0
sin u x
2
sin 0
cos
cos 0
cos u (1 x
2 2 2
1 1
1
2
sin ( x)
x 1
x c x sin ( )
0 0 1
4 4
1 x
2
2
x 1 1 1 1 2
sin ( x ) sin ( x ) x 1 x C
2
x cos xdx
2 4 4
0
2
1 1 2 1 2
sin ( x ) 2 x 1 x 1 x C
4
4
Example 2. Find x sin 1 ( x)
dx
1
sin 1 ( x dx
x )
276
Example 3. Find
xln xdx

2
x
v
2
2 2
x x 1
ln x dx
2 2 x
2
x x
ln x dx
2 2
2 2
x x
ln x K
2 4
Example 5.
x
xe dx
xe
xe
x
x
Let
u Inx
Example 4. Find
ln xdx
du 1
Let u ln x
dx x
dv
1
v x
dx
1
xln
x x dx
x
xln
x
e
1dx
x ln x x C
e
x
x
C
dx
u x
dv
dx
e
x
du
dx
1
v e
x
5. Find (i)
xsin
3x dx
(ii)
xsec
2
x dx
1
3
6. Evaluate ln( 1
x ) dx
0
2
1
7. Evaluate (i) x
2 ln xdx
(ii) 2
xe 2x
dx
8. Find (i) x
2 e
x dx (ii) ln x dx
(iii)
-1
sin (x)dx
9. Find (i) xsin x
(ii) Evaluate 0
dx
4
x e
x dx
1
1
10. Evaluate tan x
0
7.23 The mean value of a function
Consider a function over a closed interval [a,b].
to find the average (mean) value we use mean
value theorem of integral. This involves the use
of mean value of rectangle whose height is
referred as mean value of the function over the
integral defined. If we need to find the area
under the curve between a and b (As indicated
by shaded part in figure 142)
f(x)
f(b)
dx
1
EXERCISE 7 (k)
1. Evaluate 2
0
x sin 2 2xdx
f(a)
2. Show that
0
3
3
1
1
6x
tan (3x)
dx (4
3
9
3
Figure 142
x=a
x=b
x
2
3. Evaluate ln( x 4) dx
5
3
4. Evaluate
4
x cos2
2x
0
dx
277

Figure 143
Rectangle part shaded in figure 143 whose
height is f(c) crosses the curve same interval
and its area is equal to area under curve in the
interval. This is called mean value theorem.
If f(x) is continuous function on the mean value
closed interval [a, b], there exists a number C in
closed interval a and b such that area of
rectangle = [f(c)] [b-a]. but area under the curve
in figure 142 between a and b is
b
a
f xdx
hence
equating these two areas we have
f
b
a
cb
a f x
dx
b
1
b c
a
Mean value, f c f x
Example 1: find the mean value of
x
2
2 for x 1 and
Solution:
M.
V
f(b)
f(c)
f(a)
f(x)
x=a
4
x 4
dx
3 4
2 1
2
x
x dx 2 9
1
,
4 1
x
3 3
a
x=b
Example 2: The displacement x of the particle
at time t, is given by x sin t . Find the mean
value of its velocity over the interval O t
2
(i) with respect to time and (ii) respect to
displacement x
.
x
1
Solution:
dx
V cos t
dt
1
M.
V
0
2
2
sin
t 2
0
2
2
sin sin 0
2
2
1
0
2
11
ii)
v cos t
1
1
M. V
1
0
0
1 1
2 2
1
2
2
0
2
0
1
x
1
sin
o
cot dt
1
sin
2
t
1
x
t 0
2
x 0 1
1
2
1
x
2
dx
1
cos 2u
0
2
u
dx
1
2
du
cos u
Let x=x sin u
x 0 1
u 0
2
dx cos u d u
du
2
0
cos
2
2
u
du
278

