introduction to integration - the chalkface
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Introduction to Integration
Understanding Integration
If differentiation gives a meaningful answer to 0 Γ· 0 (gradient of a curve), then integration gives a meaningful answer to 0 Γ β (area under a curve).
Integration is the process of adding up an infinite number of infinitesimally small amounts. By considering what happens as small pieces shrink to nothing (and the number of them rises towards infinity), we can find exact answers to otherwise impossible questions.
Example: Finding the area of a circle. Step 1: Unwrap the circumference. This must be of length 2ππ.
Step 2: Split the circle into a number of sectors and rearrange them.
Step 3: Increase the number of slices used (they will get smaller).
Step 4: Notice that as the number of slices increases (and their size decreases), the shape increasingly resembles a rectangle. This rectangle has height π and width ππ, giving an area of:
π΄ = π Γ ππ = π ππ
Reversing Differentiation
Integration is the opposite of differentiation The integral is the opposite of the derivative
Differentiate Integrate
π(π₯) = β« πΉ(π₯)
πβ²(π₯) = πΉ(π₯)
Eg: π₯
1
(The derivative of π₯ is 1, so the integral of 1 is π₯)
π₯2
π₯3
π₯4
π₯
π₯2
π₯3
Fill in the blanks, then try to find a general rule:
The integral of π₯π is:
The Constant of Integration
Dave spends Β£300, earns Β£500, then spends a further Β£100. How much money does he have now?
A quadratic differentiates to 2π₯ β 3. Which of these could have been my quadratic? a) π₯2 β 3π₯ b) 2π₯2 β 6π₯ c) π₯2 β 3π₯ + 3 d) (π₯ + 2)(π₯ β 5)
Imagine trying to work out how much money someone has by keeping track of their income and their spending. You can tell a lot from this (how much more they have this month compared to last month, for instance), but unless you see the actual bank balance at some point you have no way of knowing the amount.
Since information is lost when we differentiate, it is not possible, without additional information, to perfectly reverse the process.
You have come across similar ideas before: If I square a number and get 9, you canβt know for sure whether my original number was 3 or β3.
The only information lost when you differentiate is the constant term (number). Integration does tell us the shape of our original function. The only thing it canβt tell us is the actual height of the graph. The curves below all belong to the same βfamilyβ of functions. They are all of the form:
π(π) = ππ β πππ β ππ + πͺ The only unknown is πΆ. The greater the value of πΆ, the higher up the curve. All of these curves have exactly the same gradient function:
πβ²(π) = πππ β ππ β π
The Format of Integration Questions
Since integration is the reverse of differentiation, often a question will provide you with a
gradient function, ππ¦
ππ₯ or πβ²(π₯) and ask for the βoriginalβ function, π¦ or π(π₯). Alternatively,
since integration is actually more powerful than simply the reverse of differentiation, it has its own symbol and notation. βThe integral of β¦ with respect to π₯β is written as β« β¦ ππ₯: All of the following are asking you to do the same thing. Your notation should reflect that used in the question, but the process you follow is essentially the same:
Question Solution
Integrate 3π₯2 β 4π₯ with respect to π₯. π₯3 β 2π₯2 + πΆ ππ¦
ππ₯= 3π₯2 β 4π₯. Find π¦. π¦ = π₯3 β 2π₯2 + πΆ
πβ²(π₯) = 3π₯2 β 4π₯. Find π(π₯). π(π₯) = π₯3 β 2π₯2 + πΆ
Find: β« 3π₯2 β 4π₯ ππ₯. π₯3 β 2π₯2 + πΆ
Find the integral of 3π₯2 β 4π₯ with respect to π₯. π₯3 β 2π₯2 + πΆ
A curve has gradient 3π₯2 β 4π₯. Find the equation of the curve. π¦ = π₯3 β 2π₯2 + πΆ
Finding the constant With no further information, your answers must include an arbitrary constant: + πΆ. However, if you are given additional information, you can calculate πΆ.
Eg: A curve with gradient ππ¦
ππ₯= 3π₯2 β 4π₯ goes through the point (1,4). Find π¦.
