Higher Higher Unit 2 What is Integration The Process of Integration ( Type 1 ) Area between to curves ( Type.
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Higher
Higher Unit 2Higher Unit 2
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What is Integration
The Process of Integration ( Type 1 )
Area between to curves ( Type 4 )
Outcome 2
Area under a curve ( Type 2 )
Working backwards to find function ( Type 5 )
Area under a curve above and below x-axis ( Type 3)
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Higher Outcome 2
Integration
1
( 1)
nn xx dx c
n
43x dx 53
5
xc
4 13
(4 1)
xc
we get
You have 1 minute to come up with the rule.
43x dx 53
5
xc
Integration can be thought of as the opposite of differentiation
(just as subtraction is the opposite of addition).
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Differentiation
multiply by power
decrease power by 1
Integration
increase power by 1
divide by new power
nx
1
1
nn xx dx
nC
Where does this + C come from?
IntegrationOutcome 2
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Integrating is the opposite of differentiating, so:
( )f x ( )f xintegrate
2
2
2
( ) 3 1
g( ) 3 4
h( ) 3 10
f x x
x x
x x
( )
( )
( )
f x
g x
h x
But: differentiate
differentiate
integrate
Integrating 6x….......which function do we get back to?
6x
IntegrationOutcome 2
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Solution:
When you integrate a function
remember to add the
Constant of Integration……………+ C
IntegrationOutcome 2
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6x dx means “integrate 6x with respect to x”
( )f x dx means “integrate f(x) with respect to x”
Notation
This notation was “invented” by
Gottfried Wilhelm von Leibniz
IntegrationOutcome 2
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Examples:
7x dx8
8x
23 2 1x x dx 3 2
3 23 2
x x
x C
3 2 x x x C
IntegrationOutcome 2
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5 3
1 4
2x dx
x x
1 352 2( 4 )
2
xx x dx
3 14 2 2
3 12 2
48
x x xC
IntegrationOutcome 2
Just like differentiation, we must arrange the
function as a series of powers of x
before we integrate.
3
2
142
1 2 8
8 3
xC
xx
3
4
1 2 8
8 3
xC
x x
Name :
34x dx 2
2
3dx
x3 x dx 3 2
1
2dx
x 3
2
3
xdx
x
( 2)(3 1) x x dx 2 3
x
dxx
41
5 x dxx
Integration techniquesArea
under curve=
Area under curve=
IntegrationIntegration
44
4
xc
1
23x dx3
2
3
2
3xc
3
22x c
22
3
xdx
12
3( 1)
xc
2
3c
x
2
3
2
xdx
1
33
2
xc
323
xx dx
2 22
6 2
x xc
2
2
1
6
xc
x
1 1
2 45 x x dx
1 5
2 4
1 5
2 4
5x xc
1 5
2 42 4x x c
23 5 2 x x dx 3 23 5
23 2
x xx c
1 1
2 22 3x x dx
4x c
23 5
22
xx x c
3 1
2 2
3 1
2 2
2 3x xc
3122
46
3
xx c
Real Application of Integration
Find area between the function and the x-axis
between x = 0 and x = 5
A = ½ bh = ½x5x5 = 12.5
5
0
A x dx52
02
x
2 25 0 250 12.5
2 2 2
Real Application of Integration
Find area between the function and the x-axis
between x = 0 and x = 4
A = ½ bh = ½x4x4 = 8
4
0
4 A x dx 42
0
42
xx
24
4 4 0 8 16 242
A = lb = 4 x 4 = 16
AT = 8 + 16 = 24
Real Application of Integration
Find area between the function and the x-axis
between x = 0 and x = 2
22
0
A x dx23
03
x
3 32 0 8 80
3 3 3 3
Real Application of Integration
Find area between the function and the x-axis
between x = -3 and x = 3
3
3
A x dx
32
32
x
2 23 ( 3) 9 90
2 2 2 2
?
Houston we have a problem !
We need to do separate integrations for above and below the x-axis.
Real Application of Integration
3
0
A x dx32
02
x
23 9 9
0 02 2 2
0
3
A x dx
02
32
x
2( 3) 9
02 2
Areas under the x-axis ALWAYS give negative values
By convention we simply take the positive value since we cannot get a negative area.
9 99
2 2TA
Integrate the function g(x) = x(x - 4) between x = 0 to x = 5
Real Application of Integration
We need to sketch the function and find the roots before we
can integrate
We need to do separate integrations for above and below the x-axis.
