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

2. 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:
kxn, with n ∈ ℝ; with 𝑛 ≠ –1
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.
Ishaak Cassim - February 2018

3. Introduction
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4. Activity 1
Activity 1
Write down at least five other functions whose derivative is 4x.
Ishaak Cassim - February 2018

5. The integral of 4x then is 2x2 + c.
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.
Ishaak Cassim - February 2018

6. Thus, to integrate 4x, we will write it as follows:
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:
Ishaak Cassim - February 2018

7. Introduction
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8. We say that 4x is integrated with respect x, i.e: 𝟒𝒙 𝒅𝒙
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 = 2 𝑥 2 + c; where c ∈ ℝ
Ishaak Cassim - February 2018

9. 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.
Ishaak Cassim - February 2018

10. Activity 2
Ishaak Cassim - February 2018

11. Activity 2
Differentiate
Integrate
axn
Ishaak Cassim - February 2018

12. The general solution of integrals of the form kxn
From Activity 2 above, the general solution of integrals of the form 𝒌 𝒙 𝒏 dx , where k and n are constants is given by:
𝒌 𝒙 𝒏 dx = 𝒌 𝒙 𝒏+𝟏 𝒏+𝟏 + c ; where 𝒏≠ −𝟏 and c ∈ ℝ
Ishaak Cassim - February 2018

13. 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 𝒙 𝟐 + c; where c ∈ ℝ
f (x) = x2
𝒙 𝟐 𝒅𝒙 = 𝟏 𝟑 𝒙 𝟑 + c; where c ∈ ℝ
f(x) = axn
𝒂𝒙 𝒏 𝒅𝒙 = 𝒂𝒙 𝒏+𝟏 𝒏+𝟏 + c; where c ∈ ℝ
f (x) = 𝒙 −𝟏 = 𝟏 𝒙
𝟏 𝒙 dx = ln 𝒙 + c; where c ∈ ℝ
f (x) = kanx
𝒌𝒂 𝒏𝒙 dx = 𝒌𝒂 𝒏𝒙 𝒏.𝒍𝒏𝒂 + c ; where c ∈ ℝ ; a > 0 and a ≠ 1
Ishaak Cassim - February 2018

14. Worked Examples: Indefinite integrals
𝑥 7 dx
= 𝒙 𝟕+𝟏 𝟕+𝟏 + c ; where c∈ ℝ
= 𝒙 𝟖 𝟖 + c ; where c∈ ℝ
Compare 𝑥 7 dx with 𝒂 𝒙 𝒏 𝒅𝒙 = 𝒂𝒙 𝒏+𝟏 𝒏+𝟏 + c,
Then: a = 1; n = 7 ; n + 1 = 8
Ishaak Cassim - February 2018

15. Worked Examples: Indefinite integralsc)
2 𝑢 2 du
= 2 𝑢 −2 du
= 𝑢 −2 du
= 2 ( 𝑢 −2+1 −2+1 ) + c ; where c∈ ℝ
= −𝟐 𝒖 + c ; where c∈ ℝ
Note: in this example, we are integrating with respect to u.
Explain the method used
Ishaak Cassim - February 2018

16. Worked Examples: Indefinite integrals
𝑥 dx
= 𝒙 𝟏 𝟐 dx =
= 𝟐 𝟑 + c; where ………..
Complete
Ishaak Cassim - February 2018

17. Activity 3 a) 4x3 b) 6x5 c) 2x d) 3x2 + 5x4 e) 10x9 – 8x7 – 1 f)
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 𝑥 2 i)
−2 5𝑥 − 𝑥
Ishaak Cassim - February 2018

18. Activty 3 - continued
2. Determine the indefinite integrals in the following cases a)
(9 𝑥 2 –4𝑥−5)𝑑𝑥 b)
(12 𝑥 2 +6𝑥+4)𝑑𝑥 c)
−5 𝑑𝑥 d)
(16 𝑥 3 − 6𝑥 2 +10𝑥−3)𝑑𝑥 e)
(2 𝑥 3 +5𝑥)𝑑𝑥 f)
(𝑥+ 2𝑥 2 )𝑑𝑥 g)
( 2𝑥 2 – 3x – 4)dx h)
(1−2𝑥− 3𝑥 2 )𝑑𝑥
3. Determine the following indefinite integrals
1 𝑥 3 dx
( 𝑥 2 – 1 𝑥 2 )dx
𝑥 𝑑𝑥
6 𝑥 𝑑𝑥
6 𝑥 𝑥 2 dx
1 𝑥 dx
Ishaak Cassim - February 2018

19. 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 𝟐
32,25
𝟑𝟐+𝟑𝟏 𝟐
31,5 31
𝟑𝟏+𝟑𝟏 𝟐
𝟑𝟏+𝟑𝟐 𝟐
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 = = 2010 cm2
Ishaak Cassim - February 2018

