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C2 Chapter 11 Integration
Dr J Frost (jfrost@tiffin.kingston.sch.uk)
Last modified: 17th October 2013
Recap
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Definite Integration
π ππ₯
π¦Suppose you wanted to find the area under the curve between and .
πΏπ₯
We could add together the area of individual strips, which we want to make as thin as possibleβ¦
Definite Integration
π₯1 π₯7π₯
π¦
π₯2 π₯3 π₯4 π₯5 π₯6
πΏπ₯
π¦= π (π₯ )
What is the total area between and ?
As β«π
π
π (π₯ )ππ₯
π π
Definite Integration
β«π
π
π (π₯ )ππ₯
You could think of this as βSum the values of between and .β
π¦=sin π₯
π₯
π¦ Reflecting on above, do you think the following definite integrals would be positive or negative or 0?
π 2π
β«0
π2
sin (π₯ )ππ₯ +β
β«0
2 π
sin (π₯ ) ππ₯
0
+β 0
β«π2
2 π
sin (π₯ ) ππ₯ +β 0
Evaluating Definite Integrals
β«π
π
π β² (π₯ )ππ₯=[ π (π₯ ) ]ππ= π (π)β π (π)
β«1
2
3π₯2ππ₯ΒΏ [π₯3 ]12
We use square brackets to say that weβve integrated the function, but weβre yet to involve the limits 1 and 2.
Then we find the difference when we sub in our limits.
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Evaluating Definite Integrals
β«1
2
2π₯3+2π₯ ππ₯ β«β2
β1
4 π₯3+3 π₯2ππ₯?
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Bro Tip: Be careful with your negatives, and use bracketing to avoid errors.
Exercise 11B
Find the area between the curve with equation the -axis and the lines and .
The sketch shows the curve with equation . Find the area of the shaded region (hint: first find the roots).
Find the area of the finite region between the curve with equation and the -axis.
Find the area of the finite region between the curve with equation and the -axis.
1
2
4
6
ace
π
ππππ
πππ
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Harder ExamplesFind the area bounded between the curve with equation and the -axis.
Sketch:(Hint: factorise!)
π₯
π¦
β1 1?
Looking at the sketch, what is and why?0, because the positive and negative region cancel each other out.
What therefore should we do?Find the negative and positive region separately. So total area is
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Harder ExamplesSketch the curve with equation and find the area between the curve and the -axis.
Adding:
-3 1 π₯
π¦
The Sketch The number crunching
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Exercise 11CFind the area of the finite region or regions bounded by the curves and the -axis.
1
2
3
4
5
113
2056
4012
113
21112
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Curves bound between two lines
π₯
πΏπ₯
π¦= π (π₯ )
π π
Remember that meant the sum of all the values between and (by using infinitely thin strips).
Curves bound between two lines
π₯
π¦=π (π₯)
π π
How could we use a similar principle if we were looking for the area bound between two lines?
What is the height of each of these strips?
π¦= π (π₯ )
? therefore areaβ¦
π΄=β«π
π
π (π₯ )β π (π₯ )?
Curves bound between two lines
π₯
π¦
π¦=π₯
π¦=π₯
(4βπ₯
)
Find the area bound between and .
Bro Tip: Always do the function of the top line minus the function of the bottom line. That way the difference in the values is always positive, and you donβt have to worry about negative areas.
β«0
3
π₯ (4βπ₯ )βπ₯ ππ₯=4.5
Bro Tip: Weβll need to find the points at which they intersect.
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Curves bound between two linesEdexcel C2 May 2013 (Retracted)
π₯=β4 ,
Area =
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y = x(
x-3)
y = 2x
A B
C
More complex areas
Bro Tip: Sometimes we can subtract areas from others. e.g. Here we could start with the area of the triangle OBC.
π¨πππ=ππππ?
Exercise 11D
A region is bounded by the line and the curve .a) Find the coordinates of the points of intersection.b) Hence find the area of the finite region bounded by and the curve.
The diagram shows a sketch of part of the curve with equation and the line with equation . The line cuts the curve at the points and . Find the area of the shaded region between and the curve.
Find the area of the finite region bounded by the curve with equation and the line .
The diagram shows part of the curve with equation and the line with equation .a) Verify that the line and the curve cross at .b) Find the area of the finite region bounded by the curve and the line.
1
3
4
9
π΄π΅
4 π΄
623
4.5
7.2
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Exercise 11D(Probably more difficult than youβd see in an exam paper, but you never knowβ¦)
The diagram shows a sketch of part of the curve with equation and the line with equation .
a) Find the area of .b) Find the area of .
Q6
π₯
π¦
7
7
π 1
π 2
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y1
y2
y3
y4
h h h
Trapezium Rule
Instead of infinitely thin rectangular strips, we might use trapeziums to approximate the area under the curve.
What is the area here?
π΄πππ=12h ( π¦1+π¦2 )+1
2h ( π¦2+π¦3 )+ 1
2h ( π¦3+π¦4 )?
Trapezium RuleIn general:
β«π
π
π¦ ππ₯ β h2
(π¦1+2 (π¦ 2+β¦+ π¦πβ1 )+ π¦π )
width of each trapezium
Area under curve
is approximately
x 1 1.5 2 2.5 3
y 1 2.25 4 6.25 9
Weβre approximating the region bounded between , , the x-axis the curve
h=0.5 π΄πππβ 8.75?
Example
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Bro Tip: You can generate table with Casio calcs . . Use βAlphaβ button to key in X within the function. Press =
0.8571?
π¨πππ=π .ππ
(π .ππππ+π (π .ππππ+π .ππππ+π .ππππ+π .ππππ )+π .ππππ )=π .πππ?
Trapezium RuleMay 2013 (Retracted)
To add: When do we underestimate and overestimate?
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