25: definite integration © christine crisp “teach a level maths” vol. 1: as core modules
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25: Definite 25: Definite IntegrationIntegration
© Christine Crisp
““Teach A Level Maths”Teach A Level Maths”
Vol. 1: AS Core Vol. 1: AS Core ModulesModules
Definite Integration
The numbers on the integral sign are called the limits of integration
e.g.1
1
2
2 23 dxx is a definite integral
Definite Integration
xx
23
3 3
Find the indefinite integral but omit C
The definite integration results in a value.
Evaluating the Definite Integral
1
1
e.g.1
1
2
2 23 dxx
Definite Integration
xx 23 e.g.1
1
2
2 23 dxx
The definite integration results in a value.
Evaluating the Definite Integral
1
2
Draw square brackets and hang the limits on the end
Definite Integration
The definite integration results in a value.
Evaluating the Definite Integral
Replace x with
• the top limit
• the bottom limit
xx 23 e.g.1
1
2
2 23 dxx1
2
Definite Integration
Evaluating the Definite Integral
Subtract and evaluate
1 1 2 3I
32 2 2 2 12I ( ) ( )
The definite integration results in a value.
xx 23 e.g.1
1
2
2 23 dxx1
2
Definite Integration
Evaluating the Definite Integral
The definite integration results in a value.
So,
1
2
2 23 dxx
xx 23 e.g.1
1
2
2 23 dxx1
2
–12–3
= 15
Definite IntegrationSUMMAR
Y
Find the indefinite integral but omit C
Draw square brackets and hang the limits on the end
Replace x with
• the top limit• the bottom limit
Subtract and evaluate
The method for evaluating the definite integral is:
Definite IntegrationEvaluating the Definite
Integral
e.g. 2 Find
1
1
2 127 dxxx
Solution:
1
1
2 127 dxxx1
1
23
122
7
3
x
xx
Indefinite integral but no C
Definite IntegrationEvaluating the Definite
Integral
e.g. 2 Find
1
1
2 127 dxxx
Solution:
1
1
2 127 dxxx1
1
23
122
7
3
x
xx
1
1 712
3 2I
Substitute for x: top limit minus bottom
limit
3 2
1
1 7 112 1
3 2I
( ) ( )( )
51 68I 5
1 615I Simplify
1
1
2 127 dxxx 1 1 I I5 5 26 6 38 15 24
Definite IntegrationExercise
s
1. Find 2
1
2 143 dxxx
2. Find
2
2
2 326 dxxx
122141212)2(2)2( 23
)2(3)2()2(2)2(3)2(1)2(2 2323
64166416 20)14(6
2
1
23
2
4
3
3
x
xx 2
11
1
2
2
23
32
2
3
6
x
xx2 1
1 1
I2 I1
I2 I–2
Definite Integration
26: Definite Integration and Areas
“Teach A Level Maths”
Vol. 1: AS Core Modules
Definite Integration
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
Definite Integration
. . . 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
Definite Integration
0 1
23 2 xy
Finding an area
So the shaded area equals 3The units are usually unknown in this type of
question
1
0
2 23 dxxSince
1
0
xx 23
A1= 3 A0= 0
Area = A1–A0 = 3
Definite IntegrationSUMMAR
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
Definite Integration
xxy 22 xxy 22
Finding an area
0
1
2 2 dxxxA area
A B
For parts of the curve below the x-axis, the definite integral is negative, sofind area
12
0
B x 2x dx and ignore the sign
Definite Integration
xxy 22
A
Finding an area
0
1
2 2 dxxxA
2
2
3
23 0
1
xx
32
0
10 1
3-1
( ) A ( )A
130 1Area
1
1
3
4Area A
Definite Integration
xxy 22
B
Finding an area
12
0
2B x x dx
2
3 1
03
xx
1
11 0
3 0 AA
2
3Area
2
3 ignore the -ve signArea B
Definite IntegrationSUMMAR
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.
Definite Integration
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
Definite Integration
B
)3)(1( xxy
A
2xy
1.
2.
Solutions: 3
1
3
1
32
3
xdxxA
3
232
3
3
2
23
xx
x
3 2
2B x 4x 3dx
( )2
Area B ignore sign3
32
338
3
)1(
3
)3(
Definite Integration
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
Definite Integration
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 )
3
2
32
1
0
32
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
Definite Integration
22 xxy
xy Instead of finding the 2 areas and then subtracting, we can subtract the functions before doing the integration.
x x dx 1
2
0
Area
Top curve – bottom curve =
Method 2
xxx 222xx
1 0
1 1Area A A 0
2 3
6
1
0 1
x x
12 3
02 3
Definite Integration
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 x
42 x2x
( y = 6 for both points )
Definite Integration
6y
22 xy
Shaded area = area of rectangle – area under curve
Area A A2 2
Area under curve
2
2
32
2
2 23
2
x
xdxx
Shaded area = area under rectangle – area under
curve 1
324 13 2310
13
8 84 4 13
3 3
Definite Integration
6y
22 xy
Area A A2 2
Area
x dx
2
2
2
4
23
8 88 8 10
3 3
Alternatively
Subtracting the functionsTop curve – bottom curve = 6 - (x2 + 2)
= 6 – x2 – 2
= 4 – x2
xx
23
2
43
Definite Integration
2xy
24 xy
,02 yx
Area of the triangle
242 xx
2x 1xor
Substitute in : 2xy31 yx
Area under the curve
1
2
31
2
2
344
x
xdxx 9
3321
(b) ; 2xy24 xy
022 xx0)1)(2( xx
Shaded area = area under curve – area of triangle
29
29
At intersection
Definite Integration
2xy
24 xy
Area 22 - x - xdx
1
2
Top curve – bottom curve = (4 – x2) - (x + 2)
Alternatively
Subtracting the functions
= 4 – x2 – x – 2
= 2 – x2 – x
A1=11/6
A–2= –31/3
Area A A 11 2 24
x xx
13 2
2
23 2
Definite Integration
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.
Definite IntegrationYou 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.
Definite Integration
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.