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50: Harder Indefinite 50: Harder Indefinite IntegrationIntegration
© Christine Crisp
““Teach A Level Maths”Teach A Level Maths”
Vol. 1: AS Core Vol. 1: AS Core ModulesModules
Indefinite Integration
Module C2
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Indefinite Integration
• add 1 to the power• divide by the new power• add C
1;1
1
nCn
xdxx
nn
Reminder:
n does not need to be an integer BUT notice that the rule is for nx
It cannot be used directly for terms such as nx
1
Indefinite Integration
e.g.1 Evaluate dxx
4
1
4
1
x
Solution: Using the law of indices,
4x
So, dxxdxx
44
1
Cx
3
3
Cx
3
3
This minus sign . . . . . . makes the term negative.
Indefinite Integration
e.g.1 Evaluate dxx
4
1
4
1
x
Solution: Using the law of indices,
4x
So, dxxdxx
44
1
Cx
3
3
Cx
3
3
Cx
33
1But this one . . . is an index
Indefinite Integration
e.g.2 Evaluate
Cx
23
23
We need to simplify this “piled up” fraction.Multiplying the numerator and denominator by
2 gives
Cx
3
2 23
Cx
2
2
23
23
We can get this answer directly by noticing that . . . . . . dividing by a fraction is the same as multiplying by its reciprocal. ( We “flip” the fraction over ).
dxx 21
dxx 21
Solution:
Indefinite Integration
Cx
23
e.g.2 Evaluate
Cx
23
We need to simplify this “piled up” fraction.Multiplying the numerator and denominator by
2 gives
Cx
2
2
23
23
We can get this answer directly by noticing that . . .
dxx 21
dxx 21
Solution: 23
2
3
. . . dividing by a fraction is the same as multiplying by its reciprocal. ( We “flip” the fraction over ).
Indefinite Integration
e.g.3 Evaluate dxx
1
xSolution:
21
x
So, dxx
dxx 2
1
11
Using the law of indices,
dxx 21
Cx
21
21
Cx 21
2
Indefinite Integration
e.g.4 Evaluate
dxx
x 1
Solution: dx
x
x 1dx
x
x
21
1
dxx 21
dx
xx
x
21
21
1
Write in index form
xSplit up the fraction
Use the 2nd law of indices:
21
211
21
xx
x
x
We cannot integrate with x in the denominator.
Indefinite Integration
e.g.4 Evaluate
dxx
x 1
Solution: dx
x
x 1dx
x
x
21
1
dxx 21
Cx 21
2
dx
xx
x
21
21
1
Instead of dividing by ,multiply by 2
332
3
2 23
x
Instead of dividing by ,multiply by 22
1
21
x
and 21
210
21
1 xx
x
The terms are now in the form where we can use our rule of integration.
Indefinite Integration
Solution:
dxx
xy 2
2 1
e.g.5 The curve passes through the point
( 1, 0 ) and
)(xfy
22/ 1
)(x
xxf Find the equation of the
curve.
( 1, 0 ) on the curve: C3
2
dxxxy 22
Cxx
y
13
13
Cx
xy
1
3
3
C 13
10
So the curve is 3
21
3
3
x
xy
It’s important to prepare all the terms before integrating any of them
Indefinite Integration
Evaluate
dxxx )1(
Exercise
dxx 3
1
Solution:
dxxdxx 3
3
1
Cx
22
1
1. 2.
Cx
2
2
dxxxdxxx )1()1( 21
dxxx 2
123
Cxx
3
2
5
2 23
25
Indefinite Integration
3. Given that , find the equation of
the curve through the point ( 2, 0 ).
2
2 1
x
x
dx
dy
Solution:
2
2 1
x
x
dx
dy dxxy 21
Cx
xy
1
1
Cx
xy 1
( 2, 0 ) on the curve: C2
120 C
2
3
So the curve is 2
31
xxy
Exercise
Indefinite Integration
Indefinite Integration
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.
Indefinite Integration
e.g.1 Evaluate dxx
4
1
4
1
x
Solution: Using the law of indices,
4x
So, dxxdxx
44
1
Cx
3
3
Cx
3
3
This minus sign . . . . . . makes the term negative.
Cx
33
1
But this one is an index
Indefinite Integration
e.g.2 Evaluate
Cx
23
23
We need to simplify this “piled up” fraction.Multiplying the numerator and denominator by
2 gives
Cx
3
2 23
Cx
2
2
23
23
We can get this answer directly by noticing that . . . . . . dividing by a fraction is the same as multiplying by its reciprocal. ( We “flip” the fraction over ).
dxx 21
dxx 21
Solution:
Indefinite Integration
e.g.3 Evaluate dxx
1
xSolution:
21
x
So, dxx
dxx 2
1
11
Using the law of indices,
dxx 21
Cx
21
21
Cx 21
2
Indefinite Integratione.g.4
Evaluatedx
x
x 1
Solution:
dx
x
x
21
1
dx
xx
x
21
21
1
Write in index form
x
Split up the fraction
We cannot integrate with x in the denominator.
Use the laws of indices: and21
211
21
xx
x
x
21
21
1 x
x
Indefinite Integration
dxx 21
Cx 21
23
2 23
x
21
x
The terms are now in the form where we can use our rule of integration.
dx
xx
x
21
21
1 So,
Indefinite Integration
Solution:
dxx
xy 2
2 1
e.g.5 The curve passes through the point
( 1, 0 ) and .
)(xfy
22/ 1
)(x
xxf Find the equation of the
curve.
( 1, 0 ) on the curve: C3
2
dxxxy 22
Cxx
y
13
13
Cx
xy
1
3
3
C 13
10
So the curve is 3
21
3
3
x
xy
It’s important to prepare all the terms before integrating any of them