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Mathematics for NST Part IA Cambridge University, 2012-2013

Notes for Integration Riemann sums and The Fundamental Theorem of Calculus: Definite integration is defined in terms of Riemann sums, and is extended to indefinite integration by means of the Fundamental Theorem of Calculus. Definite integrals: The definite integral of

f (x) is an expression of the cumulative behaviour of f ( x) with

x (the area under the curve f ( x) ). Thus, the formal definition of a definite integral can be derived by considering the area bounded by the lines

y=0 :

y

x=a , x=b and

If this area is separated into rectangles of uniform width

f ( x)

centred at

dx

xi , where

i [?][1, N ] , and with heights

f (xi )

f ( x i ) , i [?][1, N ]

x a

x i-2 x i-1 x i

dx

under the curve is approximated by the Riemann sum:

x i+1 x i+2 b

If the Riemann sum has a finite limit as

N-[?]

, then the area

and thus

dx - 0 , its limit as N - [?]

represents the area under the curve, and is equivalent to the Riemann integral: b

[?] f (x)dx= lim a

N

[?] f (x i) dx

N - [?] i=1

Indefinite integrals: By the Fundamental Theorem of Calculus, an indefinite integral

F( x ) is defined by the

reverse process of differentiation: x

Let

x+dx

F ( x )=[?] f (x)dx , thus a

F ( x+ dx )= [?] f ( x)dx

x a

x+dx

[?] F ( x +dx )=[?] f ( x) dx+ [?] f ( x) dx a

Where

x a can take any value. x+ dx

Since

Sarah Jeffreson

dx

is infinitesimally small,

[?]

x f (x) dx [?] dxf (x ) :

Mathematics for NST Part IA Cambridge University, 2012-2013

[?] F ( x +dx )=F ( x )+ dxf ( x)
[?] f ( x )=

F ( x +dx )-F(x )
dx

[?] lim f ( x)=f ( x )= lim
dx - 0

dF d
[?]
=f ( x ) [?]
dx dx

Sarah Jeffreson

[[?] ]

x f ( x ) dx =f ( x)

dx - 0

F ( x+ dx )-F ( x) dF
=
dx dx

Mathematics for NST Part IA Cambridge University, 2012-2013 Evaluation of the definite integral: Let: x

[?] f (x)dx=F (x) 0

It can be seen that: b

a

[?] f (x)dx=F ( b ) ,[?] f (x) dx=F (a) 0

0 Therefore: b

b

[?] f (x)dx=F ( b )-F ( a )=[ F ( x ) ]a a

The average of a function: A useful property of the definite integral of a function is that, as it gives the total cumulative value of the function (the area under the function) between two boundaries, it can be used to find the function's average value. The average value is equal to the total cumulative value between the two boundaries, divided by the range of

x -values over which it has accumulated:

b

1 y =
[?] y dx b-a a

Integration techniques: Evaluating infinite and improper integrals: If the range of integration for an integrand is infinite, or the integrand has singular points, the formal definition of an integral given by the area under the curve

f (x) becomes

problematic. Such integrals can be evaluated using limits, as follows: Infinite range of integration: The integrand is calculated over an infinite range:
[?]

b

[?] f (x)dx [?] lim [?] f ( x) dx a

b-[?] a

b

b

[?] f (x )dx [?] lim
-[?]

[?] f (x )dx

a--[?] a

If these limits do not exist, the integral is undefined. Improper integrals: The integrand is singular within the range of integration: Exclude the small region around the singularity and take the limit as the size of this region approaches zero:

Sarah Jeffreson

Mathematics for NST Part IA Cambridge University, 2012-2013 1

1 1 1 dx =lim [ 2 [?] x ] [?] =lim ( 2-2 [?] [?] ) =2
[?] [?]1x dx [?] lim
[?]
[?]-0 [?] [?]x
[?]-0
[?]-0 0

If these limits do not exist, the integral is undefined.

Sarah Jeffreson

Mathematics for NST Part IA Cambridge University, 2012-2013 Discontinuous integrands: The integrand contains a finite number of discontinuities within the range of integration. Break the range up into regions over which the integrand is continuous, and sum these regions, as follows:

1, x< 0 2, x> 0 f ( x )=

1 0

1 0

1 [?] [?] f (x) dx=[?] 1 dx +[?] 2 dx=[ x ]-1 + [ 2x ] 0=1+2=3
-1

0 -1

Universal substitutions for integral evaluation: Universal substitutions are substitutions which can be used to evaluate any integral of a certain form - either trigonometric or hyperbolic:

*

[?] tan x=

*

t=tan

Universal trigonometric substitution:

( x2 )

:

2t 2t 1-t 2 2 , sin x=
,cos x
=
, dx=
dt 2 2 2 1-t 1+ t 1+t 1+t 2

Universal hyperbolic substitution:

[?] sinh x=

t=e x :

t 2 -1 t 2+ 1 dt , cosh x=
, dx=
2t 2t t

Common substitutions for integral evaluation: Trigonometric substitutions: Based on the identity

