p1 chapter 13 :: integration
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P1 Chapter 13 :: Integration
jfrost@tiffin.kingston.sch.ukwww.drfrostmaths.com
@DrFrostMaths
Last modified: 26th August 2017
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Chapter Overview
1:: Find π¦ given ππ¦
ππ₯ 2:: Evaluate definite integrals, and hence the area under a curve.
This chapter is roughly divided into two parts: the first, indefinite integration, is the opposite of differentiation. The second, definite integration, allows us to find areas under graphs (as well as surface areas and volumes) or areas between two graphs.
Find the area bounded between the curve with equation π¦ = π₯2 β 2π₯ and the π₯-axis.
A curve has the gradient function
ππ¦
ππ₯= 3π₯ + 1
If the curve goes through the point 2,3 , determine π¦.
3:: Find areas bound between two different lines.
Find the points of intersection of π¦ = π₯2 β 4π₯ + 3 and π¦ = π₯ + 9, and hence find the area bound between the two lines.
Integrating π₯π terms
Integration is the opposite of differentiation.(For this reason it is also called βantidifferentiationβ)
5π₯3multiply by power reduce power by 1
15π₯2
increase power by 1divide by new power
Find π¦ when ππ¦
ππ₯= 3π₯2
Adding 1 to the power and dividing by this power give us:π¦ = π₯3
However, other functions would also have differentiated to 3π₯2:π¦ = π₯3 + 1, π¦ = π₯3 β 4, β¦
Clearly we could have had any constant, as it disappears upon differentiation.
β΄ π = ππ + π
However, thereβs one added complicationβ¦
FunctionGradient Function
π is known as a constant of integration
Textbook Error on Pg289: βWhen integrating polynomialsβ, but Example 3 is not a polynomial because it has negative and fractional powers.
Examples
Find π¦ when:
ππ¦
ππ₯= 4π₯3 π¦ = π₯4 + π
ππ¦
ππ₯= π₯5 π¦ =
1
6π₯6 + π
Exam Note: Historically βC1β penalised the lack of +π, but modules thereafter didnβt. You should always include it.
You could also write as π₯6
6. Itβs a
matter of personal preference.
ππ¦
ππ₯= 3π₯
12 π¦ = 3
2
3π₯32 + π = 2π₯
32 + π
Fro Tip: Many students are taught to write 3π₯
32
3
2
(as does textbook!). This is ugly and students then often struggle to simplify it. Instead remember back to Year 7: When you divide by a fraction, you multiply by the reciprocal.
More Fractional Examples
Find π¦ when:
ππ¦
ππ₯=
4
π₯= 4π₯β
12 π¦ = 8π₯
12 + π
When we divide by 1
2we
multiply by the reciprocal, i.e. 2. Fro Tip: I recommend doing in your head when the simplification would be simple.
ππ¦
ππ₯= 5π₯β2 π¦ = β5π₯β1 + π
ππ¦
ππ₯= 4π₯
23 π¦ = 4
3
5π₯53 + π =
12
5π₯53 + π
ππ¦
ππ₯= 10π₯β
27 π¦ = 10
7
5π₯57 + π = 14π₯
57 + π
Test Your Understanding
Find π(π₯) when:
πβ² π₯ =2
π₯7= 2π₯β7 π π₯ = β
1
3π₯β6 + π
πβ² π₯ = 3 π₯ = π₯13 π π₯ =
3
4π₯43 + π
πβ² π₯ = 33π₯56 π π₯ = 18π₯
116 + π
πβ² π₯ = 2π₯ + 7 π π₯ = π₯2 + 7π₯ + π
πβ² π₯ = π₯2 β 1 π π₯ =1
3π₯3 β π₯ + π
Note: In case youβre wondering what happens if ππ¦
ππ₯=
1
π₯= π₯β1 , the problem is that after
adding 1 to the power, weβd be dividing by 0. You will learn how to integrate 1
π₯in Year 2.