u
2
4
4
1
sin
4
1
sin
4
2u
2
0
0
2
1
sin 0
4
EXERCISE 7 (l)
1. Find the mean value power out in hour if
P 1000 4800 200t
2
2. Find mean value xx 4
for0
t 24
y in the
intervals 0 x 4
3. Find the mean value of the function
x cos x for x
4 3
2
4. Find the mean value of 2xe x for0
x 2
5. Find the mean value of sin xsec
x for
0 x
3
6. Find the mean value of
x
3
ln x
for 1 x 2
7. Find the mean value of
3 x 4
8. Find the mean value of
interval
,
6 2
x
2
2
1
for
3x 2
4cos x
3
cos
9. Find mean value of sin 1 x
with in
3
interval
0 ,
4 .
10. Find the mean value of
3 x 5
x
2
2
x
with in
1
for
4x 29
Sample Questions
Q1. (a) Find the area enclosed by the curve
y x 2 2 and x – axis and the line x = 3.
(b) Find the area of the region bounded by the
1
curve y and x – axis, and the line x
x2x
1
= 1 and x = 2.
Solution:
(a)
Figure 144
Area of element = yδx (See figure 144)
Area of lamina from x = 2 and x = 3
Hence area needed
2ydx
3
3 2
2
( x 2) dx
3
x
2x
3
3
3
2(3)
3
2
(9 6) (2)
3
2(2) 6(2)
3
3
9 4(2)
3
= 4.8856(4dps)
(b)
3
2
3
2
2(2)
3
2(2)
279

(i)
y1 2x
y
and
3
x
y2
4
y 2
Figure 145
x
2
x
y 1 – y 2
= y
y 1
Area of element y x (See figure 145)
Area of lamina from x = 1 and x = 2
2
yx
1
2
1
A
x 2
1
dx
x(2x
1)
B
x
1
1
x(2x
1)
(2x
1)
A Bx 1
A 1
1
x
2
1
B 1,
B 2
2
2
1 2
dx
x 2x
1
1
2
2 2
ln
x1
ln( 2x
1
2
1
ln 2 ln 1
ln(5) ln 3
ln( 23)
ln(5 1)
6
ln
5
0.18223
Q2. Given curves y 2 2x and x 3 4y
Find
(i) the area they enclose and
(ii) the volume generated by revolving this area
in (i) through 2 about x-axis.
Figure 146
Area of element y x (See figure 146)
2
3
x
2x
dx
4
0
4
2
2 3
2 2 x
x
3 15
0
4
2 2 2 2 2
0
3 16
8 5
1
3 3
(ii)
Figure 147
Volume = y 2 dx (See figure 147)
Volume 1 2xdx
x
2
2
x
Volume
4
2
0
2
0
2
2
2
2
0
x
4
7
x
dx
7
16
2
0
8
7
280

8 20
volume generated 4
7 7
2
NB:Becausey 1
y
2
2
y
y 2
Q3. (a) Find the capacity of water which is
needed to fill a hemispherical bowl of
height 7cm and radius of 10cm.
Figure 148
From figure 148 let depth at any point be x
below the centre and radius of cross – section
2 2 2
be so that r 10 x
Volume of the hemispherical below
10 10
2
2 2
r dx 10 x dx
7
(b) Find the formula for the volume of a sphere
of radius r
Solution
Figure 149
7
3
10
100 x
x
3
7
10
1000 343
100 x 700
3 3
1000
343
300
3
657
300 cm 3
3
r
1
2
r
2 2 2
Volume r x dx (See figure 149)
3
3
3 r
3 r
r
r
3 2
3 2 3
2r
r
3
3 3
6r 2r
3
4 r
3
3
Q4. Show that the volume generated by the
revolution about the x – axis of the area
included between that axis and the curve
5
a b
cy ( x a)(
x b)
is
2
30 c
Solution
Volume
Then
Volume
2
c
2a
2
r
r
r
r
r
y
a
b
r
2
2
x
3
x
x 3
r
3
r
2
r
dx
r
3
3
3
3 3 r
a
b
y 2 dx
x
ax
b
a
b
2
c
( x
4
bx 2ax
c
x a
2
c
a
b
b
3
x
b
2
x
a x
b
2
a
a b
c
5 3
x
2
2ab
2bx
2
3
2
2
a
x 4abx
dx
dx
5 3 2
5
1 4 a
ba
2 2 3
2
2
x
)
2
2
a b
3
a b
2
2
281