Integrate and write with + πΆ:
π¦ = π₯3 β 2π₯2 + πΆ Substitute in the known values:
π¦ = 4 π€βππ π₯ = 1 βΉ 4 = (1)3 β 2(1)2 + πΆ βΉ 5 = πΆ Rewrite now you know πΆ:
π = ππ β πππ + π
Definite Integration
Indefinite integration gives a function, usually with a + πΆ at the end. Definite integration gives a specific value β it is equivalent to the area between two limits.
Eg: Find the value of β« 3π₯2 β 4π₯4
1 ππ₯.
Integrate, then substitute in the two limit values and find the difference:
β« 3π₯2 β 4π₯4
1
ππ₯ = [π₯3 β 2π₯2]14 = {43 β 2(4)2} β {13 β 2(1)2} = {32} β {β1} = ππ
Note: + πΆ is not necessary with definite integration; if you include it, it simply cancels out.
Finding Area Using Integration While we can use integration to reverse differentiation, it actually means a lot more than that. Integration, at its most fundamental, is about adding things up.
But instead of adding a finite number of discrete values, like this summation function doesβ¦
β π
4
π=1
= 1 + 2 + 3 + 4 = 10
(The Greek letter π΄ (sigma) means βsumβ).
β¦ integration gives us a way to add up an infinite continuum of values:
β« π₯ ππ₯4
0
= 8
(The Medieval letter β« (long s) also means βsumβ).
More generally, the area under any curve can be calculated by integrating the function between two limits. This means finding the difference between the value of the integral function at the two limits:
β« π₯2 β π₯ + 3 ππ₯2
0
= [π₯3
3β
π₯2
2+ 3π₯]
0
2
= {(2)3
3β
(2)2
2+ 3(2)} β {
(0)3
3β
(0)2
2+ 3(0)} =
ππ
π
Caution: If the curve is below the π₯-axis, the integral will be negative. The size of the integral will give the area. Terminology: When a function is integrated between two limits, the process is called definite integration as opposed to general integration without limits, which is known as indefinite integration and involves an arbitrary constant, πΆ.
Area of common shapes Integration has given us, in one go, a way to find the area of any shape whose boundary can be described by an integrable function. To convince yourself that this is consistent with finding area the basic way, consider a simple shape: Area of a trapezium using integration
Specific Example General Case (to find a formula)
The area of a trapezium with apex and base 2 and 6 respectively, and height 5 is equivalent to the area under:
π¦ =4
5π₯ + 2
between 0 and 5.
The area of a trapezium with apex and base π and π respectively, and height β is equivalent to the area under:
π¦ =π β π
βπ₯ + π
between 0 and β.
π΄πππ = β«4
5π₯ + 2
5
0
ππ₯
= [2
5π₯2 + 2π₯]
0
5
= {2
5(5)2 + 2(5)} β {0} = ππ
π΄πππ = β«π β π
βπ₯ + π
β
0
ππ₯
= [π β π
2βπ₯2 + ππ₯]
0
β
= {π β π
2ββ2 + πβ} β {0} =
π + π
ππ
Integration is wasted on simple shapes like trapezia. It isnβt the most efficient way to find a formula for such simple shapes, but since it works for any polynomial curve, you can use it to show that, for instance, the area enclosed by the parabolic curve
π¦ = π2 β π₯2 is always equal to 4
3π3.
Integration Area Checklist
Skill Example
1. Area under a curve; positive I can use definite integration to find the area between a curve and the π₯-axis for everywhere-positive intervals.
Find the area bounded by the curve, the π₯-axis and the line π₯ = 1:
π¦ = π₯5 β 3π₯2 + π₯ + 5 2. Area under a curve; negative I can use definite integration to find the area bounded by a curve and the π₯-axis for everywhere-negative intervals, interpreting my answers appropriately to take into account the sign of the integral.
Find the area of the finite region bounded by the curve and the π₯-axis:
π¦ = π₯2(π₯ β 2) 3. Area under a curve; mixed I can use definite integration, reading graphs or solving equations as necessary, to find the area enclosed by a curve and the π₯-axis for a combination of positive and negative intervals, taking into account the sign of the respective integrals.
