1. Higher Maths 2 2 Integration1
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2. UNIT OUTCOME SLIDE NOTE
Higher Maths Integration 2
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average speed
Speed Time Graphs D ÷ ÷
Calculate the distance travelled in each journey. S × T 40 40 40
Speed (mph) 20
Speed (mph) 20
Speed (mph) 20 2 4 2 4 6 2 4 6
Time (hours)
Time (hours)
Time (hours)
D = 4 × 30 = 120 miles
D = 5 × 20 = 100 miles
30 miles
90 miles
+ 15 miles !
In Speed Time graphs, the distance travelled is the same as the area under the graph.
135 miles

3. UNIT OUTCOME SLIDE NOTE ! REMEMBER y x f (x)
Higher Maths Integration 3
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Reverse Differentiation !
REMEMBER D
‘rate of change of distance with respect to time’
speed = T
If we know how the speed changes, and want to find distance, we need to ‘undo’ finding the rate of change with respect to time.
In other words we need to reverse differentiate. y
f (x)
Differentiating backwards is used to find the area under a function. x

4. S S UNIT OUTCOME SLIDE NOTE y ( ) x y x S f (x) x f (x) x f (x) x
Higher Maths Integration 4
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NOTE S
means ‘the sum of...’
Estimating Area Under Curves
To estimate area under a function, split the area into vertical strips.
The area of each strip is the height, multiplied by : y
f (x) x
f (x) × x S (
Total Area
f (x) × x ) = x x
As the strips get narrower, the estimate becomes more accurate. y
f (x)
Area under the function
f (x) S
f (x) x as x x = x

5. ∫ ∫ S UNIT OUTCOME SLIDE NOTE y x f (x) f (x) d x d x f (x) x x x = =
Higher Maths Integration 5
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Integration
‘Integrate’ means ‘join together all the pieces’
The algebraic method for finding area under a function is called Integration.
Integration uses reverse differentiation to ‘undo’ finding the rate of change. y x
Area under the function
f (x) x ∫ =
f (x)
d x ∫
d x S
f (x) x as x
Expression or function to be integrated. =

6. UNIT OUTCOME SLIDE NOTE
Higher Maths Integration 6
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Differentiating Backwards
Integration involves differentiating in reverse.
f (x)
multiply by the power
reduce the power by one
f ′(x)
f (x)
divide by the power
increase the power by one
f ′(x)
How to differentiate backwards:
• increase the power of every x-term by one
• divide every x-term by the new power

7. ∫ ∫ UNIT OUTCOME SLIDE NOTE ! y y dx d x dy 8 x 3 + x 2 – 6 x
Higher Maths Integration 7
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Finding the Anti-Derivative !
How to Reverse Differentiate
The result of a differentiation is called a derivative.
• increase each power by one
• divide by each new power
The result of differentiating backwards is called the anti-derivative. ∫
d x
Example
expression or function
Find the anti-derivative for y ∫ dy
8 x 3 + x 2 – 6 x y
8 x x 2 – 6 x dx = = dx
2 x x 3 – 3 x2 1 = 3

8. UNIT OUTCOME SLIDE NOTE
Higher Maths Integration 8
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The Constant of Integration
When differentiating, part of an expression is often lost.
Example
f (x)
= x 4
All three functions have derivative
f (x)
= x 4 + 2
f ′(x)
= 4 x 3
f (x)
= x 4 – 7
f ′(x)
The anti-derivative of is
When differentiating in reverse, it is essential to remember to add back on the unknown number.
f (x)
= x 4 + c
unknown constant
This is called the constant of integration.

9. constant of integration
Higher Maths Integration 9
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Basic Integration !
How to Integrate
LEARN THIS
The result of integration is called an integral.
• increase each power by one
• divide by each new power
Example 1
• add the constant of integration ∫
f (x)
= x 6
f (x)
d x
Example 2 ∫ ∫ ∫ 4 1
x 6 =
d x
Find
d x
4 x = 2
d x x 1
= x c 1
8 x
+ c = 2 7
constant of integration
8 x + c =

10. ∫ ∫ ∫ UNIT OUTCOME SLIDE NOTE y x – f (x) x2 f (x) d x x1 x2 x1
Higher Maths Integration 10
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Integration and Area
Integration can be used to find the area ‘under’ a function between two different values of x.
Area under f (x) between x1 and x2 y
f (x) x2 ∫ =
f (x)
d x x
This is called a Definite Integral x1 x2 x1
‘Lower Limit’
‘Upper Limit’ ∫ ∫
f ( x2)
d x – =
f ( x1)
d x

11. ∫ ∫ ∫ UNIT OUTCOME SLIDE NOTE y y y x x x g (x) h (x) f (x) f (x) d x
Higher Maths Integration 11
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Area Between a Function and the x-axis
The area ‘under’ a function can be described more mathematically as the area between the function and the x-axis.
g (x)
h (x) y y y
f (x) -7 x x -6 3 8 x -2 5 5 3 8 ∫ ∫ ∫
f (x)
d x
g(x)
d x
h(x)
d x -2 -6 -7

12. Write integral inside square brackets
Higher Maths Integration 12
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Definite Integral
Evaluating Definite Integrals
Write integral inside square brackets x2
Example ∫ 3
d x ∫ [ ] 3
2 x 3 dx 1
x c x1 = 2 ( ) ( ) 1
× ( 3 ) c 1
× ( 0 ) c = – 2 2 1 = 40
units 2 2 !
Definite Integrals do not require the constant of integration.
NOTICE
The constants of integration cancel each other out.

13. [ ] ∫ ( ) ( ) ( ) ( ) UNIT OUTCOME SLIDE NOTE – – y = 9 x 2 – 2 x d x
Higher Maths Integration 13
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Evaluating Definite Integrals (continued)
y = 9 x 2 – 2 x
Example 2
Find the area below the curve between x = 1 and x = 4. 4
Write integral inside square brackets (no constant required)
[ ] 4 ∫
9 x 2 – 2 x
d x
3 x 3 – x 2 = 1 1
( )
( ) –
3 × 43 – 42
3 × 13 – 12 =
...then evaluate for each limit and subtract.
( )
( ) – =
192 – 16
3 – 1 =
174
units 2
Remember units!

14. ∫ ∫ UNIT OUTCOME SLIDE NOTE y x dx f (x) a x-axis : x- f (x) c d a dx
Higher Maths Integration 14
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Areas Above and Below the x-axis
When calculating areas by integration, areas above the x-axis are positive and areas below the x-axis are negative. ∫ dx
f (x) a b
> 0
How to calculate area between a curve and the
• draw a sketch
• calculate the areas above and below the axis separately
• add the positive value of each area (ignore negative signs)
x-axis : x-
f (x) y c d x a b ∫ dx
f (x) c d
< 0

15. ∫ ∫ ∫ ( ) UNIT OUTCOME SLIDE NOTE y x ! CAREFUL! – g (x) f (x) f (x)
Higher Maths Integration 15
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Area Between Functions
Integration can also be used to find the area between two graphs, by subtracting integrals.
Area enclosed by f (x) and g (x)
g (x)
between intersection points a and b y
f (x) b b ∫ ∫ – =
f (x)
d x
g (x)
d x x a a a b b !
CAREFUL! ∫ ( )
f (x)
g (x)
f (x)
is above
g (x) = –
d x a