5 5
4 5 3 2
b
1
a b a
2
a b
2
4
4 b
a b)
3
5 3
1 5 a 3 2 3 2 3
b b a b a b
2 3
4 4
ab ab
4
3
4
1 5 a b 1 3 2
a a b
2
c 30
6 3
4
ab 1 2 3
a b
6 3
=
1
ab
2
4
5ab b
= a b 5
2
30 c
Q5. A curve is given by the parametric
equations x 4cos 2t,
y 2sin
t .
(i) Sketch the curve for t
2 2
(ii) Find the area enclosed by curve and y –
axis.
5
1
30
5 4
3 2 2
a 5 a b 10 a b 10 a b
2
30 c
Solution:
t 0
2 4
4 2
x -4.000 0.000 4.00 0.000 -4.000
0
y -2.000 -1.414 0 1.414 2.000
y
b
5
4
3
4
Area required 2 ydx (See figure 150)
0
x
0
cos 2t
4
2sin t
8sin 2t
dt
1
x
2t
cos
4
4
4
x t
0
4
4 0
Negative will be removed by changing limits.
But sin 2t
2sin t cos t
4
Let u sin t
32 sin
2
t cos dt du
dt
0
cos t
32
4
0
16 sin t sin 2 tdt
32
2
2
0
u
2
22
2
u du
2
2
0
du
cos t
cos t
t u
0 0
2
4 2
3
2
3
2
0
2
2
= 3.7713 (4dps)
32 u
6
Figure 150
Q6. Find the area enclosed between the curves
3 2
y x x 5x
and the lines x = -2 and x
= 0 and y = 0. Hence find the area enclosed
between the curve and line y = x + 6 where -2 <
x < 0 when rotated through 90 0 about x-axis.
282

Solution:
X -3 -2 -1 0 1 2
y -15 4 9 6 1 0
266.2857243
145.6 74.66666667
24
70.01904763
Since the area is rotated through 90 0 , hence the
volume generated =
70.01904763
54.9928 cubic units
4
Figure 151
Area of element = y x (See figure 151)
Area of part required is sum of area of elements
0
x
2
Hence required area
0
3 2
= x x 5x
6 dx
1
4 1 3 5 2
x x x 6x
4 3 2
2
1 4 1 3 5 2
0 ( 2)
( 2)
( 2)
6( 2)
4 3 2
8
( 4 10
12)
3
15.33(2dpts)
Volume of area enclosed between the curve and
the line y x 6
0
2
f ( x)
xdx
2
0
2
3
0
3 2 2
x
x 5x
6 dx x
6
7 1 6 9 5 11 4
x x x x
3 5 2
13 3 2
x 30x
36x
3
3
x
6x
3
2
( x x 5 x 6) 2 ( x 6 ) 2 dx
2
36x
0
2
0
2
0
2
2
dx
Q7. Find the area enclosed y x 3 2,
x = 1, x = 2 and the x – axis.
x 0 1 2 3
Figure 152
y 2 3 10 21
The area of element y x (See figure 152).
The total area of whole lamina
x = 1 to x = 2
2
ydx
1
2
3
x 2 dx
1
2
4
x
2
4
x
1
x
16
1
4 2
4 4
283

1
8 2
4
32 8 1
4
Q8. (i) Find
x
x
2
2
2x
1
4
x 1
2x
1
When x = 1
When x = -1
x
2
2x
1
dx
4
x 1
x
2
2x
1
2 2
x
1 x 1
A
B
x
1 x
1
Cx D
2
x 1
2
2
Ax
1x
1
Bx
1x
1
2
Cx
Dx
1
1 + 2 – 1 = B(2)(2) B = 2
1
1 – 2 – 1 = A(-2)(2) A = 2
1
When x = 0
-1 = A(-1) (1) + B(1)(1) + D(-1) D = 1
Comparing coefficient of x 3
0 = A + B + C C = -1
2
x 2x
1
1 1 x 1
4
1 2 1 2 1
2
x x x x 1
1 1 x 1
2x
1 2x
1 2 2
x 1
x 1
x
(ii)
2
23
4
2x
1
1
dx dx
4
x 1
2
x
dx
2
x 1
1 1
ln
2 2
1
ln
2
1
x
dx
1
x
0
2
Let x tan u
2
dx 2 tan u sec udu
1
x
1 2x
1
1
dx
2
x 1
x
1 ln x
1
dx
2
1
x
1
tan x
K
4
0
2
tan u
2tan usec
2
1
tan u
4
2 tan
3
u sec udu
2
0
4
0
44
2
0
2
tan
(sec
4
2
sec
2
0
let t = sec u
1
3
4
Q 9. (a) Find (i)
u(secu
tan u)
du
u 1)sec
u tan udu
2
1 . 4142
2
(ii)
2
u tan u secu
du 2
2 t dt 2 sec u
0
0
2 3 1 . 4142
t 1 2 sec u
3
1
4 2 2
( 2 ) 2
1
2
3 3 ( 2 ) 2
1
1
4 2 2
( 2 ) 2 ( 2 ) 2 2
3 3 2
1
2
2 4 ) 1 . 8047
4
0
4
udu
4
0
e x dx
0
sec x dx
secu
tan udu
284