Find the total area of the two shaded regions shown below:
π¦ = (π₯ β 2)(π₯ + 1)(π₯ β 1) 4. Area between a curve and a line I can use definite integration, along with rules of area for rectangles, triangles and trapezia, to find the area enclosed by a curve and a line.
Find the area of the shaded region, enclosed by the curve and the line.
π¦ = 12π₯2 β 19π₯ β 2π₯3
Area Under Curves Exam Questions 1.
C1 Jan β05
2.
C1 Jan β08
3. The curve with equation π¦ = π₯3 + 4π₯2 + 3π₯ is sketched below.
(i) Express π₯3 + 4π₯2 + 3π₯ in the form π₯(π₯ + π)(π₯ + π) where π and π are integers.
(2 marks) (ii) State the π₯-coordinates of the three points where the curve intersects the π₯-axis.
(1 mark) (iii) Find the total area of the shaded region bounded by the curve and the π₯-axis.
(6 marks)
4.
C1 Jun β12
Integration Challenge Question The curve with equation π¦ = π₯3 β 12π₯ is shown below:
The curve crosses the π₯-axis at the origin and two other points.
a) Show that the π₯-coordinates of the other two points are 2β3 and β2β3. b) Calculate the value of the integral:
β« π₯3 β 12π₯2β3
β2β3
ππ₯
c) What does this value tell you about the area of the two shaded regions? d) By forming suitable integrals, find the total area of the shaded regions. e) Find the coordinates of the maximum point, π. f) Calculate the area of the shaded region enclosed by the curve, the tangent at π and the π¦-axis:
Combining Areas with Integration The curve with equation π¦ = π(π₯) is shown below.
The curve crosses the π₯-axis at (β5,0), (β2,0), (1.5,0) and (4,0).
Using the information provided below, answer the questions that follow:
β« π(π₯)β2
β5
ππ₯ = β5 β« π(π₯)1.5
β5
ππ₯ = β1 β« π(π₯)4
β2
ππ₯ = 1
1. Find the value of the following integrals: a) b) c)
β« π(π₯)1.5
β2
ππ₯ = β« π(π₯)4
1.5
ππ₯ = β« π(π₯)4
β5
ππ₯ =
2. Find the total area of the shaded region in each diagram: a) b) c)
π΄πππ =
π΄πππ =
π΄πππ =
Recall that integration calculates the area below a curve, meaning if the curve is actually below the π₯-axis in certain places, the integral will be negative. In some cases we want the area (eg total distance travelled on a journey), and in some cases the integral (eventual distance away from starting point).
Applications of Integration
As glacial meltwater feeds into a river, the flow rate increases by 600 litres per hour per hour, from the usual rate of 2000 litres per hour. The flow rate, π, at time π‘ hours after midnight is:
π = 600π‘ + 2000 The total volume of water that passes a certain point between 1am and 5am is given by:
β« 600π‘ + 20005
1
ππ‘
As a car brakes to reduce its speed from 30π/π (approx 60mph) to 15π/π (around 30mph), its speed, π£, at time π‘ is given by:
π£ = 30 β 5π‘ The total distance the car travels during this time is given by:
β« 30 β 5π‘3
0
ππ‘
The force, πΉ Newtons, exerted by a bungee rope on a jumper when stretched a distance π₯ metres is:
πΉ = 800π₯ The elastic energy stored in the rope between becoming taut and reaching maximum length is:
β« 800π₯30
0
ππ₯
There are many, many more examples, but most involve more complicated functions (economic growth uses ππ₯ and planetary motion involves sin π₯ and cos π₯). And this is still just 2-D. Integration can be adapted to find volumes (Core 3), and will eventually be used to solve differential equations (Core 4) which are a vital element of many physical and financial calculations.
Reversing Differentiation SOLUTIONS
Integration is the opposite of differentiation The integral is the opposite of the derivative
Differentiate Integrate
π(π₯) = β« πΉ(π₯)
πβ²(π₯) = πΉ(π₯)
Eg: π₯
1
(The derivative of π₯ is 1, so the integral of 1 is π₯)
π₯2
ππ
π₯3
πππ
π₯4
πππ
ππ
π
π₯
ππ
π
π₯2
ππ
π
π₯3
Fill in the blanks, then try to find a general rule:
The integral of π₯π is: ππ+π
π+π
Area Under Curves Exam Questions SOLUTIONS 1. i.