[?] sin M x cos N x dx ,

where

M

is an

odd number:

[?] sin M x cos N x dx , Substitution:

u=cos x , then

Reasoning: If

du du
=-sin x [?] dx =
dx
-sin x . The 'extra', odd power of

sin x is therefore negated,

leaving an expression with an even power of sine, which can then be converted, via the identity

2 2

odd number:

Substitution: u=cos x Reasoning: If

2 cos x+ sin x =1 :

2 sin x+cos x =1 , into a power of

where

N

is an

u=sin x u=sin x , then

du du
=cos x [?] dx=
dx cos x . The 'extra', odd power of

cos x

is therefore negated,

leaving an expression with an even power of cosine, which can then be converted, via the identity

2 2

sin x+cos x =1 , into a power of

cosine, which can then be expressed in

sine, which can then be expressed in terms

terms of

of

u . This substitution therefore

generates a simple polynomial of

Sarah Jeffreson

u .

u . This substitution therefore generates

a simple polynomial of

u .

Mathematics for NST Part IA Cambridge University, 2012-2013

Integrands involving the expression Substitution: Reasoning: If

x=a sin th or

[?] a2-x 2

a is a constant:

, where

x=a cos th

x=a sin th , then ( a2 -x2 ) =a2 ( 1-sin 2 th ) =a2 cos2 th , and then as

dx=a cos th dth , the resulting expression will be in terms of cosine only, which can then be solved using trigonometric identities. The same reasoning applies for Integrands involving the expression 2

Integrands involving the expression

2 2 2 ( x -a ) , where a is a constant:

(a + x ) , where a is a constant: Substitution: Reasoning: If

x=a tanth

Reasoning: If

( a2 + x 2 )=a2 ( 1+tan 2 th )=a2 sec2 th
then as

2 2

sec th

expression will be in terms of

only, which can then be solved using trigonometric identities.

Substitution: Reasoning: If

expression will be in terms of an odd power of tangent and a multiple of secant, which can be easily manipulated via a further substitution, as

:

d ( tanth )=sec 2 th . dth

x=a sinh th , then

Reasoning: If , and then

a cosh th

expression will be a function of only.

[?] x2-a2

:

x=a cosh th
x=a cosh th , then

[?] x2-a2=a [?]cosh 2 th-1=a sinh th
then as

, and

dx=asinh th dth , the resulting

expression will be a function of

a sinh th

only.

Integrands involving Substitution: Reasoning: If 2

2 Substitution:

dx=a cosh th dth , the resulting

as

2 cosh x-sinh x=1 :

Integrands involving

x=a sinh th

[?] x2 +a 2=a [?] 1+ sinh2 th=a cosh th

, and then

dx=-tanth secth dth , the resulting

as

Hyperbolic substitutions: Based on the identity

[?] x2 +a 2

x=a secth , then

( x 2-a2 ) =a2 ( sec 2 th-1 )=a 2 tan2 th

, and

dx=sec th dth , the resulting

Integrands involving

x=a secth

Substitution:

x=a tan th , then

x=a cos th .

2 2

2 a -x ,|x|a :

x=a coth th

x=a coth th , then

a2-x 2=a2 ( 1-coth 2 th )=-a csc 2 h th , and

Mathematics for NST Part IA Cambridge University, 2012-2013 2

a sech th

expression will be a function of

then as

only.

2 dx=-a csch th dth , the resulting

expression will be a function of only. a coth th

a tanhth

for

negative value,

2 a csch th

must be used instead of

|x|>a , as it gives a 2

-a csc h th , which is in

keeping with the negative value of

2 a -x

2 Integration by completing the square: Completing the square can be used to reduce a quadratic expression of the form

Q ( x )=a x2 +bx +c

Q ( x )=a x 2 +- b , such that a trigonometric or hyperbolic

to the form

substitution can then be used to evaluate the integral. The technique is typically useful for integrands of the form

[?]

1 dx=[?]
[?] x + x +1

1 Q (x)

or

1 [?](

2 2

x+

1 3
+
2 4

)

Let

[?][?]

1 [?]

u 2+

arcsinh

3 4

du=[?]

1 1
[?]Q ( x ) , as follows:

dx=[?]

1 [?]

2 u+

3 4

du ,u=x+

1 2

3 du 3 u= [?] sinh th [?] = [?] cosh th : 2 dth 2

[?]3 cosh th dth= 1 dth=th+c
[?]

[?] 3 cosh th 2 2

2u
+c
[?]3

( 12 ) + c

2 x+
arcsinh

[?]3

Integration by parts: The derivation of the expansion used in the method of integration by parts relies on the Product Rule and The Fundamental Theorem of Calculus, as follows:

d du dv ( uv )= v+ u dx dx dx
[?]u

dv d du dv d du
= (uv ) -v
[?] [?] u dx=[?] ( uv ) dx-[?] v dx dx dx dx dx dx dx

Sarah Jeffreson

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