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Mathematics Refresher
PART 1
Indices Scientific Notation Engineering Notation Logarithms Radians Exercises
PART 2
Differentiation Integration
PART 1
Indices
The power of a number is indicated by an index, sometimes called an exponent. e.g.
a4
= a x a x a x a
Index Rules
Rule Example
am
.an
= a(m + n)
a3.a
5= a
8
n
m
a
a
= a(m - n)
4
6
a
a
= a2
(am
)n
= am.n
(a4)
2= a
8
a-m =m
a
1 a-6 =
1
a6
a0 = 1 40 = 1
a1/2
= a 91/2
= 9 = 3
a1/3
= 3 a 81/3
= 3 8 = 2
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Scientific Notation
This is also known as standard form. There is one digit before the decimal point and
the remaining significant figures after it. The number is multiplied by a power of 10 to
give it its correct value. e.g.
70 = 7 x 10
700 = 7 x 100 = 7 x 10 x 10 = 7 x 102
7200 = 7.2 x 1000 = 7.2 x 10 x 10 x 10 = 7.2 x 103
Small numbers are represented using negative powers of ten. e.g.
0.006 =
6
1000
=
6
10 x 10 x 10
=
6
103
= 6 x 10-3
0.000083 =
8.3
100000
=
8.3
10 x 10 x 10 x 10 x 10
=
8.3
105
= 8.3 x 10-5
0.000000092 =
9.2
100000000
=
9.2
10 x 10 x 10 x 10 x 10 x 10 x 10 x 10
=
9.2
108
= 9.2 x 10-8
This representation of numbers is known as scientific notation.
Examples
Represent the following numbers in scientific notation
0.000000102 = 1.02 x 10-7 1.045600000000 = 1.0456 x 100 189,000,000,000,000,000,000,000 = 1.89 x 1023 2,300,000 = 2.3 x 106
Engineering Notation
Engineering Notation is similar to scientific notation except the powers are restricted
to multiples of three and symbols are used to represent these powers. i.e.
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Y Yotta 1024
Z Zetta 1021
E Exa 1018
P Peta 1015
T Tera 1012
G Giga 109
M Mega 106
k kilo 103m milli 10
-3
micro 10-6
nano 10-9
p pico 10-12
f femto 10-15
a atto 10-18
z zepto 10-21
y yocto 10-24
v vimto 10-27
Examples
Number Scientific NotationEngineering Notation
45000000 4.5 x 107
45M
80000 8 x 104
80k
189000000000000000000000 1.89 x 1023
189Z
0.00000065 6.5 x 10-7
650
Logarithms
A logarithm is a any positive number which is expressed as a power of some arbitrary
number, which is called the base.
For example. We know that 24
= 16 Instead of using the number 16, we use the index
4, and call it a logarithm (or log) of 16. Provided that we know that we are dealing
with powers of 2 (the base number) we can always turn numbers into logs and logs
into ordinary numbers. Putting this in the form of an equation.
log216 = 4
Note the base number written as a subscript to the symbol log.
Logs to the base e (where e = 2.7182818) are known as natural logarithms. Normally
loge is written as ln e.g.
y = ex ln y = x
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Examples
Index Explanation Log
22 = 4 2 is the log of 4 to base 2 log24 = 2
e2
= 7.389 2 is the log of 7.389 to base e ln 7.389 = 2
102
= 100 2 is the log of 100 to base 10 log10100 = 2
bx
= y x is the log of y to base b logby = x
Log Rules
log (m x n) = log m + log n log (m/n) = log m - log n log mn = n log m log (1/m) = -log m log 1 = 0
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Examples
In each case solve for x
i. 3 = log6xx = 6
3
x = 216
ii. 1.7 = log2xx = 2
1.7
x = 3.249
iii. -2.8 = log9xx = 9
-2.8
x = 0.00213
iv. x = log91x = 0
v. x = logb1x = 0
vi. 3x = 7x.log103 = log107
x =
log107
log103
x =
0.8451
0.4771
x = 1.771
vii. 2 = logx9x
2= 9
x = 3
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viii. z = logb1 + logb(x/y) + logb(x.y) + logb(xy)z = 0 + logb(x) - logb(y) + logb(x) + logb(y) + y.logb(x)
z = 2.logb(x) + y.logb(x)
z = (2 + y)logb(x)
z
(2 + y)
= logb(x)
x = b(z / (2+y))
Using logs with a calculator
Most calculators only have logs to the base 10 and e. To overcome this problem the
following relationship is used.
