3 Integrals

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Module 3 INTEGRALS:

Integration is an important concept in mathematics and, together with its inverse, differentiation,is one of the two main operations in calculus. Given a function of a real variablexand aninterval [a,b] of the real line, the definite integral.

An integral is an infinite sum of infinitesimal values. An easier thing to say, is to say that aDEFINITE integral represents the area underneath a curve, given by y = f(x).

Module 3.1 Basic Integration: The symbol for integration is a large stretched out S. It may alsocontain a lower or upper bound which signify the area you want to search for. Here is a generalformula for integration:

which means on the interval [a, b], integrate the function f(x). The dx is merely a notation. Itstands for differentaion in practical terms.

So, what exactly is integration doing? Its purpose is to find the area between points a and b of acurve on a graph. This (the integration) will be an approximation of the area, not the exact area.

There might also be something called an indefinite integral as denoted here:

This means a general integral and is not looking for a particular area under a curve. You will seeexamples of each of these as we proceed.

Module 3.2 Solving Indefinite Integration

An indefinite integral is one where you are not looking for a particular area under a curve but an

antiderivative of a given function. Here is the form for an indefinite integral:

The above F(x) means the antiderivative of the function f(x). The C stands for any constantbecause the derivative of F(x) is the original function so the constant term, whatever it is, willdisappear. Here is an example:

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Finding the antiderivative of f(x) will produce:

And now to reverse it, the derivative of F(x) will be:

And no matter what constant we add to the antiderivative, it will go away when we take the

derivative of the antiderivative. All of the below mean the exact same thing for the given functionf(x):

When we take the derivative of the F(x) according to each of the corresponding letters above,we get:

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So to summarize, all an indefinite integral is asking for is to find an anti-derivative to a givenfunction.

In calculus, integration is used for finding the areas under a curve on a given graph. There is alot more to it than that as you will see as these tutorials move forward. One rule of thumb that

you need is to know how to use anti-derivatives. This was reviewed in the previous section ofthis manual. Here is a list of very common derivatives of various functions used in calculus:

It is interesting to note that all trigonometric functions that begin with a "c" above, have anegative derivative. This trick may help you out quite a bit when trying to determine thederivative in an integral.

Module 3.3 Integrals are Summations: The gravitational force between two masses isdescribed by the equation:

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What if there is more than two masses? What if there were many masses named m1, m2, m3,m4,... mn and we wanted to know the force at m1? What equation do we use then? Luckily for us,we can take the vector sum of forces due to the pairs of masses, add them up and we get thetotal force on m1. So we would find the force due to the pairs m1 and m2 (F1), m1 and m3 (F2), m1and m4 (F3), etc., and add them together.

In the equation above, the right hand side can be represented by the symbol: F hich means"the sum of the forces" , the greek letter "sigma", being the mathematical symbol forsummation. So the net force is simply the sum of the individual forces.

So what happens if instead of masses, we have a continuous line of mass instead? We can stilltake the same approach but we quickly run into a serious problem, take a look at the imagebelow:

Lets say we want to find the gravitational force at P due to a line of mass of length L. If we takethe approach above and try to express the total force as a summation of forces from pointmasses on the line, we run into a serious problem, there are an infinite number of point masses.That would imply an infinite net force (if you add them up), which is clearly not correct soobviously this approach won't work. We need a new approach that can sum the forces due to aninfinite number of points on a continuous line of length L. The answer is the definite integral:

.

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So what is a definite integral? A definite integral is the sum of an infinite number of infinitelysmall continuous "elements" over a specified range. In our line example above, dx would be atiny (infinitely small (1/)) line segment with mass of dx (lamda is the mass/distance), f(x)describes how each infinitely small point mass (dm=dx) effects P (the gravitational equationfrom above describes this) and a and b would be the length of the continuous line of point

masses we are summing over, in this case L1 to L2. Notice the integral sign:

Looks like a stretched out S. This is no accident, since the stretched out S is meant to representa continuous sum, just as the greek is meant to represent a discrete sum.

