technical-mathematics-integration-17-feb_2018.pptx

INTEGRAL CALCULUS (Integration)
Technical Mathematics
Grade 12-Just In Time Training
17 February 2018 ,
Ishaak Cassim
DCES: Technical Mathematics
Ishaak Cassim - February 2018 1

Outcomes for this section
• By the end of this section, you should be able to:
• Understand the concept of integration as a summation function (definite
integral) and as converse of differentiation (indefinite integral).
• Apply standard forms of integrals as a converse of differentiation.
• Integrate the following functions:
o kxn, with n ∈ ℝ; with 𝑛 ≠ – 1
o
𝒌
𝒙
and kanx with a > 0 ; k , a ∈ ℝ
• Integrate polynomials consisting of terms of the above forms.
• Apply integration to determine the magnitude of an area included by a
curve and the x-axis or by a curve, the x-axis and the ordinates x = a
and x = b where a , b ∈ ℤ.

Introduction
The process of integration reverses the process of differentiation. In differentiation, if f(x) =
2x2
, then 𝑓′
(x) = 4x. Thus the integral of 4x is 2x2
. We can represent this process pictorially
as follows:
The situation gets a bit more complicated, because there are an infinite number of functions
we can differentiate to give 4x. Here are some of those functions:
f(x) = 2x2
+ 7 ; g(x) = 2x2
– 8 ; h(x) = 2x2
+
1
2
.

Activity 1
• Activity 1
• Write down at least five other functions whose
derivative is 4x.

Introduction
• You would have noticed that all the functions have the same
derivative of 4x, because when we differentiate the constant
term we obtain zero.
• Hence, when we reverse the process, we have no idea what
the original constant term might have been.
• So we include in our response an unknown constant, c, called
the arbitrary constant of integration.
• The integral of 4x then is 2x2 + c.

Introduction
• In differentiation, the differential coefficient
𝑑𝑦
𝑑𝑥
indicates that
a function of x is being differentiated with respect to x, the dx
indicating that it is “with respect to x”.
• In integration, the variable of integration is shown by adding
d(the variable) after the function to be integrated. When we
want to integrate a function, we use a special notation:
𝒇 𝒙 𝒅𝒙.
• Thus, to integrate 4x, we will write it as follows:

Introduction
4𝑥 𝒅𝒙 = 2x2
+ c , c ∈ ℝ.
Integral
sign
This term is called
the integrand
There must always be
a term of the form dx
Constant of
integration

• Note that along with the integral sign ( 𝑑𝑥), there is a term of the form dx,
which must always be written, and which indicates the variable involved, in our
example x.
• We say that 4x is integrated with respect x, i.e: 𝟒𝒙 𝒅𝒙
• The function being integrated is called the integrand.
• Technically, integrals of this type are called indefinite integrals, to distinguish
them from definite integrals, which we will deal with later.
• When you are required to evaluate an indefinite integral, your answer must
always include a constant of integration.; i.e:
𝟒𝒙 𝒅𝒙 = 2𝑥2
+ c; where c ∈ ℝ

Definition of anti-derivative
• Formally, we define the anti-derivative as: If f(x) is a
continuous function and F(x) is the function whose derivative
is f(x), i.e.: 𝑭′
(x) = f(x) , then:
𝒇 𝒙 𝒅𝒙 = F(x) + c; where c is any arbitrary
constant.