20. 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 × [( ) +( , )]
= 5[ 34,5+ 367,5]
= 2010 cm2
Ishaak Cassim - February 2018

21. 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”
Ishaak Cassim - February 2018

22. 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)
= [( ) + ( )]
= [ ]
= 309
Area = k × Sum of mid-ordinates
= 1,5 (309)
= 463,5 cm2
Ishaak Cassim - February 2018

23. Area trapezium = 1 2 × (sum of parallel sides) × perp. distance
The figure alongside is a trapezium with AB // CD and FE perpendicular AD.
Area trapezium = × (sum of parallel sides) × perp. distance
= × (AB + CD) × (AD)
But × (AB + CD) = mid -ordinate FE …….(1)
∴ Area of trapezium = FE × AD…………..(2) E A B C D F
Ishaak Cassim - February 2018

24. Introduction to definite integral
In the figure above, OABC is bounded by a base OC , two vertical ordinates OA and BC and a curve AB.
Ishaak Cassim - February 2018

25. 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
Ishaak Cassim - February 2018

26. 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
0,5
1,0
1,5
2,0
2,5
3,0
y = 3x – x2
1,25 2
2,25
Ishaak Cassim - February 2018

27. Corresponding y-values
Example 1
Using the mid-ordinate rule with six intervals, where the mid- ordinates are located at:
Area ≈ (0,5)[0, , , , , ,6875]
= (0,5)(9,125)
= 4,563 square units
Mid -ordinate
0,25
0,75
1,25
1,75
2,25
2,75
Corresponding y-values
0,6875
1,6875
2,1875
Ishaak Cassim - February 2018

28. Example 2: y = -x2 + 4 Area = 10 sq. units
Ishaak Cassim - February 2018

29. Example 2
NB: As the number of partitions increase the area comes closer to 9.
Ishaak Cassim - February 2018

30. Activity 4 Ishaak Cassim - February 2018 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– 𝑥 2 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?
Ishaak Cassim - February 2018

31. What about areas that fall below the x-axis?
y = x(x-2)(x +2)
Ishaak Cassim - February 2018

32. 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 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 ×[ f(–1,5) +f(-1)] × } +…+{[ 1 2 ×[ f(–1) +f(0)] × }
= 1 4 [0 +2, , , , ]
= square units OR (3,75 square units)
So, total area = 2 ( 𝟏𝟓 𝟒 ) = 𝟏𝟓 𝟐 square units OR (7,5 square units)
Ishaak Cassim - February 2018

33. 𝒂 𝒃 𝒇 𝒙 𝒅𝒙 = F(b) – F(a); where 𝑭 ′ (x) = f(x).
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.
Ishaak Cassim - February 2018

34. Evaluate: −𝟐 𝟐 𝒙 𝟑 𝒅𝒙 = 𝟏 𝟒 [𝟐 𝟒 – (–𝟐) 𝟒 ] = 0
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 𝒙 𝟑 𝒅𝒙 = 𝟏 𝟒 [𝟐 𝟒 – (–𝟐) 𝟒 ] = 0
Ishaak Cassim - February 2018

35. 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.
Ishaak Cassim - February 2018

36. Worked examples – the definite integral
Evaluate the following definite integrals 1
𝟐 𝟓 𝟖𝐱𝐝𝐱
Let I = 𝑥𝑑𝑥
= [ 𝑥 2 ] using definition
= 4[ 5 2 – 2 2 ] using fundamental theorem of calculus
= 4[ ]
= 4[21]
∴ I = 84
Ishaak Cassim - February 2018

37. Worked examples – the definite integral
Evaluate the following definite integrals 2.
𝟎 𝟐 𝟔𝐱+𝟕 𝐝𝐱
Let I = 𝟎 𝟐 𝟔𝐱+𝟕 𝒅𝒙
= [𝟑 𝒙 𝟐 +𝟕𝐱] 𝟎 𝟐
= [3 (2)2 + 7(2)] – [3(0)2 + 7(0)]
= – 0
∴ I = 26
Ishaak Cassim - February 2018

38. Worked examples- The definite integral3.
𝟎 𝟏 𝟐𝒅𝒙
Let I = 𝑑𝑥
= [2𝑥] 0 1
= [2(1) – 2(0)]
∴ I = 2
Ishaak Cassim - February 2018

39. Worked examples- The definite integral4.
𝟐 𝟒 𝟏𝟐 𝒙 dx
Let I= 𝑥 dx
= 𝑥 dx
= 𝒍𝒏 𝑥 ] 2 4
= 12(ln 4 – ln 2)
∴ I = 12ln 2
Ishaak Cassim - February 2018

