1545 integration-define

The following slides show one of the 51
presentations that cover the AS
Mathematics core modules C1 and C2.
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AS MathematicsAS Mathematics
26: Definite Integration26: Definite Integration
and Areasand Areas
© Christine Crisp

Definite Integration and Areas
Module C1
AQA Edexcel
OCR
MEI/OCR
Module C2
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0 1
23 2
+= xy
It can be used to find an area bounded, in part, by
a curve
e.g.
∫ +
1
0
2
23 dxx gives the area shaded on the graph
The limits of integration . . .
Definite integration results in a value.
Areas

. . . give the boundaries of
the area.
0 1
23 2
+= xy
a curve
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

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
3=
1
0



 xx 23
+=

SUMMARY
• 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

xxy 22
−=xxy 22
−=
Finding an area
∫−
−=
0
1
2
2 dxxxAarea
A B
∫ −−=
1
0
2
2 dxxxBarea
For parts of the curve below
the x-axis, the definite
integral is negative, so

xxy 22
−=
A
Finding an area
∫−
−=
0
1
2
2 dxxxA








−=
−
2
2
3
23
0
1
xx








−−
−
−








= 2
3
)1(
3
)1(
0






−−−= 1
3
1
1
1
3
4
=Area A⇒

xxy 22
−=
B
Finding an area
∫ −=−
1
0
2
2 dxxxB








−= 2
3
1
0
3
x
x






−





−= 01
3
1
3
2
−=
3
2
=Area B⇒

SUMMARY
 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.

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.
2
xy =
2. The area between the curve ,
the x-axis and the x = 2 and x = 3.
)3)(1( −−= xxy

A
2
xy =
1.
B
)3)(1( −−= xxy
2.
Solutions:
3
1
3
1
3
2
3∫ 





==
x
dxxA
3
2
32
3
3
2
2
3
−=





+−= xx
x
∫ +−=−
3
2
2
34 dxxxB
3
2
=⇒ B
3
2
33
8
3
)1(
3
)3(
=





−





=

Extension
The 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, becomes∫
b
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 =
3
7
=
∫
2
1
2
dyy

2
2 xxy −=
xy =
Harder Areas
e.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.
2
2 xxx −=
Solution: The points of intersection are given by
02
=−⇒ xx 0)1( =−⇒ xx
10 == xx or⇒

2
2 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 )
3
2
3
2
1
0
3
2
1
0
2
=








−=−=
∫
x
xdxxx
Area of the triangle
2
1
)1)(1(
2
1
==
Area under the curve
Required area
6
1
2
1
3
2
=−=⇒
Method 1
0 1

2
2 xxy −=
xy =
Instead of finding the 2
areas and then subtracting,
we can subtract the
functions before doing the
integration.
1
0
321
0
2
32 







−=−=
∫
xx
dxxxArea
We get
Method 2
xxx −− 2
2
2
xx −=






−





−= 0
3
1
2
1
6
1
=
0 1

6=y
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 6=y(a) ; (b) ; 2+= xy2
4 xy −=
Solution:
(a) 622
=+x
42
=x⇒
2±=x⇒
( y = 6 for both points )

6=y
22
+= xy
Shaded area = area of rectangle – area under curve
3
16
4
3
8
4
3
8
=





+
−
−





+=
Area under curve
2
2
32
2
2
2
3
2
−− 







+=+= ∫ x
x
dxx
Shaded area
3
16
24 −=
3
218=

2+= xy
2
4 xy −=
,02 =⇒−= yx
⇒ 2−=x 1=xor
Substitute in :2+= xy
31 =⇒= yx
1
2
31
2
2
3
44
−−






−=−= ∫
x
xdxx 9=
332
1
××=
(b) ; 2+= xy
2
4 xy −=
022
=−+ xx⇒
0)1)(2( =−+ xx⇒
Shaded area = area under curve – area of triangle
2
9
=
2
9
=
2
42 xx −=+⇒

3
xy =
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.
3
xy =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.

You 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.

Using 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.

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.

23 2
+= xy
. . . give the boundaries of
the area.
a curve
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

Harder Areas
e.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.
2
2 xxx −=
Solution: The points of intersection are given by
02
=−⇒ xx 0)1( =−⇒ xx
10 == xx or⇒
2
2 xxy −=
xy =

2
2 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 )
3
2
3
2
1
0
3
2
1
0
2
=








−=−=
∫
x
xdxxx
2
1
)1)(1(
2
1
==
Required area
6
1
2
1
3
2
=−=⇒

1545 integration-define

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