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The following slides show one of the 51 presentations that cover the AS Mathematics core modules C1
and C2.
Demo DiscDemo Disc
““Teach A Level Maths”Teach A Level Maths”Vol. 1: AS Core ModulesVol. 1: AS Core Modules
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AS MathematicsAS Mathematics
26: Definite 26: Definite Integration and AreasIntegration and Areas
© Christine Crisp
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Definite Integration and Areas
Module C1AQA Edexce
l
OCRMEI/OCR
Module C2
"Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages"
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Definite Integration and Areas
0 1
23 2 xy
It can be used to find an area bounded, in part, by a curvee.g.
1
0
2 23 dxx gives the area shaded on the graph
The limits of integration . . .
Definite integration results in a value.
Areas
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Definite Integration and Areas
. . . give the boundaries of the area.
The limits of integration . . .
0 1
23 2 xy
It can be used to find an area bounded, in part, by a curve
Definite integration results in a value.
Areas
x = 0 is the lower limit( the left hand
boundary )x = 1 is the upper limit(the right hand
boundary )
dxx 23 2
0
1
e.g.
gives the area shaded on the graph
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Definite Integration and Areas
0 1
23 2 xy
Finding an area
the shaded area equals 3
The units are usually unknown in this type of question
1
0
2 23 dxxSince
31
0
xx 23
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Definite Integration and Areas SUMMAR
Y
• the curve
),(xfy
• the lines x = a and x = b
• the x-axis and
PROVIDED that the curve lies on, or above, the x-axis between the values x = a and x = b
The definite integral or
gives the area between
b
a
dxxf )( b
a
dxy
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Definite Integration and Areas
xxy 22 xxy 22
Finding an area
0
1
2 2 dxxxA area
A B
1
0
2 2 dxxxB area
For parts of the curve below the x-axis, the definite integral is negative, so
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Definite Integration and Areas
xxy 22
A
Finding an area
0
1
2 2 dxxxA
2
23
23 0
1
xx
2
3)1(
3)1(
0
1
31
1
1
34
Area A
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Definite Integration and Areas
xxy 22
B
Finding an area
1
0
2 2 dxxxB
2
3 1
03
xx
0131
32
32
Area B
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Definite Integration and Areas SUMMAR
Y An area is always positive. The definite integral is positive for areas
above the x-axis but negative for areas below the axis.
To find an area, we need to know whether the curve crosses the x-axis between the boundaries.• For areas above the axis, the definite
integral gives the area.• For areas below the axis, we need to
change the sign of the definite integral to find the area.
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Definite Integration and Areas
Exercise Find the areas described in each
question.1. The area between the curve the x-axis and the lines x = 1 and x = 3.
2xy
2. The area between the curve , the x-axis and the x = 2 and x = 3.
)3)(1( xxy
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Definite Integration and Areas
A
2xy
1.
B
)3)(1( xxy
2.
Solutions: 3
1
3
1
32
3
xdxxA
3232
3
3
2
23
xxx
3
2
2 34 dxxxB
32
B
32
33
83)1(
3)3(
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Definite Integration and Areas
ExtensionThe area bounded by a curve, the y-axis and the lines y = c and y = d is found by switching the xs and ys in the formula.
So,become
sb
a
dxy
d
c
dyx
d
c
dyx
e.g. To find the area between the curve , the y-axis and the lines y = 1 and y = 2, we need
xy
37
2
1
2 dyy
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Definite Integration and Areas
22 xxy
xy
Harder Arease.g.1 Find the coordinates of the points of
intersection of the curve and line shown. Find the area enclosed by the curve and line.
22 xxx
Solution: The points of intersection are given by
02 xx 0)1( xx10 xx or
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Definite Integration and Areas
22 xxy
xy 00 yx xy Substitute
in
11 yx The area required is the area under the curve between 0 and 1 . . . . . . minus the area under the line (a
triangle )
32
32
1
0
32
1
0
2
xxdxxx
Area of the triangle 2
1)1)(1(21
Area under the curve
Required area 6
121
32
Method 1
0 1
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Definite Integration and Areas
22 xxy
xy Instead of finding the 2 areas and then subtracting, we can subtract the functions before doing the integration.
1
0
321
0
232
xxdxxxArea
We get
Method 2
xxx 222xx
0
31
21
61
0 1
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Definite Integration and Areas
6y
22 xy
Exercise Find the points of intersection of the
following curves and lines. Show the graphs in a sketch, shade the region bounded by the graphs and find its area.