Integration notation
The following notation could be used to differentiate an expression:
π
ππ₯5π₯2 = 10π₯
There is similarly notation for integrating an expression:
β« 10π₯ ππ₯ = 5π₯2 + π
The ππ₯ here means differentiating βwith respect to π₯β.
βIntegrateβ¦βββ¦this expressionβ
ββ¦with respect to π₯β(the ππ₯ is needed just as it was needed in the differentiation notation at the top of this slide)
This is known as indefinite integration, in contrast to definite integration, which weβll see later in the chapter.
It is called βindefiniteβ because the exact expression is unknown (due to the +π).
Advanced Note: For the moment you should think of the just as
β«
Examples
Find β«(π₯β3
2 + 2) ππ₯
Fro Note: The brackets are required if thereβs multiple terms.
= β2π₯β12 + 2π₯ + π
Find β«(6π‘2 β 1) ππ‘
= 2π‘3 β π‘ + π
Note the ππ‘ instead of ππ₯.
Find β«(ππ₯3 + π) ππ₯ where π and π are constants.
=1
4ππ₯4 + ππ₯ + π
Textbook (Minor) Error: βany other letters must be treated as constantsβ. Similar to the error in the differentiation chapter, it should read βany other letters, which are either constants or variables independent of π₯, can be treated as numbersβ. In β«π₯π¦ ππ₯,
if π¦ is a variable, we can only treat π¦ as a constant if it is not dependent on π₯, i.e. there is not some equation relating π¦ to π₯.
Test Your Understanding
Edexcel C1 May 2014(R) Q4b
π¦ = 2π₯5 + 6π₯β12
ΰΆ±π¦ ππ₯ =1
3π₯6 + 12π₯
12 + π
Finding constant of integration
Recall that when we integrate, we get a constant of integration, which could be any real value. This means we donβt know what the exact original function was.
π₯
π¦πβ² π₯ = 3π₯2
π₯
π¦
π π₯ = π₯3
π π₯ = π₯3 + π
π π₯ = π₯3 β 1
β«
?1,3
But if we know one point on the curve, it leaves only one possibility.The curve with equation π¦ = π(π₯) passes
through 1,3 . Given that πβ² π₯ = 3π₯2, find the equation of the curve.
πβ² π₯ = 3π₯2
β΄ π π₯ = π₯3 + πUsing the point (1,3): 3 = 13 + π β΄ π = 2
π π₯ = π₯3 + 2
Test Your Understanding
Edexcel C1 May 2014 Q10
To keep you occupied if you finish (a) quickly!
π¦ β 25 = β1
2π₯ β 4(b)
Definite Integration
So far weβve seen integration as βthe opposite of differentiationβ, allowing us to find π¦ = π π₯ when we know the gradient function π¦ = πβ² π₯ .
In practical settings however the most useful use of integration is that it finds the area under a graph. Remember at GCSE for example when you estimated the area under a speed-time graph, using trapeziums, to get the distance?If you knew the equation of the curve, you could get the exact area!
Before we do this, we need to understand how to find a definite integral:
spee
d
time
π΄πππ = πππ π‘ππππ
ΰΆ±1
5
4π₯3 ππ₯ =
These are known as limits, which give the values of π₯ weβre finding the area between.
π¦
π₯
π΄πππ = πππ π‘ππππ
1 5
π¦ = 4π₯3
?
We integrate as normal, but put expression in square brackets, meaning we still need to evaluate the integrated expression using the limits.
Write β¦ β (β¦) and evaluate the expression for each of the limits, top one first.
π₯4 15
= 54 β 14
= 624
Another Example
ΰΆ±β3
3
π₯2 + 1 ππ₯ =1
3π₯3 + π₯
β3
3
=1
33 3 + 3 β
1
3β3 3 β 3
= 12 β β12= 24
We DONβT have a constant of integration when doing definite integration. Iβll explain why later.