Activity 2
1. In each of the following determine the function f(x), if the derivative 𝑭′
(x) is given:
No. 𝑭′
(x) f(x)
1. 4x
2. x4
3. 2x2
4. 0
5.
𝒙
𝟑
𝟐
6. -2x3
7. 𝟏
𝟓
𝒙𝟏𝟎
2. Explain how the derivative 𝑭′
(x) and the function f(x)are related to each other

Activity 2
Differentiate
Integrate
axn

The general solution of integrals of the form kxn
• From Activity 2 above, the general solution of integrals of the
form 𝒌𝒙𝒏
dx , where k and n are constants is given by:
𝒌𝒙𝒏
dx =
𝒌𝒙𝒏+𝟏
𝒏+𝟏
+ c ; where 𝒏 ≠ −𝟏 and c ∈ ℝ

Table of ready to use integrals
Function, f (x) Indefinite integral 𝒇(𝒙)𝒅𝒙
f (x)= k , where k is a constant 𝒌𝒅𝒙 = kx + c ; where c ∈ ℝ
f (x)= x 𝒙𝒅𝒙 =
𝟏
𝟐
𝒙𝟐
f (x) = x2
𝒙𝟐
𝒅𝒙 =
𝟏
𝟑
𝒙𝟑
f(x) = axn
𝒂𝒙𝒏𝒅𝒙 =
𝒂𝒙𝒏+𝟏
𝒏+𝟏
f (x) = 𝒙−𝟏
=
𝟏
𝒙
𝟏
𝒙
dx = ln 𝒙 + c; where c ∈ ℝ
f (x) = kanx
𝒌𝒂𝒏𝒙
dx =
𝒌𝒂𝒏𝒙
𝒏.𝒍𝒏𝒂
+ c ; where c ∈ ℝ ; a > 0 and a ≠ 1

Worked Examples: Indefinite integrals
a)
𝑥7
dx
=
𝒙𝟕+𝟏
𝟕+𝟏
+ c ; where c∈ ℝ
=
𝒙𝟖
𝟖
Compare 𝑥7
dx with
𝒂𝒙𝒏
𝒅𝒙 =
𝒂𝒙𝒏+𝟏
𝒏+𝟏
+ c,
Then: a = 1; n = 7 ; n + 1 = 8

c) 2
𝑢2du
= 2𝑢−2du
= 2 𝑢−2du
= 2 (
𝑢−2+1
−2+1
) + c ; where c∈ ℝ
=
−𝟐
𝒖
Note: in this example, we
are integrating with
respect to u.
Explain the method used

d) 𝑥 dx
= 𝒙
𝟏
𝟐 dx
=
=
𝟐
𝟑
+ c; where ………..
Complete

Activity 3
1. Use the method outlined above to find a general expression for the function f(x) in each of the following cases.
𝑭′
(x) f(x)
a) 4x3
b) 6x5
c) 2x
d) 3x2 + 5x4
e) 10x9 – 8x7 – 1
f) –7x6 + 3x2 + 1
g) 1 – 3x-2
h) (x – 2)2 –
3
𝑥2
i) −2
5𝑥−1 +
3
5
𝑥

Activty 3 - continued
2. Determine the indefinite integrals in the following cases
a) (9𝑥2 – 4𝑥 − 5)𝑑𝑥 b) (12𝑥2 + 6𝑥 + 4)𝑑𝑥
c) −5 𝑑𝑥 d) (16𝑥3 − 6𝑥2 + 10𝑥 − 3)𝑑𝑥
e) (2𝑥3
+ 5𝑥)𝑑𝑥 f) (𝑥 + 2𝑥2
)𝑑𝑥
g) (2𝑥2– 3x – 4)dx h) (1 − 2𝑥 − 3𝑥2)𝑑𝑥
3. Determine the following indefinite integrals
a) 1
𝑥3dx b) (𝑥2 –
1
𝑥2)dx
c) 𝑥 𝑑𝑥 d)
6𝑥
2
3 𝑑𝑥
e) 6𝑥4+5
𝑥2 dx
f) 1
𝑥
dx