22 xy 6y(a) ; (b) ; 2xy24 xy
Solution:
(a) 622 x42 x
2x( y = 6 for both points
)
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Definite Integration and Areas
6y
22 xy
Shaded area = area of rectangle – area under curve
3164
384
38
Area under curve
2
2
32
2
2 23
2
xxdxx
Shaded area3
1624 3218
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Definite Integration and Areas
2xy24 xy
,02 yx
Area of the triangle
2x 1xorSubstitute in : 2xy
31 yx
Area under the curve
1
2
31
2
2
344
xxdxx 9
3321
(b) ; 2xy24 xy
022 xx0)1)(2( xx
Shaded area = area under curve – area of triangle
29
29
242 xx
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Definite Integration and Areas
3xy
The symmetry of the curve means that the integral from 1 to +1 is 0.
If a curve crosses the x-axis between the limits of integration, part of the area will be above the axis and part below.
3xy e.g. between 1 and +1
To find the area, we could integrate from 0 to 1 and, because of the symmetry, double the answer.For a curve which wasn’t symmetrical, we
could find the 2 areas separately and then add.
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Definite Integration and AreasYou don’t need to know how the formula for
area using integration was arrived at, but you do need to know the general ideas.The area under the curve is split into strips.The area of each strip is then approximated by 2 rectangles, one above and one below the curve as shown.
The exact area of the strip under the curve lies between the area of the 2 rectangles.
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Definite Integration and AreasUsing 10 rectangles
below and 10 above to estimate an area below a curve, we have . . .Greater accuracy would be given with 20 rectangles below and above . . .For an exact answer we let the number of rectangles approach infinity. The exact area is “squashed”
between 2 values which approach each other. These values become the definite integral.
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Definite Integration and Areas
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Definite Integration and Areas
The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied.For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.
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Definite Integration and Areas
23 2 xy
. . . give the boundaries of the area.
It can be used to find an area bounded, in part, by a curve
Definite integration results in a value.
Areas
The limits of integration . . .
x = 0 is the lower limit( the left hand
boundary )x = 1 is the upper limit(the right hand
boundary )
dxx 23 2
0
1
e.g.
gives the area shaded on the graph
![Page 27: 1545 integration-define](https://reader036.fdocuments.net/reader036/viewer/2022062903/58a2892c1a28ab891a8b7a89/html5/thumbnails/27.jpg)
Definite Integration and AreasSUMMAR
Y
• the curve
),(xfy
• the lines x = a and x = b
• the x-axis and
PROVIDED that the curve lies on, or above, the x-axis between the values x = a and x = b
The definite integral or
gives the area between
b
a
dxxf )( b
a
dxy
![Page 28: 1545 integration-define](https://reader036.fdocuments.net/reader036/viewer/2022062903/58a2892c1a28ab891a8b7a89/html5/thumbnails/28.jpg)
Definite Integration and Areas
xxy 22 xxy 22
Finding an area
0
1
2 2 dxxxA area
A B
1
0
2 2 dxxxB area
For parts of the curve below the x-axis, the definite integral is negative, so
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Definite Integration and Areas SUMMAR
Y An area is always positive. The definite integral is positive for areas
above the x-axis but negative for areas below the axis.
To find an area, we need to know whether the curve crosses the x-axis between the boundaries.• For areas above the axis, the definite
integral gives the area.• For areas below the axis, we need to
change the sign of the definite integral to find the area.
![Page 30: 1545 integration-define](https://reader036.fdocuments.net/reader036/viewer/2022062903/58a2892c1a28ab891a8b7a89/html5/thumbnails/30.jpg)
Definite Integration and AreasHarder
Arease.g.1 Find the coordinates of the points of intersection of the curve and line shown. Find the area enclosed by the curve and line.
22 xxx
Solution: The points of intersection are given by
02 xx 0)1( xx10 xx or
22 xxy
xy
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Definite Integration and Areas
22 xxy
xy 00 yx xy Substitute
in
11 yx The area required is the area under the curve between 0 and 1 . . . . . . minus the area under the line (a
triangle )
32
32
1
0
32
1
0
2
xxdxxx
Area of the triangle 2
1)1)(1(21
Area under the curve
Required area 6
121
32