Write out you working EXACTLY as seen here. The β¦ β (β¦ ) brackets are particularly crucial as youβll otherwise likely make a sign error.
βUse of Technologyβ Monkey says:
You can use the β«ππβ‘ button on your calculator to
evaluate definite integrals. But only use it to check your answer.
Problem Solving
Given that π is a constant and β«152ππ₯ + 7 ππ₯ = 4π2, show that there are two
possible values for π and find these values.
ΰΆ±1
5
2ππ₯ + 7 ππ₯ = ππ₯2 + 7π₯ 15
= 25π + 35 β π + 7= 24π + 28
β΄ 24π + 28 = 4π2
π2 β 6π β 7 = 0π + 1 π β 7 = 0π = β1 ππ 7
Remember: π is a constant, so just treat it as a number.
Exercise 13D
Pearson Pure Mathematics Year 1/ASPage 297
(Classes in a rush may want to skip this exercise and go to the next section, which continues definite integration, but in the context of areas under graphs).
Extension
[MAT 2009 1A] The smallest value of
πΌ π = ΰΆ±0
1
π₯2 β π 2 ππ₯
as π varies, is what?
π° π = ΰΆ±π
π
ππ β ππππ + ππ π π
=π
πππ β
π
ππππ +
π
πππ
π
π
=π
πβπ
ππ + ππ
We then use differentiation or completing the square to find that the minimum value
of π
πβ
π
ππ + ππ is
π
ππ.
[MAT 2015 1D] Let
π π₯ = β«01π₯π‘ 2 ππ‘ and π π₯ = β«0
π₯π‘2 ππ‘
Let π΄ > 0. Which of the following statements are true?
A) π π π΄ is always bigger than π π π΄
B) π π π΄ is always bigger than π π π΄
C) They are always equal.
D) π π π΄ is bigger if π΄ < 1, and π π π΄
is bigger if π΄ > 1.
E) π π π΄ is bigger if π΄ < 1, and π π π΄
is bigger if π΄ > 1.
(Official Sln) Evaluating: π π =ππ
πand π π =
ππ
π. Hence
π π π¨ =π¨π
ππand π π π¨ =
π¨π
ππ. Hence answer is (B).
1
2
Areas under curves
Earlier we saw that the definite integral β«πππ(π) π π gives
the area between a positive curve π¦ = π(π₯), the π-axis, and the lines π₯ = π and π₯ = π.(Weβll see why this works in a sec)
π¦
π₯
π΄πππ = πππ π‘ππππ
π π
π¦ = π(π₯)
π΄πππ
Find the area of the finite region between the curve with equation π¦ = 20 β π₯ β π₯2 and the π₯-axis.
π¦
π₯
Factorise in order to find roots:
ππ β π β ππ = πππ + π β ππ = ππ + π π β π = ππ = βπ ππ π = π
Therefore area between curve and π-axis is:
ΰΆ±π
π
ππ β π β ππ π π = πππ βπ
πππ β
π
πππ
βπ
π
= ππ β π βππ
πβ βπππ β
ππ
π+πππ
π
=πππ
π
β5 4
Just for your interestβ¦
Why does integrating a function give you the area under the graph?
Part 1:Youβre already familiar with the idea that gradient is the rate at which a quantity changes, and we consider an infinitesimally small change in the variable.You could consider the gradient as the little bit youβre adding on each time.
Hereβs some practical examples using formulae you covered in your younger years!
ππ If you wanted to consider the rate at which the area of a circle increases with respect to the radius, consider a small change ππ in the radius.The change in area is an infinitely thin ring, which looks like youβve drawn a circumference.So what is this rate?
π΄ = ππ2
β΄ππ΄
ππ= 2ππ
OMG THAT IS THE CIRCUMFERENCE
π
So to draw a shaded circle (which has area!), itβs a bit like repeatedly drawing the circumference of a circle with gradually increasing radius.Since the circumference is what the area is increasing by each time, the circumference is the gradient of the area of the circle, and conversely (by the definition of integration being the opposite of differentiation), the area is the integral of the circumference.