Module 3.4 Area Under The Curve

The value of an integral is often described as "the area under the curve". The curve in thisexpression is the function f(x) being integrated. I said above that the Integral is a summation, sowhy should a summation be equal to the area under a curve? To answer this, lets take a look ata simple function.

For a simple function, say f(x)=3, we would would have the following values:

f(x)=3f(1)=3f(2)=3f(3)=3f(4)=3f(5)=3

etc.

Graphically we can represent this with a straight line like the one below:

So lets say we're interested in summing f(x) over the range x=0 to x=5 above. We see that thevalue of f(x) = 3 for all values of x, so the sum is 3 + 3 + 3 + 3 + 3 = 15. So "area under thecurve" means the same thing as summing since the "height" of the function f(x) is simply it'svalue at a particular point.

Unfortunately, most functions f(x) aren't so easy to add up since they are varying continuously.Take a look at this example:

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The curve in the example above is changing continuously, so for any given x we will have aslightly different f(x). This makes adding the values of the function up very difficult, even for thesmallest of ranges. So we use a clever trick. Since we can add up discrete values, like in ourf(x)=3 function above, why don't we just chop up the curve into discrete values that we can addup? The more we chop it up, the closer we'll get to an accurate sum. So the most accurate sumwould be if we chopped our function up an infinite number of times, which would require infintelysmall cuts. You can see in the example above, using limits as a way of getting to "infinitely smallcuts" that's exactly what Integrals do.

An Example

Take a look at the example below. The relation between Electric Potential and charge is:

V = k Q / r

where V is the Electric Potential, k is 1/40 (the electrostatic constant), Q is charge, and r isdistance from the charge. That's for a point charge, but what if you have a line charge likebelow?

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The trick is to divide the line into little "charge elements" dq and add them up from -a to b. Wecan express dq = dx. The lamda () is simply charge per length, which, when multiplied by a

infinitely small length (dx) will give an infinitely small charge (dq). By summing up all thecontributions of those small charges we can get the total potential due to the line of charges

Module 3.5 Integration as an Inverse Process of Differentiation: Integration is the inverseprocess of differentiation. Instead of differentiating a function, we are given the derivative of a function

and asked to find its primitive, i.e., the original function. Such a process is called integration orantidifferentiation.

Let us consider the following examples:

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For sake of easiness the following table will represent the above equation in terms of:

Module 3.6 Definite Integrals

The formal definition of a definite integral is stated in terms of the limit of a Riemann sum.Riemann sums are covered in the calculus lectures and in the textbook. For simplicity's sake,we will use a more informal definiton for a definite integral. We will introduce the definite integral

defined in terms of area.

Let f(x) be a continuous function on the interval [a,b]. Consider the area bounded by the curve,the x-axis and the lines x=a and x=b. The area of the region that lies above the x-axis should betreated as a positive (+) value, while the area of the region that lies below the x-axis should betreated as a negative (-) value.

The image below illustrates this concept. The positive area, above the x-axis, is shaded greenand labelled "+", while the negative area, below the x-axis, is shaded red and labelled "-".

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The integral of the function f(x) from a to b is equal to the sum of the individual areas boundedby the function, the x-axis and the lines x=a and x=b. This integral is denoted by

where f(x) is called the integrand, a is the lower limit and b is the upper limit. This type ofintegral is called a definite integral. When evaluated, a definite integral results in a real number.It is independent of the choice of sample points (x, f(x)).

Properties of Definite Integrals

The following properties are helpful when calculating definite integrals.

Module 3.7 Properties of Indefinite Integration

There are some useful properties of indefinite integrals:

Property 1: Constants

Property 2: Exponents

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Property 3: Logarithms

Property 4: Exponential Functions

Property 5: Arctan Function

Property 6: Arcsin Function

Properties 3, 5 and 6 will be shown in greater detail when we get to the techniques ofintegration. For now, we will see some general examples of indefinite integration.

Module 3.8 Examples of Indefinite Integration

Here are a few examples of indefinite integration as some relate to the above properties.

