Integration
description
Transcript of Integration
Integration
Learning Objectives:
In this chapter, you will learn about
• the concept of indefinite integral
Learning Outcomes:
•Determine integrals by reversing differentiation
2y x
2dy
dx
2 3y x
2dy
dx
2 4y x
2dy
dx
y mx c
2 dx 2x c2 dx
2x c
2 dx
2x c
2y x c
2y x c 2y x c
( ), ( )dy
if f x then y f x dxdx
3.1.2 Integration of algebraic expressions
m dx mx c
8dx 8x c
3.5 dx 3.5 x c
Integrate (a) 8 (b) 3.5 (c) 3
4
3
4dx
3
4x c
3.1.2 (a) Integral of Constant
3.1.2 Integration of algebraic expressions
3.1.2 (b) Integral of nax
Differentiation Integration
2dy
xdx
22dy
xdx
32dy
xdx
ndynx
dx
2x dx 22
2
xc 2x c
22x dx 32
3
xc
32x dx 42
4
xc 41
2x c
42x dx 52
5
xc
During differentiation, we carry out two operations on each term in x: multiply the term with the index, and reduce the index by 1.
ndyax
dx
1nnax
42dy
xdx
nax dx 1
1
naxc
n
Integrate each of the following with respect of x:
(a) 8x8x dx
8 1
8 1
xc
9
9
xc
(b) 4
6
x
4
6dx
x
4 16
4 1
xc
46x dx
36
3
xc
32x c
3
2c
x
If the derivative of a function is given as find the function y. 3
1,
9
dy
dx x
3
1,
9
dy
dx x
31,
9
dyx
dx
31
9x
3 11
9 3 1
xc
2
18
xc
2
1
18c
x
2y x
2dy
dx
2 3y x
2dy
dx
2 4y x
2dy
dx
y mx c
2 dx2x c
2 dx
2x c
2 dx
2x c
2y cx 2y cx 2y cx
3.1.3 Determine the constant of Integration
If and y=5 when x=3, find the value of y when x=5
23 6 4dy
x xdx
23 6 4y x x dx 3 23 6
43 2
x xy x c
3 23 3 4 (1)y x x x c
Subsitute x=3 and y=5 into (1)
3 25 3(3) 3(3) 4(3) c
5 27 27 12 c 5 12 c
7 c
3 23 3 4 7y x x x
,therefore
5when x 3 23(5) 3(5) 4(5) 7y
63y
3.1.4 Equations of curve from functions of gradients
Find the equation of curve.
The gradient of a curve passing through the point (-1, 2) is given by 4 5dy
xdx
sin 4 5,dy
ce xdx
by integration,
4 5y x dx 245
2
xy x c
22 5y x x c
The curve passing through the point (-1, 2)x=-1 when y=2
22 2( 1) 5( 1) c
2 2 5 c 5 c
22 5 5y x x
The equation of the curve is 22 5 5y x x
The gradient function of a curve passing through the point (-1, 2) and (0,k) is 23 10x x . Find the value of k.
y
x
2
2
4
4
6
6
8
8
10
10
– 2
– 2
– 4
– 4
– 6
– 6
– 8
– 8
– 10
– 10
2
2
4
4
6
6
8
8
10
10
– 2
– 2
– 4
– 4
– 6
– 6
– 8
– 8
– 10
– 10
(-1, 2)
(0, k)
23 10 ,dy
x xdx
23 10y x x dx 3 23 10
3 2
x xy c
3 25y x x c
The curve pass through (-1, 2)
,Therefore3 22 ( 1) 5( 1) c
8 c
Therefore, the equation of the curve is
3 25 8y x x
At point (0, k),
3 20 5(0) 8k
8k
Exercise 3-1-09 t0 6-1-09
1. Given that and that y=5 when x= -1, find the value of y
when x=2
2 7dy
xdx
2. Given that and that v=2 when t = 1, find the value of v
when t = 2.