You might be rightly upset that you canβt add a length to an area (only an area to an area, innit).But by considering the infinitely thin width ππ of the circumference youβre drawing, it does have area!
ππ΄
ππ= 2ππ β ππ΄ = 2ππ ππ
i.e. the change in area, ππ΄, is 2ππ Γ ππIf we βroll outβ the added area weβre adding each time, this forms a rectangle, whose length hopefully corresponds to the circumference of the circle:
?
ππ
If the added area is 2ππ ππ and the thickness is ππ, then the
length is 2ππ ππ
ππ= 2ππ as expected.
Since π π₯ = π΄β²(π₯), the area function π΄(π₯) is the integral of π(π₯). Thus:
ΰΆ±π
π
π π₯ ππ₯ = π΄ π₯ ππ = π΄ π β π΄(π)
Part 2:It gets better!
Consider the volume of a sphere:
π =4
3ππ3
ππ
ππ= 4ππ2
OMG THATβS THE SURFACE AREA
This works for a similar reason to before.
π₯
π(π₯)
π΄(π₯)
And the same principle applies to area under a graph. If π΄(π₯)gives the area up to π₯, then what weβre βadding on each timeβ (i.e. the gradient) is sort of like drawing a line of length π(π₯) at the right-most end each time (or more accurately, an infinitely thin rectangle of area π π₯ Γ β). Thus if π(π₯) is the gradient of the area, then conversely the area is the integral of π π₯ .
π₯ + β
This gives us an intuitive sense of why it works, but we need a formal proof:
β
Using the diagram above, the area up to π₯ + β, i.e. π΄(π₯ + β), is approximately the area up to π₯ plus the added rectangular strip:
π΄ π₯ + β β π΄ π₯ + (π π₯ Γ β)Rearranging:
π π₯ βπ΄ π₯ + β β π΄ π₯
βIn the limit, as the rectangle becomes infinitely thin, this becomes an equality:
π π₯ = limββ0
π΄ π₯ + β β π΄ π₯
βBut we know from differentiating by first principles that:
π΄β² π₯ = limββ0
π΄ π₯ + β β π΄ π₯
ββ΄ π π₯ = π΄β²(π₯)
And thus we have proven that the gradient of the area function is indeed π(π₯) (and hence the integral of π(π₯) the area).
But weβre missing one final bit: Why does
β«πππ π₯ ππ₯ give the area between π₯ = π and π₯ = π?
π π
The area between π and π is the area up to π, i.e. π΄(π), with the area up to π, i.e. π΄ π , cut out. This gives π΄ π β π΄ π as expected.Because we obtained π΄(π₯) by an integration, we have a constant of integration +π. But because of the subtraction π΄ π β π΄(π), these constants in each of π΄(π) and π΄ πcancel, which explains why the constant is not present in definite integration.
This is known as the Fundamental Theorem of Calculus.
Exercise 13E
Pearson Pure Mathematics Year 1/ASPages 299-300
Extension
[MAT 2007 1H] Given a function π(π₯), you are told that
ΰΆ±0
1
3π π₯ ππ₯ + ΰΆ±1
2
2π π₯ ππ₯ = 7
ΰΆ±0
2
π π₯ ππ₯ + ΰΆ±1
2
π π₯ ππ₯ = 1
It follows that β«02π(π₯) ππ₯ equals what?
Because the area between π = π and 2 is the sum of the area between 0 and 1, and between 1 and 2, it follows that
β«πππ π π π = β«π
ππ π π π + β«π
ππ π π π. Also
note that β«ππ π π π = πβ«π(π) π π
Letting π = β«πππ π π π and π = β«π
ππ π π π
purely for convenience, then:ππ + ππ = ππ + π + π = π
Solve, π = π, π = βπ, β΄ π + π = π
[MAT 2011 1G] A graph of the function π¦ = π π₯ is sketched on the axes below:
What is the value of β«β11π(π₯2 β 1) ππ₯?