Mid-ordinate rule :Revision
Example 1: The area of an irregular metal plate needs to be calculated. The ordinates are drawn 5 cm apart across the surface of the
metal plate. The lengths of the ordinates in cm’s are: 29; 32; 33; 32,5; 32; 31; 31; 32; 33; 35; 37; 39; 40.
Solution:
Ordinate Calculation Mid-ordinate
29 𝟐𝟗 + 𝟑𝟐
𝟐
30,5
32 𝟑𝟐 + 𝟑𝟑
𝟐
32,5
33 𝟑𝟑 + 𝟑𝟐, 𝟓
𝟐
32,75
32,5 𝟑𝟐, 𝟓 + 𝟑𝟐
𝟐
32,25
32 𝟑𝟐 + 𝟑𝟏
𝟐
31,5
31 𝟑𝟏 + 𝟑𝟏
𝟐
31
31 𝟑𝟏 + 𝟑𝟐
𝟐
31,5
32 𝟑𝟐 + 𝟑𝟑
𝟐
32,5
33 𝟑𝟑 + 𝟑𝟓
𝟐
34
35 𝟑𝟓 + 𝟑𝟕
𝟐
36
37 𝟑𝟕 + 𝟑𝟗
𝟐
38
39 𝟑𝟗 + 𝟒𝟎
𝟐
39,5
40
Sum of mid-ordinates 402
Area of sheet
metal = k sum of
mid-ordinates
= 5 402
= 2010 cm2

Area of irregular shapes- Revision
• We can verify, the accuracy of our work in example 1 above,
by using the following formula:
• Area = k × [(average of first & last ordinate)+ (sum of rest
of ordinates)]
= 5 × [(
29+40
2
) +( 32+33+32,5+32+31+31+32+33+35+37+39)]
= 5[ 34,5+ 367,5]
= 2010 cm2

Revision: Area of irregular shapes
• Thus, when we are using the ordinates:
We use the mid-ordinate rule:
Sum of mid-ordinates = (Average of first and last ordinates) +
(Sum of rest of ordinates), and
Area = k × Sum of mid-ordinates, where k is the “width”

Example 2: Area of irregular shape
Example 2: Determine the area of an irregular metal plate. Ordinates are drawn
1,5 cm apart. The lengths of the ordinates in cm are: 0;30;42;49;56;46;38;30;18;0
Solution:
Sum of mid-ordinates = (average of first and last ordinates) + (sum of other
ordinates)
= [(
0+0
2
) + (30 + 42+49+56+46+38+30+18)]
= [ 0 + 309]
= 309
Area = k × Sum of mid-ordinates
= 1,5 (309)
= 463,5 cm2

The figure alongside is a trapezium with AB // CD and
FE perpendicular AD.
Area trapezium =
1
2
× (sum of parallel sides) × perp.
distance
=
1
2
× (AB + CD) × (AD)
But
1
2
× (AB + CD) = mid -ordinate FE …….(1)
∴ Area of trapezium = FE × AD…………..(2) E
A
B
C
D
F

Introduction to definite integral
O
x
A
C
B
D
I
E
H
G
F
J
In the figure above, OABC is bounded by a base OC , two
vertical ordinates OA and BC and a curve AB.

Introduction to definite intrgral
• To determine the area below the curve, divide the base OC into any
number of equal parts, each with length k units.
• Each of the strips, OADT, TDES, SEFR, RFGP, PGHN, NHIM, MIJL
and LJBC resemble trapezia (the plural of trapezium).
• Halfway, between each of the ordinates we draw mid-ordinates, i.e the
average of any two consecutive ordinates, as denoted by the dotted lines.
• We can thus calculate the area of each trapezium using the formula
developed above,
• Area of each strip = k × length of mid-ordinate, where k is the length of
the base of each trapezium

Example 1
• Consider the following example: Plot the graph of y = 3x –x2,
by completing a table of values of y from x = 0 to x = 3.
• Determine the area enclosed by the curve, the x-axis and
ordinates x = 0 and x = 3 using the mid-ordinate rule.
x 0 0,5 1,0 1,5 2,0 2,5 3,0
y = 3x – x2 0 1,25 2 2,25 2 1,25 0

Example 1
• Using the mid-ordinate rule with six intervals, where the mid-
ordinates are located at:
Area ≈ (0,5)[0,6875 + 1,6875 + 2,1875 + 2,1875 + 1,6875 +
0,6875]
= (0,5)(9,125)
= 4,563 square units
Mid -ordinate 0,25 0,75 1,25 1,75 2,25 2,75
Correspondin
g y-values
0,6875 1,6875 2,1875 2,1875 1,6875 0,6875

Example 2: y = -x2 + 4
Area = 10 sq.
units

Example 2
NB: As the
number of
partitions
increase the
area comes
closer to 9.