β« π₯3 β 3π₯2 β 6π₯ + 8 ππ₯ =ππ
πβ ππ β πππ + ππ + πͺ
ii.
β« π₯3 β 3π₯2 β 6π₯ + 81
β2
ππ₯ = [π₯4
4β π₯3 β 3π₯2 + 8π₯]
β2
1
= (14
4β 13 β 3(12) + 8(1)) β (
(β2)4
4β (β2)3 β 3(β2)2 + 8(β2)) = (4.25) β (β16) = ππ. ππ
2. i.
β« π₯3 β 7π₯ β 63
β1
ππ₯ = [π₯4
4β
7π₯2
2β 6π₯]
β1
3
= (34
4β
7(32)
2β 6(3)) β (
(β1)4
4β
7(β1)2
2β 6(β1)) = (β
117
4) β (
11
4) = βππ
ii. π΄πππ = |β32| = ππ
3. i.
π₯3 + 4π₯2 + 3π₯ = π₯(π₯2 + 4π₯ + 3) = π(π + π)(π + π) ii.
π = π, π = βπ and π = βπ iii.
πΉπππ π‘ π πππ‘πππ (ππππ£π π₯ β ππ₯ππ ): β« π₯3 + 4π₯2 + 3π₯β1
β3
ππ₯ = [π₯4
4+
4π₯3
3+
3π₯2
2]
β3
β1
= ((β1)4
4+
4(β1)3
3+
3(β1)2
2) β (
(β3)4
4+
4(β3)3
3+
3(β3)2
2) = (
5
12) β (β
9
4) =
32
12=
8
3
ππππππ π πππ‘πππ (πππππ€ π₯ β ππ₯ππ ): β« π₯3 + 4π₯2 + 3π₯0
β1
ππ₯ = [π₯4
4+
4π₯3
3+
3π₯2
2]
β1
0
= ((0)4
4+
4(0)3
3+
3(0)2
2) β (
(β1)4
4+
4(β1)3
3+
3(β1)2
2) = (0) β (
5
12) = β
5
12
πππ‘ππ ππππ = |8
3| + |β
5
12| =
ππ
ππ
4. i.
π₯2 β 3π₯ + 5 = π₯ + 5 βΉ π₯2 β 4π₯ = 0 βΉ π₯(π₯ β 4) = 0 βΉ π₯ = 0 ππ π₯ = 4 π₯ = 0 ππππππ πππ‘π π‘βπ πππππ‘ π΄: (0,5). π₯ = 4 βΉ π¦ = (4) + 5 = 9 βΉ π©: (π, π)
ii.
β« π₯2 β 3π₯ + 5 ππ₯ =ππ
πβ
πππ
π+ ππ + πͺ
iii.
π΄πππ π’ππππ ππππ =π + π
2β =
5 + 9
2(4) = 28
π΄πππ π’ππππ ππ’ππ£π = β« π₯2 β 3π₯ + 54
0
ππ₯ = [π₯3
3β
3π₯2
2+ 5π₯]
0
4
= (43
3β
3(42)
2+ 5(4)) β (0) =
52
3
π΄πππ ππ ππππππ π = (π΄πππ π’ππππ ππππ) β (π΄πππ π’ππππ ππ’ππ£π) = 28 β52
3=
ππ
π
Integration Challenge Question SOLUTIONS The curve with equation π¦ = π₯3 β 12π₯ is shown below:
The curve crosses the π₯-axis at the origin and two other points.
a) Show that the π₯-coordinates of the other two points are 2β3 and β2β3. π₯3 β 12π₯ = 0 βΉ π₯(π₯2 β 12) = 0 βΉ π₯ = 0 ππ π₯2 β 12 = 0
π₯2 = 12 βΉ π₯ = Β±β12 = Β±πβπ b) Calculate the value of the integral:
β« π₯3 β 12π₯2β3
β2β3
ππ₯ = [π₯4
4β 6π₯2]
β2β3
2β3
= {(2β3)
4
4β 6(2β3)
2} β {
(β2β3)4
4β 6(β2β3)
2} = {β36} β {β36} = π
c) What does this value tell you about the area of the two shaded regions? The area of the part above the π₯-axis is equal to that below the π₯-axis. d) By forming suitable integrals, find the total area of the shaded regions.
