Mathl15

Notes

The Integration

Power Rule

Of

all

the

integration

rules,

this

is the

most used

ffid

,

the

most

important

to understand.

This

rule

takes

the

form

of

Ju'

du:

u'*I/(n*1)

+

C,

for any

power where

rL+-l.In

that case,

The

integral

become

s

t

au

t u= lnlul+C,

a

very special

form,

ln is the natural

log.

The

tricky

part of

the

power de is

making

sure

all of au

is

present before

we integrate. If

it is not

fi6*t

ir; it

entirety,

we need to

make *1d;*t

"nt

to

make sure

we

have it.

However, the only

fupe

of

adjustment

that

can be

made

is

by multiplying

the inside

of the

integral

by

a Constant

tirin

dividing

the

outside

by

the same

constant.

That way,

the

integral

is not

altered and

the

answer

is

not

changed.

This seems

to

be

difficult

for

most students

to

understand.

Any

type

of

variable

adjustment

above

is

not allowed-

Once

the

correct

adjustment

is made

(assuming it can

be

made),

we

can

proceed to the answer

given by

the

rule.

Note that

in the

answer

the

du is absent,

it simply

gets

absorbed

back

into the

expression

In

the

cases

where

a constant

adjustment

can

not be made,

the indefinite

integral can not be

found.

Fortunately,

most integrals

of this

type are

definite

integrals

(i.e.,

limits on

the integral

symbol)

and

can

be

approximated

very nicely

by

our calculator.

(these

are

just

numbers).

In the

case

where

u=r*q

where

a is any

constant,

du:dx,

and no

adjustment

is

necessary

since

dx

is the

correct

du automatically.

We

called

these forms

the simple

cases

of the

power

rule.

Examples:

1)

I<r*-5)'dx.

Since

u:7x-5, du:7dx

(the

differential

of

an expression

is the

derivative

times

the differential

of

the variable

of integration).

So,

we need to

multiply the

inside

by

7 and

divide

the outside

by

7

(or

multiply

the outside

by

ll7).

We

then have,

(ll7)[1tx-5)3 7dx

:{ll7)(7x'l)-

t+

*C,

or

(ll2l)(7x-5)4+C

Note

that

the

du=7dx

is absent

in the answer.

(it

is absorbed back

into

the

problem).

Z)

!lx'dx.

Since

the

u=r,

the du:dx

and no

adjustment

is necessary.

However,

we need to

eliminate

the

3 from

the

inside. So,

simply'move

it to

the outside

(constants

have free

passage).

We

get,

3lr'dx:

(3/6)x6+C

or

(ll2)x6+C.

This

is an example

of

the simply case.

3)

Be awarethat

many

problems are

"ridded

up" so

that the du can

be

present. For example,

take

Jtzx2

+3x+4)s(6+8x)dx.

Note

that u:2x2

+3x+4

and du{4x+3)dx.

By

factoring

out a2 and

rearranging

the

part

in front of

dx,

we have,

2[{z*'

+3x+4)8(4x+3)dx.

The integral is

not

in

perfect

form.

The

answ er

is

(219(2x2

+3x+4)'g+C.

Note that u

is unchanged though out this

process

and

the du

part

disappears