(b) Find (i)
Solution
e x
0
(ii)
(a) (i) dx =
3
x
dx
2
x x 20
1
2
x
2
x
1x
4 dx
1
1
x
e
1
= 0
1
= 1
1
(ii) sec x dx=
dx
cos x
x
Let t = tan
2
dt
dx = 2
2
1
t
1 2
dx = 1
t
2dt
cos x
2 2
1
t 1
t
2
=
1
t 1
t
0
Using partial fraction
A B 2
2
1
t 1
t 1
t
A = 1 and B = 1
2 1 1
dt dt dt
2
1
t 1
t 1
t
ln (1 + t) – ln (1 – t) + c
1
t
= ln C
1
t
x
1 tan
ln 2 C
x
1 tan
2
(b) (i)
3
x
21x
20
dx
2
x x
2
20 x x 20
21x
20 A B
x 5 x 4 x 5 x
x
1
dx
4
dt
21x – 20 = A(x – 4) + B(x + 5)
When x = 4 B =
9
64
135
When x = -5 A =
9
3
x
dx
x
1dx
2
x x 20
135 64
dx dx
9( x 5) 9( x 4)
(ii)
2
2
x 135
x ln( x 5)
2 9
64
ln( x 4)
C
9
2
2
x
2
x 1 x 4
1
2
x A Bx c
but
2
2
x
1 x 4 x 1
x 4
x 2 = A (x 2 + 4) + (Bx + c) (x – 1)
When x = 1
1
1 = 5A A =
5
Coeff. Of x 2
1 = A + B
4
B =
5
Const. O = 4A - C
4
C =
5
x
2
x
1x
4
1
2
2
1 1
dx dx
5 x 1
1
2
4 x 1
dx
5 2
x 4
1
1
2
ln
x
1
5 1
2 2 2
ln
x
4
1
5
2
1
x
tan
2
1
285

1 1 2
ln 1 ln 2 ln 8
2 2 5
2 1
ln 5 tan 1
tan
5
1 2 8
2 8
ln
5 5
0 . 5856
Solution
1
Cos 3x
0
1
2
3
2
Q 10 (a) Evaluate 1 dx
3
Cos 3x
0
3
2
1 dx
3
2
2 3x
Cos 3x
1 Cos dx
0
3
3
1dx
2Cos3xdx
0 0
3
1
1
Cos6xdx
2
ln 2
2
3
2
3
ln
5
1
2
0 . 3466
0
ln 2
x
0
sin 3 x 0
3
0 . 3976
0 . 2390
x 1
sin 6
2
x
o
0
12
2
o
sin
sin 0
3 3
3
1
0 6
0
6 12
2
1
3 6
6 6 2
5
3
4
3
0 . 1475
(b) Evaluate
4
2
d
2 2
3sin
Cos
Solution
4
d
2 2
3sin
Cos
2 2
4
2
sec
d
2
3tan 1
22
2
sec dt
=
2 2
1 3tan
sec
dt
Let t = tan d
sec
1
2
= 1
t 1
dt
2 2
1
3t
1 t
1
1
dt
2
1
3t
1 1 1
tan t 3
3
1 1
1 1
tan 3 tan
3
3
1 1
3 3 3 2
Q 11. Evaluate
2
dx
(a)
3 5cos x
0
2
3
(b)
dx
1
sin x
0
2
4
1 5
3 6
286