Activity 4
1. Using rectangles, calculate the areas described in each of the following cases. Use a partition with a sensible number
of rectangles. Make sure each rectangle has the same width. Make a rough sketch for each case.
a) f(x) = 2x + 7 between the x-axis, and x = – 2 and x = 3.
b) k(x) = x2 bounded by the x –axis and x = – 4 and x = – 1.
c) j(x) =
𝟏
𝒙
, with x > 0, bounded by the x-axis and x = 1 and x = 4.
d) m(x) = x2 + 2, between the x-axis and x = – 5 and x = 3. Is this area an under- or overestimate of the true area?
e) p(x) = 25– 𝑥2 between the x-axis and the two x-intercepts. Can you work out the exact area under this curve? How
accurate was your partitioning method?
f) h(x) = 2 sinx, between the x-axis and x =
𝝅
𝟒
and x = 𝜋.
2. Create an Excel spreadsheet to approximate the area bounded by the curve
t(x) = –x2 + 4 on the interval [–1 ; 2] for 3 different rectangle widths.
3. Determine the area under the function g(x) =
–12
𝑥
, for x ∈ [– 1 ; 6], by dividing the given interval into five trapeziums.
Let the height of each trapezium be 1 unit. Use the following formula to assist you to calculate the area of each
trapezium:
Area =
𝟏
𝟐
× (sum of parallel sides) × height
3.1 How does this method compare with method of dividing the given interval into rectangles?

What about areas that fall below the x-axis?
y = x(x-2)(x +2)

Area below the x-axis
• Whether you use the rectangle or trapezium method to calculate the area bounded by the curve,
the x-axis, between x = –2 and x = 2, you would use all the y-values as positive lengths in your
calculations. Alternatively, the total area is double the area under the curve from x = – 2 to x = 0.
• Using the trapezium rule, with partitions of
1
2
unit wide, the area of the left-hand half under the
curve, between x = –2 and x = 0 is:
Area = sum of areas of the trapeziums
= {[
1
2
× [ f(–2) +f(-1,5)] ×
1
2
} + {[
1
2
× [ f(–1,5) +f(-1)] ×
1
2
} +…+{[
1
2
× [ f(–1) +f(0)] ×
1
2
}
=
1
4
[0 +2,625 + 2,625+ 3 + 3+ 1,875 + 1,875 + 0]
=
15
4
square units OR (3,75 square units)
• So, total area = 2 (
𝟏𝟓
𝟒
) =
𝟏𝟓
𝟐
square units OR (7,5 square units)

The definite integral
• In the previous section we developed a method (a long method!) of finding the
area between the curve and the x-axis.
• The general notation for the area under the curve is 𝒂
𝒃
𝒇 𝒙 𝒅𝒙. This is known
as the definite integral of f(x).
• We define the definite integral of a function f(x) as :
𝒂
𝒃
𝒇 𝒙 𝒅𝒙 = F(b) – F(a); where 𝑭′
(x) = f(x).
• We call this a definite integral because the result of integrating and evaluating
is a number. (The indefinite integral has an arbitrary constant in the result).
• The numbers a and b are called the lower limit and upper limit, respectively.
• We can see that the value of a definite integral is found by evaluating the
function (found by integration) at the upper limit and subtracting the value of
this function at the lower limit.