40. a) −1 1 𝑥−1 𝑥+2 𝑑𝑥 −1 1 𝑥−1 𝑥+2 𝑑𝑥 b) 0 1 2𝑥−3 𝑑𝑥 c) 0 1 3𝑥+2 𝑑𝑥 d)
Activity 5 1.
Evaluate the following definite integrals, using the method outlined above. a)
−1 1 𝑥−1 𝑥+2 𝑑𝑥 −1 1 𝑥−1 𝑥+2 𝑑𝑥 b)
0 1 2𝑥−3 𝑑𝑥 c)
0 1 3𝑥+2 𝑑𝑥 d)
–1 0 5𝑥 4 dx e)
1 2 𝑑𝑥 𝑥 2 f)
𝑥 𝑥 2 dx g)
𝑛𝑥 dx h)
−2 3 ( 𝑥 2 +5)𝑑𝑥
Ishaak Cassim - February 2018

41. Some properties of integrals
Here are some simple properties of the integral that are often used in computations. Throughout take f and g as continuous functions. 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
Ishaak Cassim - February 2018

42. Comment on your findings.
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:
f(x) = x2 and k = 2
f(x) = 2x and k = –0,5
f(x) = 𝟏 𝒙 𝟐 and k = –1,5
Comment on your findings.
Ishaak Cassim - February 2018

43. Investigate the validity of the following property:
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].
You will need to draw graphs of the two functions on the same set of axis.
You will need to show all working details.
Comment on your findings.
Ishaak Cassim - February 2018

44. Activity 8 1. Show that: 𝟏 𝟑 𝟒𝒙+𝟏 𝒅𝒙 = – 𝟑 𝟏 𝟒𝒙+𝟏 𝒅𝒙 2.2.
Show that: 𝟎 𝟐 (𝟑 𝒙 𝟑 −𝟔𝒙−𝟗)𝒅𝒙 = 𝟑 𝟎 𝟐 ( 𝒙 𝟑 −𝟐𝒙−𝟑)𝒅𝒙 3.
Show that: 𝟏 𝟒 𝟐 𝒙 −𝟐 𝒅𝒙 = 𝟏 𝟐 𝟐 𝒙 −𝟐 𝒅𝒙 + 𝟐 𝟒 𝟐 𝒙 −𝟐 𝒅𝒙 4.
Show that: −𝟑 𝟒 (− 𝒙 𝟐 +𝟓𝒙−𝟔)𝒅𝒙 = −𝟑 𝟒 − 𝒙 𝟐 𝒅𝒙 + −𝟑 𝟒 ( 𝟓𝒙−𝟔)𝒅𝒙
Ishaak Cassim - February 2018

45. 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!!
Ishaak Cassim - February 2018

46. 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 = 𝑥 2 𝑑𝑥
= [ 𝑥 3 3 ] 1 3
= –
A = 𝟐𝟔 𝟑 square units
Ishaak Cassim - February 2018

47. Area under a curve
Find the area between the curve y = x2 + 4x and the x-axis from:
x = – 2 to x = 0
x = 0 to x = 2
x = – 2 to x = 2 A2 A1
Ishaak Cassim - February 2018

48. Worked example 2 Ishaak Cassim - February 2018
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 𝒙 𝟐 +𝟒𝒙 𝒅𝒙
= [ 𝑥 𝑥 2 ] −2 0
= 0 –[ − ( −2) 2 ]
= 0 – [ ]
= – …..the minus sign tells us the area is below the x-axis
Thus, A1 = 𝟏𝟔 𝟑
Calculate A2 , as follows:
A2 = 𝑥 2 +4𝑥 𝑑𝑥 = (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.
Ishaak Cassim - February 2018

49. With your partner, attempt the following.
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.
Ishaak Cassim - February 2018

50. Find the area enclosed between the curve
Challenge
Find the area enclosed between the curve
y = x2 – 2x –3 and the straight-line y = x + 1.
Ishaak Cassim - February 2018

51. Activity 9 1.
Use the properties of definite integrals which you have studied thus far, to evaluate the following definite integrals. a)
− 𝑡 2 +1 𝑑𝑡 b)
0 1 2𝑥+1 𝑥+3 𝑑𝑥 c)
2 4 5𝑥−4 𝑑𝑥 d)
− 𝑥 3 +2𝑥 𝑑𝑥 2.
Calculate the areas of the following:
Ishaak Cassim - February 2018

52. 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 = 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.
Calculate area of the region, above the x-axis, bounded by the x-axis and the curve.
Calculate area of the region, below the x-axis, bounded by the x-axis and the curve.
Ishaak Cassim - February 2018

53. Activity 9
Ishaak Cassim - February 2018

54. Activity 9 10. Calculate the areas of the following: a) y = 1 + x2 forb)
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 15 𝑥 2 dx = Determine the numerical value of a
13.
Given that: 1 𝑝 (8 𝑥 3 +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
Ishaak Cassim - February 2018

55. Conclusion
Ishaak Cassim - February 2018