(a)
Solution
2
dx
3 5cos x
0
let t = tan 2
dx =
2dt
1
t
1
2dt
2 2
31 t
51 t
2
2
8 2t
dt
2
4 t
1 A B
2
4 t 2 t 2 t
1 A2
t
B2
t
A 11
and B 11
4
4
1 1 1
dt dt
42
t
4 2 t
1 1
ln 2 t ln 2 t C
4 4
1 2 t
ln C
4 2 t
22
dx
3 5cos x
0
x
4
1 2 tan x 4
ln 2
4 2 tan x x
2
0
1 1 1 1
ln 3
ln 2 ln 1
ln
4 2 4 4
1 3
ln
4 8
2
3 5
1
1
t
t
2
2
2
2dt
1 t
2
2
287
2
3
(b) dx let t = tan
x
1
sin x 22
0
dt 1 sec
2 x
dx 2 22
2dt
dx
2
1
t
Q 12. Find
(a)
(b)
(a)
1
3 2dt
2t
2
1
0
1
t
2
1
t
1
dt
6
2
1
2t
t
0
6
1
1
0
3
dt
t
Solution
2
1
6
1
t
1
0
1
- 6
1
2
3
6
x
0
1
1
x
x
dx
3
x
dx
3
x
1
1
2
x
x
t
0
1
dx
using x = sin
using x = 3 cos 2

(b)
dx
cos
dx cos
d
d
1
sin
cos
d
2
1
sin
x
1
sin d
cos
K
sin
1
x
x 1
dx
1
x
x 1
1
1
x
2
K
dx 6sin 2 d
12 sin cos
d
1
x
1
x
1
cos 2
12 sin cos d
1
cos 2
2cos
12 sin cos d
2
2sin
cos
12 sin cos d
sin
2
12
cos d
2
dx
3 3cos 2
12 sin cos d
3 3cos 2
1
12
1
cos 2
2
6
3sin 2
K
3cos
Q 13. Evaluate
1
2
x
3
3
9 x
2
K
4
1
dx
2
1
sin x
0
4
4
2
1
sec x
dx
dx
2
2 2
1 sin x sec x tan x
0
0
Let tan x = u
let
1
x
u
0 0
0
Q 14. Evaluate
du
dx =
2
sec
du
2
1
tan x
du
2
1
u
4
1
u
2
1
u
0
1
du
2
1 2u
0
2
u
2
1
0
x
du
1
u
2
2u
tan
1 2
d
sec d
2
4
1 2
1
tan d
2
1 1 2
1
tan d
2
1
tan 2
u
1 tan 1
2
1
2
2
d
1
2
1
tan
2
0
4
tan
2 0. 6755
1 1
288

(a)
(b)
(c)
(a)
ln 2
0
e
1
e
x
dx
Solution
ln 2
dx
x 2
0 e
1
let y = e x
dy = e x dx
dy
dx =
y
1
3
x ln x
2
2
dx
1
e x 22
1 dx
2
dy
yy
1
1
2
A
y
B
C
2
y
1 y 1
y
1
2
Ay
1 Byy
1 Cy
y
1
1
y 0 A 1
Ceoff.
of
2
y
y 1
C 1
1
0 A B B 1
2
dy
2
yy
1
1
2
1 1 1
dy dy
y y 1
2
y 1
1
2
2
2 1
ln
y ln
y 1
1 1
1
y 1
1 1
ln 2 ln 1
ln 3 ln 2
3 2
ln
4 1
3
6
6
2
(b)
(c)
3
x
x
3
e
3
x
ln
2
x
dx
1
dv 3
let x
dx
4
x
V
4
2
U ln
x
du 2
ln x
dx x
2 x 2 x
ln
x dx ln
x
4 4
1 4 2 x
x ln
x
4
4
1 4 2 x
x ln
x ln x
4
8
x
4
e
1
32
ln x
4
2
ln x
dx
2
4
2
1 0
4
4
5 e 1
1
let
e
1
4
4
4
x
8
1
e x 2
dx
let
e
8
1
1
ln x
x = ln y
e x = y
dy dx
y
1
dy
y
2
1
y 2
1
1 y
y u
e
32
1
2
2
4
1
e
1
u
dy 2u du
4
2
ln
x
x
3
ln xdx
2
3
x
dx
8
4
x
32
1
32
e
1
dx
289