3 2dv
tdt
3.1.5 Integrate by substitution
Find the integration by substitution
4(2 5)x dx2 5let u x
, 2du
Hencedx
1
2dx du
4 1.2
u du
4 1.2
u du 51
2 5
uc
51
10u c
51(2 5)
10x c
3.1.5 Integrate by substitution
Find the integration by substitution
2
1
(3 2)dx
x 3 2let u x
, 3du
Hencedx
1
3dx du
2
3 2x dx
2
1
(3 2)dx
x
2 1.3
u du
11
3 1
uc
1
3c
u
1
3(3 2)c
x
3.1.5 (a) Integral of ( )nax b
( )nax b dx 1( )
( 1)( )
nax b
n a
4(2 5)x dx4 1(2 5)
(4 1)(2)
x
5(2 5)
10
xc
51(2 5)
10x c
The slope of a curve at any point P(x, y) is given by . Find the equation of the curve given that its passes through the point ( 4, -3)
5(3 )x
The gradient of the curve, 5(3 )
dyx
dx
Integrate with respect to x, we have
5
3y x dx 63
6( 1)
xy c
Since the curve passes through (4, -3)
61(3 )6
y x c
613 (3 4)
6c
17
6c
,Therefore
The equation of the curve is
61 17(3 )6 6
y x or
66 (3 ) 17y x
Area under a curve
3.2.2(b) Area under a curve bounded by x= a and x=b
A
The area A under a curve by y = f(x) bounded by the x-axis from x=a to x=b is given by
b
aA y dx
Integration as Summation of Area
3.2.2(b) Area under a curve bounded by x= a and x=b
A
2 2y x
2
1A y dx
1 2
2 2
12A x dx
23
1
23
xA x
23 2
1
2 12(2) 2(1)
3 3A
8 14 2
3 3A
213
3A unit
Step (1) Find x-intercept
a b
On the x-axis, y =0
2 2x x y
2 2x x 02 2 0x x
( 2) 0x x 0x 2and x
0 2
2 2
02A x x dx 23 2
0
2
3 2
x xA
23
2
03
xA x
23 2
2 2
0
2 02 0
3 3A
24
3A unit
24
3A unit
Area under a curve bounded by 2 4 4y x x
2( 1)y x Area under a curve bounded by
The area under a curve which is enclosed by y = a and y = b is
b
aA x dy
The area under a curve is 2 2
12A y ydy
1
2
2 2x y y
23 2
1
2
3 2
y yA
3 32 23 13 1
3 3A
19 9 1
3A
20
3A
22
3A unit
A
Area under a curve bounded by curve
Area under a curve bounded
Exercise 19-2-09 23-2-09
1
2
Exercise Text Book Page 71 23-2-09
10- ( a) ( b ) ( c )
11- ( a) ( b ) ( c )
12- ( a) ( b )
Exercise Text Book Page 72
13- ( a) ( b ) ( c )
14
17 ( a ) (b ) ( c )
Volume of Revolutions
The resulting solid is a cone
To find this volume, we could take slices (the yellow disk shown above), each dx wide and radius y:
The volume of a cylinder is given by
V = πr2h
Because radius = r = y and each disk is dx high, we notice that the volume of each slice is:
V = πy2dx
Adding the volumes of the disks (with infinitely small dx), we obtain the formula:
y = f(x) is the equation of the curve whose area is being rotated
a and b are the limits of the area being rotated
dx show that the area is being rotated abount the x-axis.
Example 2
Find the volume if the area bounded by the curve y = x3 + 1, the x-axis and the limits of x = 0 and x = 3 is rotated around the x-axis..
When the shaded area is rotated 360° about the x-axis, we again observe that a volume is generated:
2y x
2 2
1V y dx
2 2
1(2 )V x dx
2 2
14V x dx
23
1
4
3
xV
3 34(2) 4(1)
3 3V
32 4
3 3V
28
3V
3193
V unit