logbx =
logyx
logyb
Example
Calculate log255
log255 =
log1055
log102
log255 =
1.74036
0.30103
log255 = 5.78136
Example
Calculate log430
log455 =
log1030
log104
log455 =
1.47712
0.60205
log455 = 2.4534
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Radians
An angle of one radian is formed when the length of the arc on the circumference of a
circle is equal to the radius of the circle. i.e.
= 1 rad
The constant is defined as the ratio of the diameter of a circle to its circumference.
i.e.
=c
d
= 3.141592654
where c is the circle's circumference and d is the circle's diameter.
The circumference of the circle can be defined as
c = 2r
Where r is the radius of the circle and c is the circumference
There are therefore 2. radius lengths in circumference of the circle. i.e
c
r
= 2.= 6.283185307
It therefore follows that there are 2. radians in a circle. i.e.
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Converting between radians and degrees
Consider the following two identical angles. The angle (d) in the left circle is
measured in degrees, while the angle (r) in the right circle is measured in radians.
It must hold that the ratio ofd to the complete circle is the same as the ratio ofr to
the complete circle. i.e
d
360
=
r
2
Example
Convert 32o
to radians.
r=32 x 2
360
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r= 0.5585 rad
Example
Convert 2.12 radians to degrees.
d=
2.12 x 360
2d= 121.5
o
Sinusoids
The term sinusoidal is used for waveforms defined by
f(t) = A.sin(.t + )
where A is the Amplitude, is the angular velocity, t is the independent time
variable, is the phase angle.
Exercises
1. IndicesSimplify
i. a5.a4ii. (a2.4)1.82
iii. 4m4.1 x 2.3m8.4iv. a-2.a4v.
15c4
3c5
vi. (3a3)2vii.
(3y2)
4
9y2
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viii. n3 x n-1 x n-2ix. (a4)x. 3 (27b6)
2. Scientific NotationRepresent the following numbers in scientific notation
i. 0.000000000006789ii. 0.1
iii. 4iv. 7800000000000000v. 30.2000000000
vi. 0.006734vii. 8000
viii. 0.00000356ix. 120100000000x. 2000.34
3. Engineering NotationRepresent the following numbers in Engineering Notation
i. 0.1ii. 8.67x103
iii. 0.707x102iv. 1.141x10-7v. 0.000000000006789
vi. 4vii.
1
6.78x105
ix.1
50x10-4
x. 7800000000000000xi. 30.2000000000
4. Logarithmsi. Evaluate.
a. log10 100b. log10 1000c. log10 1000000
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ii. For each of the expressions below, write the equivalent in logarithmicform.
a. 23 = 8b. 54 = 625c. 30 = 1d. 361/2 = 6e. 5-3 = 0.0008f. 25-1/2 = 0.2
iii. For each of the expressions below, write the equivalent in index form.a. log264 = 6
b. log31 = 0c. log55 = 1d. log464 = 3
iv. Simplify.a. log104 + log10 2
b. log1010 - log102c. 2.log103d. 0.5 log1016e.