Q 15. (a) Evaluate
(i)
(ii)
(b)
(a) (i)
Find
Solution
let
u
2u du
2
u 1
2
2u
1
du 2 1
du
2
u 1
2
u 1
1
2 du
1 du du
u 1 u 1
u 1
2 u ln
u 1
e
1
2
xsin
2 2 x dx
0
e
1
2 1 e
2 1 e
ln x
dx
x
ln x
x
1
x 2
x
u = ln x
du
1
dx x
dx x du
x
1
2
2
u 1
ln
2
u 1
x tan 1
u
1 0
e 1
ln
1
xdx
e
e
x
x
1
1
2
1
2
(ii)
(b)
1
u
x
2 2
0
1
0
2
1
2
1
2
u
x du
2
sin
2 2 x dx
x
0
xx
2
x
1
cos 4x
dx
2
0
2
x 1
dx xcos 4x dx
2 2
0
2 2
x 2 1 x 1
sin 4x
sin 4x dx
4
2 4 8
0
2
2 2
2
x 2 x 2 1
sin 4x
2
sin 4x dx
4
8
0
0 8
0
2
1
2
sin 2 cos 4x
2
16 16 32 0
2
1 1
cos 2 cos 0
16 32 32
2
1 1
16 32 32
2
16
1
x tan
x
dx
0
290

1
x tan
1
u tan
x
du 1
dx 2
1
x
dv
x
dx
2
x
V
2
x
2
x 1
tan
2
dx
x
2
x 1
dx
2 2
1
x
(b) (i)
2
Q 16. (a) Evaluate
dx
x
e 1
let y = e x
dy x
(ii)
e
dx
dy
dxdx
y
y
1
(b) Find (i)
(ii)
x sec x dx
3
x
dx
2
1
x
Solution
(a)
dx
x
e 1
dx 1 dy
x y
e
2
x
tan
2
2
x
tan
2
1
1
1
x
x
y
1
dy
y -1
1 1
2 1 x
x
tan
2
y
1
2
dx
x C
291
y
Let
1
1
1 Ay B
B 1
A B
y y 1
y
ln
and
y 1
u = x
du
1
dx
dv
sec
dx
v tan x
A 1
2
dx
2
x
ln e 1
x
e 1
1
1
x
x
2
e
1
ln e
1 2
1
1.854587 0.541325 1
0.3133 (4dps)
2
sec x
dx
2
x
2
1
= x tan x – (- ln cos x) + C
= x tan x + ln cos x + C.
3
x
dx
2
1 x
Let x = sin u
dx = cos u du
3
sin u
cos
u du
2
1
sin u
sin 3 u
cos u du
cos u
sin 3 u du
sin 2 u sin
u du
dy
y 1 y
1 1
dy
y 1 y
ln y 1 ln y
x
x
e 1 ln e C
ln
x sec 2 x dx x tanx tan x . 1 dx

cos u cos u C
cos u (cos 2 u – 1)
1
Q17: (a) Find
3 cos d
using t = tan
2
d
(b) Find
4 3sin
using t = tan
2
(a)
1 cos 2
u sin u du
sin u du
2
cos usin
u du
3
2 2
1
x 1
x 1
2 2
x 1
x
Solution
1
3 cos d
let t = tan
2
1 2
dt sec d
2 2
2 dt
d
2
1 t
1
3 cos d
2dt
1
2 2
1
t
1
t
3
2
1
t
2dt
2 2
3 3t
1
t
2
dt
2 4t 2
1
dt
1
2t 2
292
Q18 (a) Show that
sec
2
u du
21
tan
1
2
du
1
u
2
1 1
tan t
2
2 C
1 1
tan 2 tan C
2
2
(b)
d
4 3sin
let t = tan
2
2dt
d
2
1
t
1 2dt
2t 2
4 3
1
t
2
1
t
2dt
dt
2
2
4t
4 6t
2t
3t
1
dt
2
7
16 3
2 t
1
16
7 4
8 1
2
7 16 3
t
1
7 4
let tan u
4 7
7
4 7
7
1
2
0
1 x
1 x
2
4 3
t
7 4
2
sec u du
2
1 tan u
tan
1
1
2
u
1
dx
2 3
7
4
sec
2
2
2
u
4 3
tan C
7 2 4
3
du dt