The definite integral
• The definite integral, 𝒂
𝒃
𝒇 𝒙 𝒅𝒙 can be interpreted as the
area under the curve of y = f(x) from x = a to x = b, and in
general as a summation.
• NB: If we asked to evaluate the definite integral, without
mention of calculating area, one would proceed as follows:
• Evaluate: −𝟐
𝟐
𝒙𝟑
𝒅𝒙 =
𝟏
𝟒
[𝟐𝟒
– (– 𝟐)𝟒
] = 0

The fundamental Theorem of Calculus
• If f(x) is continuous on the interval 𝒂 ≤ 𝒙 ≤ 𝒃 and if F(x) is any
indefinite integral of f(x), then:
𝑎
𝑏
𝑓 𝑥 𝑑𝑥 = [𝐹 𝑥 ]𝑎
𝑏
= F(b) – F(a)
• This theorem is showing us how to evaluate an integral once the
integration process has taken place.
• NB: When evaluating the definite integral there is no need to
consider the arbitrary constant.

Worked examples – the definite integral
Evaluate the following definite integrals
1
𝟐
𝟓
𝟖𝐱𝐝𝐱 Let I = 2
5
8𝑥𝑑𝑥
=
8
2
[𝑥2 ]2
5
using definition
= 4[52 – 22] using fundamental
theorem of calculus
= 4[ 25 - 4]
= 4[21]
∴ I = 84

Worked examples – the definite integral
Evaluate the following definite integrals
2.
𝟎
𝟐
𝟔𝐱 + 𝟕 𝐝𝐱 Let I = 𝟎
𝟐
𝟔𝐱 + 𝟕 𝒅𝒙
= [𝟑𝒙𝟐
+ 𝟕𝐱]𝟎
𝟐
= [3 (2)2 + 7(2)] – [3(0)2 + 7(0)]
= 12 + 14 – 0
∴ I = 26

Worked examples- The definite integral
3.
𝟎
𝟏
𝟐𝒅𝒙 Let I = 0
1
2𝑑𝑥
= [2𝑥]0
1
= [2(1) – 2(0)]
∴ I = 2

Worked examples- The definite integral
4.
𝟐
𝟒 𝟏𝟐
𝒙
dx Let I = 2
4 12
𝑥
dx
= 12 2
4 1
𝑥
dx
= 12 𝒍𝒏 𝑥 ]2
4
= 12(ln 4 – ln 2)
∴ I = 12ln 2

Activity 5
1. Evaluate the following definite integrals, using the method outlined above.
a)
−1
1
𝑥 − 1 (𝑥 + b)
0
1
2𝑥 − 3 𝑑𝑥
c)
0
1
3𝑥 + 2 𝑑𝑥 d)
–1
0
5𝑥4
dx
e)
1
2 𝑑𝑥
𝑥2
f)
1
2 𝑥4+1
𝑥2 dx
g)
2
3
10𝑛𝑥
dx h)
−2
3
(𝑥2
+ 5)𝑑𝑥

Some properties of integrals
1.
𝑎
𝑏
𝑓 𝑥 𝑑𝑥 = – 𝑏
𝑎
𝑓 𝑥 𝑑𝑥
2.
𝑎
𝑏
𝑓 𝑥 𝑑𝑥 + 𝑏
𝑐
𝑓 𝑥 𝑑𝑥 = 𝑎
𝑐
𝑓 𝑥 𝑑𝑥
3 Constants may be factored through the integral sign:
𝒂
𝒃
𝒌. 𝒇 𝒙 𝒅𝒙 = k. 𝒂
𝒃
𝒇 𝒙 𝒅𝒙
4. The integral of a sum (and /or difference) is the sum (and /or difference) of the integrals:
𝒂
𝒃
[𝒇 𝒙 ± 𝒈 𝒙 ]𝒅𝒙 = 𝒂
𝒃
𝒇 𝒙 𝒅𝒙 ± 𝒂
𝒃
𝒈 𝒙 𝒅𝒙
5. The integral of a linear combination is the linear combination of the integrals:
𝒂
𝒃
[𝒌. 𝒇 𝒙 + 𝒎. 𝒈 𝒙 ]𝒅𝒙 = k. 𝒂
𝒃
𝒇 𝒙 𝒅𝒙 + m. 𝒂
𝒃
𝒈 𝒙 𝒅𝒙
Here are some simple properties of the integral that are often used in computations. Throughout take f
and g as continuous functions.