log103 + 2.log107f. 3.log104 + 2log105 - log1020
v. Solve for xa. logx9 = 2
b. log28 = xc. 4x = 5d. 0.25x = 9
5. Radiansi. Convert the following angles to Radians
a. 50ob. 360oc. 120od. 6790oe. 180o
ii. Convert the following angles to degreesa. 5 Rad
b. /2 Radc. 3. Radd. 42 Rade. 73.89 Rad
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PART 2(a)
Differentiation
An Introduction ToAn Introduction ToAn Introduction ToAn Introduction ToDifferentiationDifferentiationDifferentiationDifferentiation
Reference:MathsDirectMathsDirectMathsDirectMathsDirect
Differentiation is the maths of changing variables. In science, what is reallyimportant, is not what value something has at the moment, but what value it will
have. This is where differentiation comes in. By differentiating a variable, we findthe rate at which it is changing. To begin with, we consider the gradients ofcurves, but soon move on, to apply differentiation to quantities changing withtime.
The Gradient Of A CurveThe Gradient Of A CurveThe Gradient Of A CurveThe Gradient Of A Curve
For a straight line,we defined thegradient as
This gradient wasconstant
If our graph is curved however, thenthe gradient will constantly bechanging.
The curves on the right are the same,but you can see that their gradients atthe two points are very different.
We define the gradient of a curve at a point, as being the gradient of the tangent to
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the curve at that point.
Clearly, this will change, depending on where you are on the curve. In otherwords, the gradient will be a function of x.
There is a special notation for
differentiation
If the curve has anequation connecting y &x
then the gradient iswritten
The most common type of expressionthat you will be asked to differentiate, isof the form
e.g.
There is avery simplerule, thatworks for anyvariation ofthis
=> where a & n are constants
i.e. You multiply by the power and thenreduce the power by one
Below are some very simple examples
=>
=>
=>
There are 2 special cases for this rule:
If the power of x is 1, then you are justleft with the number in front of it.
e.g. =>
=>
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If you have a number, on it's own, thisvanishes when you differentiate
e.g. =>
=>
The power of x does not have to bepositive integer
=>
=>
It is important thatbefore you applythe differentiationformula, you makesure that yourexpression iswritten in the
form:
You may need to re-arrange yourexpression, before you differentiate.
e.g.
Differentiate
First re-arrange
Then differentiate
Differentiate =>
Re-arrange
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Below are a few more examples of this basic formula
Differentiate =>
Differentiate =>
Differentiate =>
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If youdifferentiate
you get
Notice that the power of e does notchange after differentiation. Yousimply multiply by the number infront of the x
Examples
Differentiate =>
Differentiate =>
Differentiate =>
Stationary points:
There are threetypes of gradient a
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curve can have
In this case thecurve is increasing
Alternatively thecurve could bedecreasing
However, the mostinteresting case is.
At this point thecurve is stationarystationarystationarystationary.
This allows you tofind the maximumor minimum valueof a function, sincethese always occurwhere the gradientis 0
Stationary pointscan come in 3varieties
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A Minimum
A Maximum
A Point ofInflection
Below areexamples offinding the 3 typesof stationarypoints.
To find thestationary points,differentiate andsay that the resultmust equal 0.
Solve this equation,to find the value ofx at which thestationary pointoccurs.
Put this value of xinto yyyy.
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Therefore, thecurve has astationary point at
To find the natureof the stationarypoint, you look atthe gradient oneither side of the
point.
Gradient at x = -2
Gradient at x = 0
So the gradient isnegative to the leftof the point andpositive to theright.
This can berepresented by asketch
The point is aminimum.
Differentiate andsay that the resultis 0.
Solve the equationto find x.
Put this value intoy.
So the stationarypoint is at
To find the natureof the point, look atthe gradient oneither side of thepoint
The gradient ispositive to the leftand negative to theright
The point is amaximum
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Differentiate andsay that the resultis 0
Solve to find x andthe correspondingy.
Look at thegradient on eitherside of the point.
The gradient ispositive on bothsides of the point
This is a point ofinflection.