We first need to reflect on what part of the function π weβre actually using. π in the integral varies between -1 and 1, thus ππ β π, i.e. the input of π, varies between -1 and 0.Weβre therefore only using the left half of the graph, and thus π π = π + π
β΄ ΰΆ±βπ
π
π(ππ β π) π π = ΰΆ±βπ
π
ππ β π + π π π
=π
πππ
βπ
π
=π
πβ β
π
π=π
π
12
βNegative Areasβ
Sketch the curve π¦ = π₯(π₯ β 1)(π₯ β 2) (which expands to give π¦ = π₯3 β 3π₯2 + 2π₯).
Now calculate β«02π₯ π₯ β 1 π₯ β 2 ππ₯. Why is this result surprising?
ΰΆ±π
π
ππ β πππ + ππ π π =π
πππ β ππ + ππ
π
π
= π
So the total βareaβ is 0!
π₯
π¦
1 2ππ₯
π(π₯) Integration β« π π₯ ππ₯ is just the sum of areas of infinitely thin rectangles, where the current π¦ value (i.e. π(π₯)) is each height, and the widths are ππ₯.i.e. The area of each is π π₯ Γ ππ₯
The problem is, when π(π₯) is negative, then π π₯ Γ ππ₯ is negative, i.e. a negative area!The result is that the βpositive areaβ from 0 to 1 is cancelled out by the βnegative areaβ from 1 to 2, giving an overall βareaβ of 0.So how do we resolve this?
Fro Note: This explains the ππ₯ in the β«π π₯ ππ₯, which effectively
means βthe sum of the areas of strips, each of area π π₯ Γ ππ₯. So the ππ₯ is not just part of the β« notation, itβs behaving as a physical quantity! (i.e. length)
Example
Find the total area bound between the curve π¦ = π₯(π₯ β 1)(π₯ β 2) and the π₯-axis.
π₯
π¦
1 2
Strategy: Separately find the area between π₯ = 0and 1, and between 1 and 2. Treat any negative areas as positive.
π π β π π β π = ππ β πππ + ππ
ΰΆ±π
π
ππ β πππ + ππ π π =π
πππ β ππ + ππ
π
π
=π
π
ΰΆ±π
π
ππ β πππ + ππ π π =π
πππ β ππ + ππ
π
π
= βπ
π
Treating both as positive:
π¨πππ =π
π+π
π=π
π
πΌ π
Exercise 13F
Pearson Pure Mathematics Year 1/ASPages 301-302
Extension
[MAT 2010 1I] For a positive number π, let
πΌ π = ΰΆ±0
π
4 β 2π₯2ππ₯
Then ππΌ
ππ= 0 when π is what value?
1
Hint: Itβs not actually even possible to
integrate 2π₯2, but we can still sketch the
graph. Reflect on what ππΌ
ππactually means.
π₯
π¦
π
π° π represents the area from π = π up to π.
π¦ = 4 β 2π₯2
π π°
π πrepresents the rate of change of area as π
increases. Thus if π π°
π π= π, the area is not
changing. This must happen at the π-intercept of the graph, because once the curve goes negative, the total area will start to decrease.
π β πππ= π β π = π
The answer is π = π.
[STEP I 2014 Q3]The numbers π and π, where π > π β₯ 0, are such that
ΰΆ±π
π
π₯2 ππ₯ = ΰΆ±π
π
π₯ ππ₯
2
(i) In the case π = 0 and π > 0, find the value of π.(ii) In the case π = 1, show that π satisfies
3π3 β π2 β 7π β 7 = 0Show further, with the help of a sketch, that there is only one (real) value of π that satisfies the equation and that it lies between 2 and 3.
