HIGHER MATHEMATICS
Unit 2 – Topic 2
Integration
Differentiation Remember that DIFFERENTIATION is a measure of a curve’s gradient.
For a function f(x) = axn
f’(x) = na xn-1
Remember that the derivative of any constant will always be zero.
f(x) = 2x5 + 7x3 + 9x2 + 5
f’(x) = 10x4 + 21x2 + 18x
y = 2x5 + 7x3 + 9x2 + 5
= 10x4 + 21x2 + 18xdy
dx
Newton’s Notation
Leibniz’s Notation
Differential Equations
If the gradient of a tangent to a
parabola is given by = 12x.dydx
y = 6x2
y = 6x2 + 1
y = 6x2 – 8
…………
There are infinite possibilities since the derivative of any constant is ZERO!!
In General the equation of the parabola is y = 6x2 + c, where c is a constant.
To find the particular solution we need to be given a point
What are the possible equations of the parabola?
Notation:
= 12x – Differential Equations.
y = 6x2 + c - General solution.
y = 6x2 + 4 - Particular solution.
dy
dx
Example 1:
y = 2x2 + c
b) Now find the particular solution if itpasses through the point (3 , 15).
y = 2x2 + c
15 = 2(3)2 + c
15 = 18 + c
c = -3
So particular solution is y = 2x2 - 3
a) Find the general solution for thedifferential equation: dy
dx= 4x
Exercises from MIA book:
Page 126 Ex 1 All Qu’s
Exercises from Heinemann book:
Page ? Ex ? Qu ??
Integration The general solution of a differential equation is called the Anti-derivative, because you get it from “undoing” differentiation!!
The process of calculating an Anti-derivative is called INTEGRATION.
Leibniz invented a useful notation for Integration as follows:
f(x) dx = F(x) + c where f(x) = F’(x)
e.g. 6x + 5 dx = 3x2 + 5x + c
The Anti-derivative, F(x), is called the INTEGRAL.
c is called the CONSTANT of INTEGRATION.
dx is read as “with respect to x”.
Like differentiation there is a rule for Integration:
Differentiation: For y = xn
To Integrate you add 1 to the power & divide by this NEW power!!
Integration:
xn dx = + cxn+1
n + 1
= n xn-1dy
dx
For y = xn
Remember, to DIFFERENTIATE you MULTIPLY by the power and take 1 from the power to get the NEW power!!
Also INTEGRATION is the reverse of Differentiation. So:
Remember, like Differentiation you cannot Integrate roots or fractions, these must be expressed as
indices prior to Integration!!!!!!!!!!!!!!!!!!!!!!!!
Rules: kf(x) dx = k f(x) dx
f(x)+g(x) dx = f(x) dx + g(x) dx Example 2:
2x dx =2x2
2+ c
= x2 + c
3x2 dx =3x3
3+ c
= x3 + c
4x2 + 5x dx = 4x3
3+ + c5x2
2
8x-3 + 7x4 dx = 8x-2
-2+ + c7x5
5
= -4x-2 + + c7x5
5
8x¾ + 6x-3 + 2 dx = 32x7/4
7- + 2x + c3x-2
Exercises from MIA book:
Page 127 Ex 2A/B & 3Odd Qu’s
Exercises from Heinemann book:
Page ? Ex ? Qu ??
Area under a curve
Remember that differentiation gives the GRADIENT of a curve at a point.
Well Integration is used to find the AREA UNDER A CURVE bounded by x = a, x = b and the x-axis.
x
y
Oa b
y = f(x)
The blue Area opposite would be expressed as:
dxxfAreab
a )(
This is called a DEFINITE INTEGRAL
To Solve a Definite Integral we do:
dxxfAreab
a )( b
a )(xF )()( aFbF
i.e. Integrate, sub in each boundary and subtract!
Example 3:
x
y
O3 6
y = 2x - 6 dxxArea 6
3 62
6
3
26xx
)3(63)6(6622
)9(0 2
units 9
Notice we ignored the c in the integral as it will simply disappear when we do the subtraction.