Note that a pointof inflection couldbe negative on bothsides of the point.
The Chain Rule
The function on the rightcould be considered acomposite function
where
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In differentiation we callthis "a function of afunction." The function ycould be written
where u is a function of x.
To differentiate such afunction, w.r.t. x, we usethe chain rulechain rulechain rulechain rule.
the du cancel out.
For example, to
differentiate the caseabove.
we say
where
We now differentiate thetwo expressionsseparately.
Now combine these, usingthe formula
For this example, we caneasily check that this iscorrect.
We can just expand thebrackets and differentiateas normal.
So the two methods agree.
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PRODUCT RULE
To differentiatea function
in the form
You should
use theformula
Example
Differentiate
First define u & v
Now differentiate u & vseparately
Now combine your results
Simplify
ResultResultResultResult =>
Quotient rule
To differentiatea function
in the form
You shoulduse the
formula
Example
Differentiate
First define u & v
Now differentiate u & vseparately
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Now combine yourresults
Simplify
ResultResultResultResult =>
IMPLICIT FUNCTION
So far we have differentiated functions in the form
so that y is expressed as a function of x. This is called explicit differentiation. Wenow want to look at differentiating expressions such as
where y is not given explicitly. This is called implicit differentiationimplicit differentiationimplicit differentiationimplicit differentiation....
To differentiate the above equation, w.r.t. x, you just differentiate each termseparately. This is straightforward for the x2 term and the 4, but how do youdifferentiate y2 w.r.t x?
In fact you use the chain rule. We can say that
Then applying the chain rule
we get
and we can work out that
so we have
We can now differentiate our original equation
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We could in this case, have re-arranged to make y the subject, then differentiated,but this would have been much more complicated, and in any case, there will beoccasions when it is not possible to make y the subject.
Let's consider a second example- Differentiate, w.r.t. x
Using the chain rule
so
The general rule for differentiating a y term w.r.t. x is, Diff the term w.r.t. y andDiff the term w.r.t. y andDiff the term w.r.t. y andDiff the term w.r.t. y andmultiply by dy/dxmultiply by dy/dxmultiply by dy/dxmultiply by dy/dx.
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Parametric differentiation
It is not always easy to write equations in the form
It is sometimes easier to write the relationship between x & y in terms of a thirdvariable, called a parameter. This parameter is usually represented by the letter t.
For example ;
So , when t = 3 ;
To find the gradient of a curve given parametrically, you have two options.
1 Rewrite the equation in cartesian form.
2 Differentiate parametrically, using the chain rule.
The problem with the first method is that equations are generally givenparametrically, specifically because they are hard to write in a cartesian form.
The example above, however is easy to convert, as an example.
Make t the subject of the xequation.
Substitute into the yequation
Differentiate as usual
In general, however, you should not use this method.
Parametric DifferentiationParametric DifferentiationParametric DifferentiationParametric Differentiation
To differentiate equations given parametrically, you should use the chain rule,
That is, you differentiate y and x separately, with respect to the parameter t. Thiscan be rewritten
In the example above
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This is the same answerthat we got before.
Another example
Find the gradient of the curve described by the parametric equations
This is an ellipse
In this case it would be difficult to write in a cartesian form. Differentiatingseparately gives.
Combining these gives
On the next page there is an example of finding the equation of a tangent to acurve given parametrically.
Trigonometric functions
This, and the derivatives of the other main trig functions are standard and may bequoted.
You will find them in the formula book, but you should try to learn them, as thiswill make solving problems muchmuchmuchmuch easier. The main derivatives are given below.
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The chain rule can be applied to all the trigonometric functions. Two examples aregiven below.
Define u
Differentiate u and y separately.
Combine the results and remove u.
Remember that this means all cubed.
Define u.
Differentiate u and y separately.
Combine the results and remove the u.
This is clearly a tedious procedure. The main results from using the chain rule ontrig functions are standard and can therefore be quoted. The main results are givenbelow.