(iii) Show that 3π2 + π2 = 3π2π, where π = π + πand π = π β π, and express π2 in terms of π.
Deduce that 1 < π β π β€4
3
Guidance for this problem on next slide.
2
[STEP I 2014 Q3] The numbers π and π, where π > π β₯ 0, are such that β«πππ₯2 ππ₯ = β«π
ππ₯ ππ₯
2
(i) In the case π = 0 and π > 0, find the value of π.(ii) In the case π = 1, show that π satisfies
3π3 β π2 β 7π β 7 = 0Show further, with the help of a sketch, that there is only one (real) value of π that satisfies the equation and that it lies between 2 and 3.
(iii) Show that 3π2 + π2 = 3π2π, where π = π + π and π = π β π, and express π2 in terms of π.
Deduce that 1 < π β π β€4
3
Guidance for Extension Problem 2
Areas between curves and lines
π¦ = π₯π¦
?How could we find the area between the line and the curve?
Start with the area under π = π π β π up to the point of intersection, then subtract the area of the triangle to βcut it outβ.
Click for Fro-animation >
Determine the area between the lines with equations π¦ = π₯ 4 β π₯ and π¦ = π₯
Find point of intersection:π₯ 4 β π₯ = π₯β΄ π₯ = 0 ππ π₯ = 3
Area under curve:
ΰΆ±0
3
π₯ 4 β π₯ ππ₯ = 2π₯2 β1
3π₯3
0
3
= 9
Area of triangle =1
2Γ 3 Γ 3 =
9
2
β΄ Shaded area = 9 β9
2=
9
2
A Harder One
π
πΆ
π΄ π΅
[Textbook] The diagram shows a sketch of the curve with equation π¦ = π₯ π₯ β 3 and the line with equation π¦ = 2π₯.Find the area of the shaded region ππ΄πΆ.
What areas should we subtract this time?Start with triangle πΆπ©πͺ and subtract the area under the curve π¨πͺ.
First find points of intersection:π π β π = ππ β π = π ππ π = π
When π = π, π = ππ β πͺ π, ππAlso need to find the point π¨:
π π β π = π β π¨(π, π)
β΄ Area of triangle πΆπ©πͺ =π
πΓ π Γ ππ = ππ
Area under π¨πͺ: β«πππ π β π π π = β¦ =
ππ
π
β΄ Shaded area = ππ βππ
π=
ππ
π
Test Your Understanding
Edexcel C2 May 2012 Q5
π΄ 2,8 , π΅ 9,1
571
6or
343
6
Alternative Method:If the top curve has equation π¦ = π π₯and the bottom curve π¦ = π π₯ , the area between them is:
ΰΆ±π
π
π π₯ β π π₯ ππ₯
This means you can integrate a single expression to get the final area, without any adjustment required after.
a
b
Exercise 13G
Pearson Pure Mathematics Year 1/ASPages 304-306
Extension
[MAT 2005 1A] What is the area of the region bounded by the curves π¦ = π₯2 and π¦ = π₯ + 2?
π
π
[MAT 2016 1H] Consider two functionsπ π₯ = π β π₯2
π π₯ = π₯4 β πFor precisely which values of π > 0 is the area of the region bounded by the π₯-axis and the curve π¦ = π π₯ bigger than the area of the region bounded by the π₯-axis and the curve π¦ = π π₯ ?(Your answer should be an inequality in terms of π)
1
2
(Official solution)The area bounded by the π-axis and the curve π = π π , π¨π is equal to
π¨π = ΰΆ±π
π
π(π) π π =π
ππππ
whilst the area bounded by the π-axis and the curve π = π π , π¨π is equal to
π¨π = ΰΆ±βπ π
π π
π π π π =π
ππππ
We require an π such that π¨π > π¨π soπ
ππππ >
π
ππππ
πππππ > πππ
ππ
πππ >
π
πand so the answer is (e).
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