Real Application of Integration
4
0
( 4) A x x dx 43
2
0
23
xx
3
24 64 322 4 0 32
3 3 3
5
4
( 4) A x x dx 3 3
2 25 4 25 32 2 5 2 4
3 3 3 3
Since under x-axis
take positive value
32
3
7
3
Real Application of Integration
4 5
0 4
( 4) ( 4) TA x x dx x x dx 32 7 13
3 3
Find upper and lower limits.
Area between Two Functions
12
0
( ) A x x dx 12 3
02 3
x x
2 31 1 1
0 2 3 6
1 6
2x x2 0x x
( 1) 0x x 0 1x x
then integrate
top curve – bottom curve.
y x2y x
Find upper and lower limits.
Area between Two Functions
12 2
1
( 1) ( -1) A x x dx
2 21 1x x 22 2 0x
2( 1)( 1) 0x x 1 1x x then integrate
top curve – bottom curve.
22( 1) 0x
12
1
( 2 2) x dx
1
2
1
2 ( 1) x dx
Take out common factor
Area between Two Functions
12
1
2 ( 1) x dx
13
1
23
xx
3 3(1) ( 1)2 1 ( 1)
3 3
1 12 1 1)
3 3
22 2
3
8
3
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Higher
2If f '( ) 3 4 1 and f (2) 11, find f ( ).x x x x To get the function f(x) from the derivative f’(x)
we do the opposite, i.e. we integrate. 2( ) (3 4 1)f x x x dx
3 2
3 43 2
x xCx
3 22x x Cx
3 2
(2) 11
2 2.2 2 11C
f
8 8 2 11 C
3 C
3 2( ) 2 3f x x x x Hence:
IntegrationOutcome 2
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IntegrationOutcome 2
5
12x dx
52
1
2
2
x
2 25 1 24
2 3
14x dx
24
2
4
4
x
4 4 2 ( 2) 6
Example :
52
1x
24
2x
Calculus Revision
Back NextQuit
Integrate 2 4 3x x dx 3 24
33 2
x xx c
3 212 3
3x x x c
Integrate term by term
simplify
Calculus Revision
Back NextQuit
Integrate 33 4x x dx4 23 4
4 2
x xc
4 23
42x x c
Integrate term by term
Calculus Revision
Back NextQuit
Evaluate4
1
x dx1
2
4
1
x dx 43
2
1
2
3x
4
1
32
3x
3 32 24 1
3 3
16 2
3 3
14
3
2
34
Straight line form
Calculus Revision
Back NextQuit
Evaluate
2
21
dx
x2
2
1
x dx 21
1x 1 12 1
11
2
1
2
Straight line form
Calculus Revision
Back NextQuit
Integrate3
2x dxx
1
32 2x x dx 3 222
3
2
2
xx c
3222
3x x c
Straight line form
Calculus Revision
Back NextQuit
Integrate3 1x dx
x
13 2x x dx
14 2
1
24
x xc
14 21
42x x c
Straight line form
Calculus Revision
Back NextQuit
Integrate
3 5x xdx
x
1 1
2 2
3 5x xdx
x x
5 1
2 25x x dx 7 3
2 22
7
10
3cx x
Straight line form
Calculus Revision
Back NextQuit
Integrate
324
2
x xdx
x
3
2
1 1
2 2
4
2 2
x xdx
x x
1
2 1
22x x dx
3
2 24 1
3 4x x c
Split into separate fractions
Calculus Revision
Back NextQuit
Integrate2 5x
dxx x
3
2
2 5xdx
x
3 3
2 2
2 5xdx
x x
1 3
2 25x x dx
3 1
2 2
3 1
2 2
5x xc
3 1
2 22
310x x c
Straight line form
Calculus Revision
Back NextQuit
Find p, given
1
42p
x dx 1
2
1
42p
x dx 3
2
1
2
342
p
x
3 3
2 22 2
3 3(1) 42p
2 233 3
42p 32 2 126p
3 64p 3 2 1264 2p 1
12 32 16p
Calculus Revision
Back NextQuit
Integrate (3 1)( 5)x x dx 23 14 5x x dx
3 23 145
3 2
x xx c
3 27 5x x x c
Multiply out brackets
Integrate term by term
simplify
Calculus Revision
Back NextQuit
Integrate2(5 3 )x dx
3(5 3 )
3 3
xc
31
9(5 3 )x c
Standard Integral
(from Chain Rule)
Calculus Revision
Back NextQuit
Integrate 2 2
2
2 2, 0
x xdx x
x
4
2
4xdx
x
4
2 2
4xdx
x x 2 24x x dx
3 14
3 1
x xc
31
3
4x c
x
Split into
separate fractions
Multiply out brackets
Calculus Revision
Back NextQuit
Evaluate22
2
1
1x dxx
1
222
1x x dx
Cannot use standard integral
So multiply out
4 22
12x x x dx
25 2 1
1
1
5x x x 5 2 5 21 1 1 1
5 2 5 12 2 1 1
32 1 1
5 2 54 64 40 20 2
10 10 10 10 82
10 1
58
Calculus Revision
Back NextQuit
The graph of
32
1 1
4
dyx
dx x
( )y g x passes through the point (1, 2).