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PART 2(a)
Integration
What Is Integration?What Is Integration?What Is Integration?What Is Integration? ReferenceMathsDirectMathsDirectMathsDirectMathsDirect
Suppose you were told that a function had a gradientgiven by
Could you work out that function?
By referring back to differentiation, you would notice that this gradient was givenby the function
Is this then our function?
Not necessarily! We would get the same gradient for the function
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This process of finding the function from it's gradient, is called IntegrationIntegrationIntegrationIntegration.
This is only the most basic use of integration. Together with differentiation, it isprobably the most important area of maths.
The basic method of integration, is to reverse differentiation.
To differentiate, wemultiplied by the power,then subtracted 1 from thepower
=>
To integrate, we reversethis exactly.
First add 1 to the power,then divide by the newnewnewnewpower.
=>
Notice that there is a constant added onto the function. This is because, as in thecase above, we do not know whether or not a number must be added to ourfunction.
ExampleExampleExampleExample
Find the curve whosegradient is
Add 1 the power an divideby new power.
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SimplifyImportantImportantImportantImportant If you leave offthe c, your answer is wrongand you willwillwillwill lose a mark
Clearly it is not very satisfactory to have these unknowns in our answers.
If we are told a point that the graph passes through, then we can evaluate theconstant.
For example
A curve has a gradient function
and passes through the point
Find the equation of the curve
Add 1 to the power and
divide by new power.Simplify
Substitutein
So the curve is
THE NOTATION OF INTEGRATION
Although integration hasbeen introduced as thereverse ofdifferentiation, it is animportant mathematicaloperation in it's ownright.
As such, it has it's ownspecial notation.
The symbol for
"integrate" isSo to integrate x, youwrite
This reads, "integrate xwith respect to x."
It is essential that youinclude the dx, as thistells you which variableyou want to integrate.The solution to theabove problem thenwould be
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If the expression thatyou are integratingcontains more than oneterm, then you mustbracket the terms, withthe dx outside thebrackets. e.g.
Another example
As with differentiation,it is important that allterms are in the correctform, before youintegrate. This mayinvolve some re-arranging.
DEFINITE INTEGRAL
Imagine a curve,whose gradient isgiven by
To find the differencein y, between x=1 andx=4, you would firstneed to integrate
When x=1
and when x=4
The change in y is,therefore
Notice that theunknown constant hasvanished.
When you find thedifference between anintegral at 2 values of
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x, you can ignore theconstant and are leftwith a number.
This is called DefiniteDefiniteDefiniteDefiniteIntegrationIntegrationIntegrationIntegration.
If you are going toevaluate a definiteintegral, you need tospecify limits. Thatis, you need to give 2values of x tointegrate between.
You give these limits,by writing the valuesat the top and bottomof the integral sign.The example above is
written:
The result of theintegration is writtenin square brackets,
with the limits to theright.
You then substitutethe limits into theintegral, the top limitfirst. Write the 2 partsin curly brackets, tokeep the separate.
Once you have workedthese out, subtract thesecond bracket fromthe first.
If you study thefollowing examples,the process shouldbecome clear.
Integrate 2x+3,between 0 and 5.
Evaluate thefollowing definiteintegral.
The terms are firstwritten in the form
ln1=0
ln 4 = ln 22 = 2 ln 2
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Evaluate thefollowing definiteintegral
Some Standard IntegralsSome Standard IntegralsSome Standard IntegralsSome Standard Integrals
Treating Integration as the reverse of differentiation,leads to some standard results. These results can all bequoted without any proof or derivation.
Below are some simple
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examples of thesestandard integrals.
Example 1Example 1Example 1Example 1
Example 2Example 2Example 2Example 2
Example 3Example 3Example 3Example 3
Example 4Example 4Example 4Example 4
Example 5Example 5Example 5Example 5
Example 6Example 6Example 6Example 6
Integration Using IdentitiesIntegration Using IdentitiesIntegration Using IdentitiesIntegration Using Identities
Sometimes before integrating an expression, you needto re-arrange it. There are two main types of questionlike this.
