higher maths 2 2 integration1. speed time graphs d st × ÷÷ 20 40 0 024 time (hours) speed (mph)...
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Higher Maths 2 2 Integration
1
Speed Time Graphs
D
S T×÷÷
20
40
00 2 4
Time (hours)
Sp
eed
(mp
h)
Calculate the distance travelled in each journey.
20
40
00 2 4 6Time (hours)
Sp
eed
(mp
h)
20
40
00 2 4 6Time (hours)
Sp
eed
(mp
h)
D=4 × 30= 120
miles
average speed
D=5 × 20= 100
miles
30 miles 90 miles + 15
miles 135 miles
In Speed Time graphs, the distance travelled is the same as the area under the graph.
Higher Maths 2 2 Integration
2
Reverse Differentiation
D
Tspeed
=
‘rate of change ofdistance with respect to time’
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.Differentiating backwards is used to find the area under a function.
f (x)
Higher Maths 2 2 Integration
3
Estimating Area Under Curves
To estimate area under a function, split the area into vertical strips.
f (x)
x
f (x)
x
The area of each strip is the height, multiplied by :x
Total Area
f (x) x×
f (x) x×( )=
As the strips get narrower, the estimate becomes more accurate.
Area under the function=
f (x)
f (x) x as 0
means ‘the sum of...’
Higher Maths 2 2 Integration
4
x
Higher Maths 2 2 Integration
5
IntegrationThe algebraic method for finding area under a function is called Integration.
‘Integrate’means ‘join together all the pieces’
Integration uses reverse differentiation to ‘undo’ finding the rate of change.
∫Area under the function
=
f (x)
f (x) x
d x
xas 0
f (x)=
x
∫ d x
Expression or function to be integrated.
Higher Maths 2 2 Integration
6
Differentiating BackwardsIntegration involves differentiating in reverse.
multiply by the power
reduce the power by one
f (x) f ′(x)
divide by the power
increase the power by one
f (x) f ′(x)
• divide every x-term by the new power
How to differentiate backwards:• increase the power of every x-term by one
Higher Maths 2 2 Integration
7
Finding the Anti-DerivativeThe result of a
differentiation is called a
derivative.
• divide by each new power
How to Reverse Differentiate• increase each power by one
The result of differentiating
backwards is called the anti-derivative.
Example
dydx
= 8 x 3 + x
2 – 6
x
Find the anti-derivative
for
y
y = ∫ dx8 x 3 + x
2 – 6 x
=2 x
4 + x 3
– 3
x2
13
∫ d x expression or function
Higher Maths 2 2 Integration
8
The Constant of IntegrationWhen differentiating, part of an expression is
often lost.Example
f (x)= x 4 + 2
f (x) = x 4
f (x) = x 4 – 7
f ′(x) = 4 x 3
All three functions have derivative
The anti-derivative of
f ′(x) is
f (x)= x 4 + c
unknown
constant
When differentiating in
reverse,
it is essential to remember
to add back on the unknown
number.This is called the constant of
integration.
Higher Maths 2 2 Integration
9
Basic IntegrationThe result of
integration is called
an integral.
f (x)
Example 1
∫ d xf (x)= x 6
• divide by each new power
How to Integrate
• increase each power by one
=
17= x
7 + c
∫ d xx 6
• add the constant of integration
constant of integration
Example 2
∫ d x4x
Fin
d
= d x4 x12∫
= 8 x +
c
= 8 x12 + c
Higher Maths 2 2 Integration
10
Integration and AreaIntegration can be used to find the area ‘under’ a
function between two different values of x.
f (x)
x1 x2
∫ d xf (x)
x1
x2
= ∫ d xf ( x2) – ∫ d xf ( x1)
Area under f (x) between x1
and x2
=
‘Upper Limit’
‘Lower Limit’
This is called a Definite Integral
The area ‘under’ a function can be described more mathematically as the area between the function and the
x-axis.
Area Between a Function and the x-axis
Higher Maths 2 2 Integration
11
f (x)
∫ d xf (x)
5
-2
5-2
g (x)
∫ d xg(x)
3
-6
3-6
h (x)
8-7
∫ d xh(x)
8
-7
Higher Maths 2 2 Integration
12
Evaluating Definite Integrals
=∫ dx2 x 3
0
3
[ x 4 + c1
2 ]0
3
= ( × ( 3 ) 4 + c1
2 ) ( × ( 0 ) 4 + c1
2 )–
= 4012
The constants of integration cancel each other out.
Definite Integrals do not require the constant of integration.
∫ d xx1
x2
Definite Integral
Write integral inside square
brackets
units 2
Example
[
]
Higher Maths 2 2 Integration
13
Evaluating Definite Integrals (continued)
Example 2
4
1∫ d x
4
19 x
2 – 2 x 3 x 3 – x
2=
( )3 × 43 –
42
= – ( )3 × 13 –
12
( )192 – 16
= – ( )3 – 1
= 174
units
2
Find the area below the curve
between x = 1 and x = 4.
y = 9 x 2 – 2 x
Write integral inside square
brackets...
(no constant required)
...then evaluate for
each limit and subtract.
Remember units!
When calculating areas by integration, areas above
the x-axis are positive and areas below the x-axis are negative.
Areas Above and Below the x-axis
Higher Maths 2 2 Integration
14
bac d
∫ dxf (x)
a
b
> 0
∫ dxf (x)
c
d
< 0
f (x)
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-
Area Between FunctionsIntegration can also be used to find the area between
two graphs, by subtracting integrals.
Higher Maths 2 2 Integration
15
f (x)
g (x)
Area enclosed by f (x) and g
(x)
∫ d xf (x)
=
= – ∫ d xg (x)
∫ f (x) d xg (x)( – )
ba
b
a
b
a
b
a
between intersection points a
and b
is above
f (x) g (x)
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