(b)
(a)
Find the integral
2
5x
16
dx.
4
x 16
Solution
1
1
2
1
1
x 2
dx
1
x
0
Let x cos 2θ
dx = -2 sin 2θ dθ
x
0
1
2
1
2
1
1
x 2
dx
1
x
0
66
4
4
6
44
6
1
1
cos 2
2
1
1
cos 2
2
1
2
cos
4sin cosd
sin
4
4
4 cos d
66
4
6
2
1
4 1
cos 2 d
2
1
1
4
2
sin 2
4
6
6
6
2sin 2
d
The limits are interchanged
because of
the negative
b)
Q19
(b)
2
4 6
2 3
6 2
2
5x
16
dx
4
x 16
9 9
dx
8 x 2 8
Solution
d
(a) (i)
sin
=
d
2sin
cos
2 2
1
dx
x 2
1 1
dx
2 2
x 4
9 9
ln x 2 ln x 2
8 8
1 1 x
tan C
4 2
x 2 1
ln tan
8 x 2 4
d
Find (a) (i)
sin
d
(ii)
cos
d
(iii)
a bcos
x
2e
(i)
2x
e 1
dx
x
e
(ii)
dx
2
2
e
x 1
sec
2
2
d
2 2
tan
2
22
dt
where t tan
t
2
9 1
2
sin sin
2 3
1
2 3 2 3
2
x
C .
2
293

log
2dt
(ii) Let t tan d
2
2 1
t
2
d
1
t 2dt
2
dt
2 2
t t
2
cos
1 1 1
t
1 1
dt dt
1
t
1
t
1
t
ln C
1
t
ln C
4 2
d
t
2
a b cos 2
t 2
t
a b 1
1
2
t
(iii)
1
dt
2
a b ( a b)
t
2
a b a b
2
tan
1 tan
C
a b a b a b 2
If a c b, the integral involves a logarithm and is
effected by partial fractions.
x
2e
(b) (i)
e
2x 1
Let y = e x
dy
x
e
dx
dy
dx =
y
2y
dy
2
y 1
y
1
2 dy
2
1 y
1
2 tan y C
e
e
t log
e
x
2 1
2
e
2 tan
x
1
tan C
2
x
e
C .
dt
(ii)
Q 20
(i)
x
2e
2x
e 1
dx
Let y = e x
dy
dx =
y
2y
dy
2
y 1
y
2
2
y 1
dy
1 A B
2
y 1
y 1
y 1
1 = A(y – 1) + B(y + 1)
1
B =
1 and A
2 2
1
1
2 dy
2y 1 2y
1
ln y 1 ln y 1 C
=
y 1
ln C
y 1
Evaluate
1
(i)
dx
2
x 4
2
x
1
(ii)
dx
x ln x
Solution
1
dx
x
2 4 x
2
Let x = 2 sin
dx = 2 cos d
1 1
2cos
d
2
4sin
2
4 4sin
4
x
2 e
e
dx ln
2 x
e 1
e
1
cos ec d cot C
4
1 2
x
x
1
C
1
294

(ii)
1
x ln
x dx
u ln x
du 1
dx x
dx xdu
1
du
x .x
du
u C
ln x C
MISCELLANEOUS EXERCISE 7:
1(a) When the area bounded by the curve
y = 5x (2 – x) and the x – axis is rotated
through 2 radians about the x – axis.
Find the volume of the solid generated
(b) The area enclosed by the curve
y = x 2 + 2 and line y = 2 for 2 x 0
rotated about y = 2.
Find the volume generated.
2(a) Find the enclosed by the x–axis, the line
8
x = 4 and the y = x
2
x
If this area is now rotated about the x–axis
through 2 , determine the volume the solid
generated, correct to 3 sig. fig
(b) Determine the volume of the solid
generated when the region bounded by the
curve y = cos2x and the x–axis for the value of
x between 0 and 4
is rotated about x–axis
3. A shell is formed by rotating the portion of
the parabola y 2 = 4x for 0 ≤ x ≤ 1 through
two right angles about its axis. Find the
volume of the solid formed.
1
4. Find (i) dx and
1
cos x
is
(ii)
sin .
7 xdx
5. Given that curve y = x -
x
(i) determine the turning point
(ii) find intercept
(iii) sketch the curve and find the area enclosed
by curve, x = 3 and x-axis
2
2x
x 14
6. Express y = as partial
2
4x
1
x 3
fractions Hence evaluate
1
Π
2
7. Evaluate (i) sin 2x cosx dx and
0
2
3
3
(ii)
dx
5 4cos x
0
3
x 3
8. Express as partial fractions.
2
x 1
x 2
Hence or otherwise find
9. Show that
2 x 2
x ln x 4 2 ln A 2
x 2
1
10. Find
d 3 2cos
3 2
3x
2x
6x
2
11. Express f ( x)
as
2 2
x x 2 x 2
partial fraction and find
12. Find the volume generated by rotating the
curve y = x 2 – 1 from x = -1 to x = 1
through 360 0 about the x- axis
8
2
ln x 4dx
2
2
ydx
3
x 3
2
x
1 x 2
x
f xdx
dx
295