Find the shaded area below.
Example 4: Find the shaded area below.
x
y
O-3 -1
y = -x2 – 4x
dxxxArea
1
3
24
1
3
23
23
xx
23
23
)3(23
)3()1(2
3
)1(
)9(3
5
2units
322
Exercises from MIA book:
Page 133 Ex 5 Even Qu’s
Exercises from Heinemann book:
Page ? Ex ? Qu ??
Example 5: Find the shaded area below.
x
y
O
y = x2 – 6xPrior to Integrating to find the Area we must firstly find the boundaries.
In this case the points where the curve cuts the x-axis.
Cuts x-axis when y = 0:
x2 – 6x = 0x(x – 6) = 0
x = 0 or x – 6 = 0x = 0, 6
6
Note there is no need to find the y coordinate.
Now we need to find the area under the curve.
dxxxArea 6
0
26
6
0
23
33
xx
23
23
)0(33
)0()6(3
3
)6(
036
36
The answers is a negative as the area being found is BELOW the x-axis. Since an area cannot be negative we ignore the minus sign!
units 362
x
y
O3 6
y = 2x - 6 dxxArea 6
0 62
6
0
26xx
)0(60)6(6622
002
units 0
We can clearly see that the Area is not zero!!
We get this answer because part of the area is below the x-axis which give’s a –’ve answer.
We must calculate this type of Qu. in parts.
Example 6: Find the shaded area below.
dxxArea 3
0 621
3
0
26xx
)0(60)3(6322
092
units 9
dxxArea 6
3 622
6
3
26xx
)3(63)6(6622
90 2
units 9
Area = 9 + 9
= 18 units2
x
y
O3 6
y = 2x - 6
Exercises from MIA book:
Page 135 Ex 6 All Qu.
Exercises from Heinemann book:
Page ? Ex ? Qu ??
y
O xa b
y = f(x)
The blue Area opposite is:
dxxfAreab
a )(
O
y = g(x)
a b
y = f(x) The green area opposite is:
dxxgAreab
a )(
O
y = f(x)
a b
y = g(x)
Look at this curve and write down its area between a and b:
Now look at this curve and write its area between a & b
Area between 2 curves
How could we find the area between the two curves:
The area between the two curves is:
dxxfAreab
a )( dxxgb
a )(dxxgxfb
a )()(
What can you say about the boundary points a & b?
The boundary points are the Points of Intersection of the 2 curves. To find them we make the 2 equations equal to each other and solve!!!
Example 7: Find the shaded area below.
x
y
O
y = -x2 – 3x
y = 2
Points of Intersection:
232
xx
0232
xx
0)23(2
xx
0)1)(2( xx
1 ,2 xx
Area:
42
3
3
7
2units
6
1
dxxxA
1
2
223
1
2
2
23
3
23
xxx
)2(2)2(
3
)2()1(2)1(
3
)1( 2
23
32
23
3
46
3
82
2
3
3
1
6
24
6
9
6
14
Points of Intersection:
2244 xx
0822
x0)4(2
2x
0)2)(2( xx
2 ,2 xx
O
y = 4 - x2
y = x2 - 4
x
yExample 8: Find the shaded area below.
Area:
323
32
3
64
dxxA 2
2
282
2
2
3
83
2
xx
)2(8
3
)2(2)2(8
3
)2(233
16
3
1616
3
16
3
96
3
32 2
units 3
64
Area:
Find the area between these curves:
y=x2+2x-3 &
y=2x2-5x-3
Points of Intersection:
3235222
xxxx
072
xx0)7( xx7 ,0 xx
6
1029
6
686
6
343
dxxxA 7
0
27
7
0
23
27
3xx
02
)7(7
3
)7(23
2
343
3
343
Example 9:
2units
6
343
Exercises from MIA book:
Page 137 Ex 7 & 8 All Qu.
Exercises from Heinemann book:
Page ? Ex ? Qu ??