Activity 6
Property 3: 𝒂
𝒃
𝒌. 𝒇 𝒙 𝒅𝒙 = 𝒌. 𝒂
𝒃
𝒇 𝒙 𝒅𝒙
• With your partner, explore the validity of property 3 using
the following functions:
o f(x) = x2 and k = 2
o f(x) = 2x and k = –0,5
o f(x) =
𝟏
𝒙𝟐 and k = –1,5
• Comment on your findings.

Activity 7
Investigate the validity of the following property:
• 𝒂
𝒃
𝒇 𝒙 + 𝒈 𝒙 𝒅𝒙 = 𝒂
𝒃
𝒇(𝒙)𝒅𝒙 + 𝒂
𝒃
𝒈(𝒙)𝒅𝒙,
if f(x) = –3x and g(x) = x2 + 6 for the interval [–4 ;–1].
o You will need to draw graphs of the two functions on the same set
of axis.
o You will need to show all working details.
• Comment on your findings.

Activity 8
1. Show that: 𝟏
𝟑
𝟒𝒙 + 𝟏 𝒅𝒙 = – 𝟑
𝟏
𝟒𝒙 + 𝟏 𝒅𝒙
2. Show that: 𝟎
𝟐
(𝟑𝒙𝟑
− 𝟔𝒙 − 𝟗)𝒅𝒙 = 𝟑 𝟎
𝟐
(𝒙𝟑
− 𝟐𝒙 − 𝟑)𝒅𝒙
3. Show that: 𝟏
𝟒
𝟐𝒙−𝟐𝒅𝒙 = 𝟏
𝟐
𝟐𝒙−𝟐𝒅𝒙 + 𝟐
𝟒
𝟐𝒙−𝟐𝒅𝒙
4. Show that: −𝟑
𝟒
(−𝒙𝟐 + 𝟓𝒙 − 𝟔)𝒅𝒙 = −𝟑
𝟒
−𝒙𝟐 𝒅𝒙 + −𝟑
𝟒
( 𝟓𝒙 − 𝟔)𝒅𝒙

Area under a curve
• Use the properties of definite integrals developed in the
previous section to calculate the area under a curve.
• NB: If asked to evaluate a definite integral without explicitly
asked to calculate the area, then proceed to evaluate the
definite integral without concern about the answer-i.e the
answer can be 0 or a negative value!!

Area under a curve- Worked example
• Find the area under the curve y = x2 , between x = 1 and x =
3.
Let the required
area be A , then:
A = 1
3
𝑥2𝑑𝑥
= [
𝑥3
3
]1
3
=
33
3
–
13
3
A =
𝟐𝟔
𝟑
square
units

Area under a curve
Find the area between the curve y = x2 + 4x and the
x-axis from:
ox = – 2 to x = 0
ox = 0 to x = 2
ox = – 2 to x = 2
A1
A2

Worked example 2
For the interval [– 2 ; 2], we observe that the required area comprises of two parts – one part below the x-axis and the other part
above the x-axis.
Let A1 = −𝟐
𝟎
𝒙𝟐
+ 𝟒𝒙 𝒅𝒙
= [
𝑥3
3
+ 2𝑥2
]−2
0
= 0 –[
−23
3
+ 2(−2)2
]
= 0 – [
16
3
]
= –
16
3
…..the minus sign tells us the area is below the x-axis
Thus, A1 =
𝟏𝟔
𝟑
• Calculate A2 , as follows:
A2 = 0
2
𝑥2
+ 4𝑥 𝑑𝑥 =
32
3
(The reader should verify the accuracy)
• So, the area over the interval [–2 ; 2] is obtained by adding A1 and A2, i.e:
A = A1 + A2
=
𝟏𝟔
𝟑
+
𝟑𝟐
𝟑
= 16 square units.