express y in terms of x. If
3 2 1
4
dyx x
dx
4 1 1
4 1 4
x xy x c
simplify
4 1 1
4 4
xy x c
x Use the point
41 1 1
2 14 1 4
c
3c Evaluate c41 1
4 4
13y x x
x
Calculus Revision
Back NextQuit
A curve for which 26 2
dyx x
dx passes through the point (–1, 2).
Express y in terms of x.
3 26 2
3 2
x xy c 3 22y x x c
Use the point3 22 2( 1) ( 1) c 5c
3 22 5y x x
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Further examples of integration
Exam Standard
IntegrationOutcome 2
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The integral of a function can be used to determine the area between the x-axis and the graph of the
function.
x
y
ba
( )y f x
Area ( )abf x dx
( ) ( ) ( ) ( ) ( )a
baf x f x dx F Fdx F bx If then
NB: this is a definite integral.
It has lower limit a and an upper limit b.
Area under a CurveOutcome 2
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Examples:
5
1(3 )x dx
52
1
32
xx
25 115 3
2 2 24
2 2
0(3 1)x dx
23
0x x 3 32 2 0 0 6
Area under a CurveOutcome 2
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( )From the definition of it follows that:b
a
f x dx
( ) ( )b af x dx f x dx
a b
( ) ( ) ( ( ) ( ) )F b F a F a F b
Conventionally, the lower limit of a definite integral
is always less then its upper limit.
Area under a CurveOutcome 2
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a bc d
y=f(x)
( ) 0b
a
f x dx ( ) 0d
c
f x dx
Very Important Note:
When calculating integrals:
areas above the x-axis are positive areas below the x-axis are negative
When calculating the area between a curve and the x-axis:• make a sketch
• calculate areas above and below the x-axis separately
• ignore the negative signs and add
Area under a CurveOutcome 2
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a b
( )y f x
( )y g x
The Area Between Two Curves
To find the area between two curves we evaluate:
( )top curve bottom curveArea
( ( ) ( )b
a
Area f x g x dx
Area under a CurveOutcome 2
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Higher Example:
2 2
2, 4
4
Calculate the area enclosed by the lines
and the curves and
x x
y x y x
4 42 2 2
2 2
[ (4 )] (2 4)Area x x dx x dx 44 3
2
2 2
2(2 4) 4
3
xx dx x
128 16( 16) ( 8)
3 3112
83
88
3
2y x
24y x
2x
4x
y
x
Area under a CurveOutcome 2
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Higher Complicated Example:
The cargo space of a small bulk carrier is 60m long. The shaded part of the diagram represents the uniform cross-section of this space.
2
, 6 64
1 9.
xy x
y y
It is shaped like a parabola with ,
between lines and
Find the area of this cross-section and hence find the volume of cargo that this ship can carry.
Area under a CurveOutcome 2
9
1
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1y
9y
2
4
xy
x
y
ss
The shape is symmetrical about the y-axis. So we calculate the area of one of the light shaded rectangles and one of the dark shaded wings. The area is then
double their sum.t t
The rectangle: let its width be s2
14
2ss
y
The wing: extends from x = s to x = t
2
694
ty t
The area of a wing (W ) is given by:
2
(9 )4
t
w
s
xA dx
Area under a Curve
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6 2
2
(9 )4w
xA dx
63
2
(9 )12
xx
218
3
The area of a rectangle is given by:
(9 1) 16R s
The area of the complete shaded area is given by:
2( )A R W
2 2082(16 18 )
3 3A
The cargo volume is:3208
60 60. 20.208 413
60A m
Area under a CurveOutcome 2
3 36 2(9 6 ) (9 2 )
12 12
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Exam Type QuestionsOutcome 2
At this stage in the course we can only do
Polynomial integration questions.
In Unit 3 we will tackle trigonometry integration
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Higher Outcome 2
Are you on Target !
• Update you log book
• Make sure you complete and correct
ALL of the Integration questions in
the past paper booklet.
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