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 ??