Own work
With your partner, attempt the following.
The given sketch is a graphical representation of the function
defined by: f(x) = x3 – 4x2 + 3x.
Determine the area between the curve and the x-axis from x = 0
to x = 3.

Challenge
• Find the area enclosed between the curve
y = x2 – 2x –3 and the straight-line y = x + 1.

Activity 9
1. Use the properties of definite integrals which you have studied thus far,
to evaluate the following definite integrals.
a)
−2
2
6𝑡2
+ 1 𝑑𝑡 b)
0
1
2𝑥 + 1 𝑥 + 3 𝑑𝑥
c)
2
4
5𝑥 − 4 𝑑𝑥 d)
−3
3
6𝑥3 + 2𝑥 𝑑𝑥
2. Calculate the areas of the following:
a)
−2
2
6𝑡2
+ 1 𝑑𝑡 b)
0
1
2𝑥 + 1 𝑥 + 3 𝑑𝑥
c)
2
4
5𝑥 − 4 𝑑𝑥 d)
−3
3
6𝑥3
+ 2𝑥 𝑑𝑥

Activity 9
3. Find the area under the curve y = x2 from x = 0 to x = 6.
4. Find the area under the curve y = 3x2 + 2x from x = 0 to x = 4.
5. Find the area under the curve y = 3x2 -2x from x = – 4 to x = 0.
6. Find the area enclosed by the x-axis and the following curves and straight
lines.
a) y = x2 + 3x ; x = 2 ; x =5 b) y =
1
8
x3 + 2x ; x = 2 ; x =4
c) y = (3x–4)2 ; x = 1 , x =3 d) y = 2 – x3 ; x = –3 ; x = – 2
7.a) Sketch the curve y = x(x +1)(x –3), showing where it cuts the x-axis.
b) Calculate area of the region, above the x-axis, bounded by the x-axis and the
curve.
c) Calculate area of the region, below the x-axis, bounded by the x-axis and the
curve. Ishaak Cassim - February 2018 52

Activity 9
8. The diagram shows the region under
y = 4x +1 between x =1 and x = 3. Find
the area below the graph between x = 1
and x= 3, using:
a) The formula for the area of a trapezium
b) integration
c) How do the results in a) and b) compare
with each other? Explain.
9. The diagram alongside shows the region
bounded by y =
1
2
𝑥 – 3, by x =14 and the
x-axis. A(r ; 0) is the x-intercept of the
straight line graph.
a) Determine the numerical value of r.
b) Determine area of the shaded region
using:
i) the formula for the area of a triangle.
ii) integration
1 3
y = 4x +1
A (r ;0) 14
y =
1
2
𝑥 – 3
x

Activity 9
10. Calculate the areas of the following:
a) y = 1 + x2 for
0 < x < 2
b) y = 5x – x2 ; for x ∈ [0 ; 5]
11. Evaluate 0
2
𝑥 𝑥– 1 𝑥– 2 𝑑𝑥 and explain your answer
with reference to the graph of
y = x(x–1)(x–2).
12. Given that: −𝑎
𝑎
15𝑥2
dx = 3430. Determine the
numerical value of a
13. Given that: 1
𝑝
(8𝑥3
+ 6𝑥)𝑑𝑥 = 39. Determine two
possible values of p. Use a graph to explain why there
are two values.
14. Show that the area enclosed between the curves
y = 9 –x2 and y = x2 – 7 is
128 2
3

Conclusion

technical-mathematics-integration-17-feb_2018.pptx

Recommended

Recommended

More Related Content

Similar to technical-mathematics-integration-17-feb_2018.pptx

Similar to technical-mathematics-integration-17-feb_2018.pptx (20)

Recently uploaded

Recently uploaded (20)

technical-mathematics-integration-17-feb_2018.pptx