Trigonometric expressions.
Partial Fractions.
You will often needto usetrigonometricidentities beforeintegrating. Forexample
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For this example,you first need theidentity.
This is now astandard integral
So the solution is.
A second example
First expand thebrackets.
Use an identity.The integral of sec2xcan be quoted.
The integral is
therefore.
To integrate somequotients, you needto split theexpressions intopartial fractions.
For example
which can bewritten as
Each part can beintegratedseparately.
Therefore theintegral is
A second example
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Split the expressioninto partial fractions.
Cross multiply.
Collect terms.
Equate thenumerators.
Form simultaneousequations and solveto find A and B.
Therefore thepartial fractionsare.
The integral cantherefore be written
So the answer is.
Integration By SubstitutionIntegration By SubstitutionIntegration By SubstitutionIntegration By Substitution
How do you integrate
Notice that one part isthe derivative of theother part.
If we make asubstitution,
then,
which re-arranges to
The integral thenbecomes.
which can beintegrated, w.r.t. u,
and then substituteback the xs.
This method ofsubstitution can be
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used for any integralwhere one part is thederivative of the other.
The two main formsare
and
For these cases yousubstitute as follows
Below are someexamples
Integrate
Use the substitution,
which gives.
The integral thenbecomes.
Integrate w.r.t. u
Substitute back thexs.
Integrate
Use the substitution,
which gives.
So the integralbecomes,
Integrate,
and substitute back thexs.
Integrate
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Use the substitution,
which gives
So the integralbecomes.
Integrate,
and replace the xs.
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Integration By PartsIntegration By PartsIntegration By PartsIntegration By Parts
How do you integratea product?
We saw that if one ofthe terms was thederivative of the other,i.e.
then we could use asubstitution
If this is not the case,however, you must usea method called
Integration by partsIntegration by partsIntegration by partsIntegration by parts.
This can be derivedfrom the product rulein differentiation.
Differentiate w.r.t. x.
Now integrate w.r.t. x
The left hand side justreturns to y, which wecan write as uv.
If we now re-arrange,we get.
This is the formula forintegration by parts.Note that one of theterms of the product istaken to be aderivative.
To see how this worksin practice, look at theexample below.
Integrate
First you must decidewhich term is thederivative and whichterm you are going todifferentiate. Ingeneral, differentiatethe x term.
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For the formula, youwill need tointegrate/differentiatethe terms separately.
Now just substitutethe above terms intothe formula.
You get anotherintegral, but this timeit is simple. Integrateit to get the finalanswer.
It may not always beso straight-forward
For example, integrate
Decide which terms tointegrate/differentiate
Substitute the termsinto the formula
We have been leftwith anotherproduct tointegrate. Thepower of x has,however, reducedby 1, and one moreintegration byparts will removeit.
You now need tointegrate
For the secondintegration, again
decide which terms tointegrate/differentiate.
You mustdifferentiate the sameterm as before,
otherwise you willjust return to whereyou started.
Put the above resultsinto the formula.
This time we get asimple integral.
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Now put this resultinto the originalintegral.
to get the finalanswer
One example ofintegration by partsthrows up a particulardifficulty.
Integrate.
In this case there is noobvious choice ofwhich term todifferentiate/integrate.
In fact, it does notmatter which way youdo it.
Using the abovechoices gives
We do not appear tohave gottenanywhere.
You now need topersevere, and performthe second integrationby parts.
You mustmustmustmustdifferentiate/integratethe same terms asbefore.
Again this does notappear to have helped.However, if wecombine the tworesults,
we get
Removing the
brackets, we see thatwe have the sameintegral on bothsides.
Re-arrange, to get
All h i i