β« π₯3 β 12π₯0
β2β3
ππ₯ = [π₯4
4β 6π₯2]
β2β3
0
= {(0)4
4β 6(0)2} β {
(β2β3)4
4β 6(β2β3)
2} = {0} β {β36} = 36
βΉ π΄πππ = 2 Γ 36 = ππ e) Find the coordinates of the maximum point, π.
ππ¦
ππ₯= 3π₯2 β 12 & ππ‘ππ‘. ππ‘ βΉ
ππ¦
ππ₯= 0 βΉ 3π₯2 β 12 = 0 βΉ π₯ = Β±2
(π¦)π₯=2 = 23 β 12(2) = β16 & (π¦)π₯=β2 = (β2)3 β 12(β2) = 16 π2π¦
ππ₯2= 6π₯ βΉ (
π2π¦
ππ₯2)
π₯=β2
= β12 < 0 βΉ π΄ππππππ ππ (βπ, ππ)
f) Calculate the area of the shaded region enclosed by the curve, the tangent at π and the π¦-axis:
π΄πππ = 2 Γ 16 β β« π₯3 β 12π₯0
β2
ππ₯
= 32 β ({0} β {β20}) = ππ
Combining Areas with Integration SOLUTIONS The curve with equation π¦ = π(π₯) is shown below.
The curve crosses the π₯-axis at (β5,0), (β2,0), (1.5,0) and (4,0).
Using the information provided below, answer the questions that follow:
β« π(π₯)β2
β5
ππ₯ = β5 β« π(π₯)1.5
β5
ππ₯ = β1 β« π(π₯)4
β2
ππ₯ = 1
1. Find the value of the following integrals: a) b) c)
β« π(π₯)1.5
β2
ππ₯ = 4 β« π(π₯)4
1.5
ππ₯ = β3 β« π(π₯)4
β5
ππ₯ = β4
2. Find the total area of the shaded region in each diagram: a) b) c)
π΄πππ = |β5| = π
π΄πππ = 5 + |4| = π
π΄πππ = 9 + |β3| = ππ
Recall that integration calculates the area below a curve, meaning if the curve is actually below the π₯-axis in certain places, the integral will be negative. In some cases we want the area (eg total distance travelled on a journey), and in some cases the integral (eventual distance away from starting point).
Applications of Integration SOLUTIONS
As glacial meltwater feeds into a river, the flow rate increases by 600 litres per hour per hour, from the usual rate of 2000 litres per hour. The flow rate, π, at time π‘ hours after midnight is:
π = 600π‘ + 2000 The total volume of water that passes a certain point between 1am and 5am is given by:
β« 600π‘ + 20005
1
ππ‘ = [300π‘2 + 2000π‘]15
= (17500) β (2300) = πππππ ππππππ
As a car brakes to reduce its speed from 30π/π (approx 60mph) to 15π/π (around 30mph), its speed, π£, at time π‘ is given by:
π£ = 30 β 5π‘ The total distance the car travels during this time is given by:
β« 30 β 5π‘3
0
ππ‘ = [30π‘ β5π‘2
2]
0
3
= (135
2) β (0) = ππ. π ππππππ
The force, πΉ Newtons, exerted by a bungee rope on a jumper when stretched a distance π₯ metres is:
πΉ = 800π₯ The elastic energy stored in the rope between becoming taut and reaching maximum length is:
β« 800π₯30
0
ππ₯ = [400π₯2]030
= (360000) β (0) = πππ, πππ π±πππππ
There are many, many more examples, but most involve more complicated functions (economic growth uses ππ₯ and planetary motion involves sin π₯ and cos π₯). And this is still just 2-D. Integration can be adapted to find volumes (Core 3), and will eventually be used to solve differential equations (Core 4) which are a vital element of many physical and financial calculations.