Teach a level maths integration

1. 50: Harder Indefinite50: Harder Indefinite
IntegrationIntegration
© Christine Crisp
““Teach A Level Maths”Teach A Level Maths”
Vol. 1: AS Core ModulesVol. 1: AS Core Modules

2. Indefinite Integration
Module C2
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3. Indefinite Integration
• add 1 to the power
• divide by the new power
• add C
1;
1
1
−≠+
+
=
+
∫ nC
n
x
dxx
n
n
Reminder:
n does not need to be an integer BUT notice that the
rule is for
n
x
It cannot be used directly for terms such as n
x
1

4. Indefinite Integration
e.g.1 Evaluate dx
x
∫ 4
1
=
4
1
x
Solution: Using the law of indices, 4−
x
So,
∫∫
−
= dxxdx
x
4
4
1
C
x
+
−
=
−
3
3
C
x
+−=
−
3
3
This minus sign . . .
. . . makes the
term negative.

5. Indefinite Integration
e.g.1 Evaluate dx
x
∫ 4
1
=
4
1
x
Solution: Using the law of indices, 4−
x
So,
∫∫
−
= dxxdx
x
4
4
1
C
x
+
−
=
−
3
3
C
x
+−=
−
3
3
C
x
+−=
3
3
1But this one . . .
is an index

6. Indefinite Integration
e.g.2 Evaluate
C
x
+
2
3
2
3
We need to simplify this “piled up” fraction.
Multiplying the numerator and denominator by 2 gives
C
x
+=
3
2 2
3
C
x
+×
2
2
2
3
2
3
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
∫ 2
1
=
∫ dxx 2
1
Solution:

7. Indefinite Integration
C
x
+=
2
3
e.g.2 Evaluate
C
x
+
2
3
We need to simplify this “piled up” fraction.
Multiplying the numerator and denominator by 2 gives
C
x
+×
2
2
2
3
2
3
We can get this answer directly by noticing that . . .
dxx
∫ 2
1
=
∫ dxx 2
1
Solution:
2
3
2
3
. . . dividing by a fraction is the same as multiplying
by its reciprocal. ( We “flip” the fraction over ).

8. Indefinite Integration
e.g.3 Evaluate dx
x∫
1
=xSolution: 2
1
x
So,
∫∫ = dx
x
dx
x 2
1
11
Using the law of indices, ∫
−
= dxx 2
1
C
x
+=
2
1
2
1
Cx += 2
1
2

9. Indefinite Integration
e.g.4 Evaluate dx
x
x
∫
+1
Solution: dx
x
x
∫
+1
dx
x
x
∫
+
=
2
1
1
dxx
∫ += 2
1
dx
xx
x
2
1
2
1
1
+=
∫
Write in index form
xSplit up the fraction
Use the 2nd
law of indices:
2
1
2
11
2
1
xx
x
x
==
−
We cannot integrate with x in the denominator.

10. Indefinite Integration
e.g.4 Evaluate dx
x
x
∫
+1
Solution: dx
x
x
∫
+1
dx
x
x
∫
+
=
2
1
1
dxx
∫ += 2
1
Cx +2
1
2
dx
xx
x
2
1
2
1
1
+=
∫
Instead of dividing by ,multiply by2
3
3
2
+=
3
2 2
3
x
Instead of dividing by ,multiply by 22
1
2
1−
x
and
2
1
2
10
2
1
1 −−
== xx
x
The terms are now in the form where we can use
our rule of integration.

11. Indefinite Integration
Solution: dx
x
xy
∫ += 2
2 1
e.g.5 The curve passes through the point
( 1, 0 ) and
)(xfy =
2
2/ 1
)(
x
xxf +=
Find the equation of the curve.
( 1, 0 ) on the curve: C=⇒
3
2
dxxxy
∫
−
+= 22
⇒
C
xx
y +
−
+=
−
13
13
⇒ C
x
x
y +−=
1
3
3
⇒
C+−= 1
3
1
0⇒
So the curve is
3
21
3
3
+−=
x
x
y
It’s important to prepare all the terms
before integrating any of them

12. Indefinite Integration
Evaluate
dxxx )1( +
∫
Exercise
dx
x∫ 3
1
Solution:
dxxdx
x ∫∫
−
= 3
3
1
C
x
+−= 2
2
1
1. 2.
C
x
+
−
=
−
2
2
dxxxdxxx )1()1( 2
1
+=+
∫∫
dxxx
∫ += 2
1
2
3
C
xx
++=
3
2
5
2 2
3
2
5

13. 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
∫
−
+=⇒ 2
1
C
x
xy +
−
+=⇒
−
1
1
C
x
xy +−=⇒
1
( 2, 0 ) on the curve: C+−=⇒
2
1
20 C=−⇒
2
3
So the curve is
2
31
−−=
x
xy
Exercise

14. Indefinite Integration

15. Indefinite Integration
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For most purposes the slides can be printed
as “Handouts” with up to 6 slides per sheet.

16. Indefinite Integration
e.g.1 Evaluate dx
x
∫ 4
1
=
4
1
x
Solution: Using the law of indices, 4−
x
So,
∫∫
−
= dxxdx
x
4
4
1
C
x
+
−
=
−
3
3
C
x
+−=
−
3
3
This minus sign . . .
. . . makes the
term negative.
C
x
+−=
3
3
1
But this one
is an index

17. Indefinite Integration
e.g.2 Evaluate
C
x
+
2
3
2
3
We need to simplify this “piled up” fraction.
Multiplying the numerator and denominator by 2 gives
C
x
+=
3
2 2
3
C
x
+×
2
2
2
3
2
3
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
∫ 2
1
=
∫ dxx 2
1
Solution:

18. Indefinite Integration
e.g.3 Evaluate dx
x∫
1
=xSolution: 2
1
x
So,
∫∫ = dx
x
dx
x 2
1
11
Using the law of indices, ∫
−
= dxx 2
1
C
x
+=
2
1
2
1
Cx += 2
1
2

19. Indefinite Integration
e.g.4 Evaluate dx
x
x
∫
+1
Solution:
dx
x
x
∫
+
=
2
1
1
dx
xx
x
2
1
2
1
1
+=
∫
Write in index form
x
Split up the fraction
We cannot integrate with x in the denominator.
Use the laws of indices: and2
1
2
11
2
1
xx
x
x
==
−
2
1
2
1
1 −
= x
x

20. Indefinite Integration
dxx
∫ += 2
1
Cx +2
1
2+=
3
2 2
3
x
2
1−
x
The terms are now in the form where we can use
our rule of integration.
dx
xx
x
2
1
2
1
1
+∫So,

21. Indefinite Integration
Solution: dx
x
xy
∫ += 2
2 1
e.g.5 The curve passes through the point
( 1, 0 ) and .
)(xfy =
2
2/ 1
)(
x
xxf +=
Find the equation of the curve.
( 1, 0 ) on the curve: C=⇒
3
2
dxxxy
∫
−
+= 22
⇒
C
xx
y +
−
+=
−
13
13
⇒ C
x
x
y +−=
1
3
3
⇒
C+−= 1
3
1
0⇒
So the curve is
3
21
3
3
+−=
x
x
y
It’s important to prepare all the terms
before integrating any of them