13. The curve y = (x – 1) 2 (2x – x 2 ) cuts x–axis
at three different points A, B and C
(i) Find the co-ordinates of A, B and C.
(ii)
Find the turning points of the curve,
and determine their natures
(iii) Sketch the curve
(iv) Find the area bounded by curve and
x–axis.
3
14. Evaluate (i) sin 3x
dx
6
0
Π
(ii)
6 sin x sin 3x dx.
0
1
3
x
(iii)
2
1
x
0
6x
15. Express f(x) = into partial
2
x 4 x 2
fractions. Hence find f ( x)
dx
16. A triangle A(0,0), B(3,3) and C(1,2) is
rotated about the x – axis. Find the volume
of the solid generated by that rotation.
2x
17. (a) Find dx
2
4 x
2
cos x
(b) Evaluate dx
2
1
sin x
0
(c)
Find
xsec
2
xdx
18. The area enclosed by curve y = 15 – 3x 2
and chord joining the points (0, 15) and (2, -3)
is rotated through 360 0 about the x – axis to
generate solid.
(i) Find the area enclosed between the chord
and the solid
(ii) Find the volume of the solid generated
19. (a) Prove by both integration by-parts and
using double-angle that
1
sin
2 xdx x
cos
xsin
xC.
2
1
x
dx tan C
1 cos x 2
(b) Show that
1
x 3
20. Evaluate (i)
dx
3 x
0
1
2x
3
(ii)
.
2
4
dx
x
0
6 5
21. (a) Prove that cos xdx
32
0
3x
11
(b) Find
dx
2
x 4x
5
4
2x
17
x 1
22. Express f(x) = as partial
2
x 2 x 5
fractions, hence find
f
xdx
2
dx 1
23. Show that
dx
1
2 4 2
1 x x
12
2
2
x
x 3
8x
24. Find (i) x16 dx and (ii) e dx
4
5
11x
12
25. Evaluate
dx
2
2x
3x
x 6
4
4dx
26. Find
2 2
9sin x cos x
1
2
1
27. Evaluate dx by using
2
1
x 1
x
0
x = cos
4
4 3
1
(b) Show that sin xdx
32 4
0
296

3
2
2x
1
28. (a) Evaluate
2
x
1 x
2 dx
2
(b) Show that the area enclosed by curve y = e x ,
y axis and the line y = 2 is 2 ln 2 – 1.
29. (a) Integrate x tan -1 (x) with respect to x.
1
1
Hence evaluate x tan x dx
0
(b)
1
1
tan x Evaluate
dx
2
1
x
0
30. (a) Use the substitution of x =
1
u
to
2
dx
evaluate
2
x x
1
1
(b) Express
3 2
3x
2x
6x
2
f x
2 2
x
x 2 x 2
as
partial fractions hence find f
xdx.
31. (a) Show that
(b) Find xsec
x xdx
, hence evaluate
2
4
2
x sec x x
dx
0
4 2 2
2cos x sin x
32. (a) Evaluate
dx
2
1
cos x
0
x
(b) Find dx
1
2x
2
4x
33. (a) Integrate (i)
6
1
x
2
4x
x 1
(iii) (iii)
6 2
1 x x x x 3
(iv) tan 3 x with respect to x
0
4
dx ln 3
3
5sin x
2
297