Example 1

This clearly shows property 1 above so the final answer is:

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Example 2

This clearly shows property 2 above so the final answer is:

Example 3

We note the function and it's antiderivative below:

We can do some cleaning up in the above antiderivative and get an answer of:

Example 4

We note the function and it's antiderivative below:

Therefore, we get the following result:

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The above example now leads to some integrals that can be found easily just by using basicproperties of antiderivatives. These are listed below:

Example 5

We note the above property 4 and get a final result of:

Recall that the derivative of ex is itself so the antiderivative produces the same result.

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Example 6

Making use of another property above, we get a result of:

Example 7

This example may seem tricky at first but recall that there is still a constant term in theexponent; here it is a -1. So therefore, use the same property as the above and get a result of:

Example 8

I can probably bet you said "how on earth do I solve this one?" Well, just look at one of the aboveproperties for the arctan function. This is actually in the form of an arctan with the value of a

being 3 (3 squared produces 9). You get answer of:

Module 3.9 The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus defines the relationship between the processes ofdifferentiation and integration. That relationship is that differentiation and integration are inverseprocesses.

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The Fundamental Theorem of Calculus :

If f is a continuous function on [a,b], then the function denoted by

is continuous on [a,b], differentiable on (a,b) and g'(x) = f(x).

If f(t) is continuous on [a,b], the function g(x) that's equal tothe the area bounded by the u-axis and the function f(u) andthe lines u=a and u=x will be continuous on [a,b] anddifferentiable on (a,b). Most importantly, when we differentiate

the function g(x), we will find that it is equal to f(x). The graphto the right illustrates the function f(u) and the area g(x).

The Fundamental Theorem of Calculus :

If f is a continuous function on [a,b], then

where F is any antiderivative of f.

If f is continuous on [a,b], the definite integral with integrand f(x) and limits a and b is simplyequal to the value of the anti-derivative F(x) at b minus the value of F at a. This property allowsus to easily solve definite integrals, if we can find the antiderivative function of the integrand.

Parts one and two of the Fundamental Theorem of Calculus can be combined and simplifiedinto one theorem.

Let f be a continuous function on [a,b].

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Module 3.10 Table of Indefinite Integrals

The following tables list the formulas for antidifferentiation. These formulas allow us todetermine the function that results from an indefinite integral. Since the formulas are for themost general indefinite integral, we add a constant C to each one. With these formulas and the

Fundamental Theorem of Calculus, we can evaluate simple definite integrals.

The next table lists indefinite integrals involving trigonometric functions.

Module 3.11 The Total Change Theorem

The total change theorem is an adaptation of the second part of the Fundamental Theorem ofCalculus. The Total Change Theorem states: the integral of a rate of change is equal to thetotal change.

If we know that the function f(x) is the derivative of some function F(x), then the definite integralof f(x) from a to b is equal to the change in the function F(x) from a to b.

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Module 3.12 The Substitution Rule

Suppose that we have an integral such as

With our current knowledge of integration, we can't find the general equation of this indefiniteintegral. There are no antidifferentiation formulas for this type of integral. However, from ourknowledge of differentiation, specifically the chain rule, we know that 4x3 is the derivative of thefunction within the square root, x4 + 7. We must also account for the chain rule when we areperforming integration. To do this, we use the substitution rule.

The Substitution Rule states: if u = g(x) is a differentiable function and f is continuous on therange of g, then

Note: Recall that if u = g(x), then du = g'(x)dx. If we substitute u into the left side of the equationfor g(x) and du for g'(x)dx, then we get the integral on the right side of the equation.

From our previous example, if we let u = (x4+7), then du = 4x3dx. If we substitutite these valuesinto the integral, we get an integral that can be solved using the antidifferentiation formulas.

However, this answer is still in terms of u. We must substitute u = (x4+7) into the resultingfunction, so that it is a function of x, rather than u.

The substitution rule also applies to definite integrals. The Substitution Rule for DefiniteIntegrals states: If f is continuous on the range of u = g(x) and g'(x) is continuous on [a,b], then

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Module 3.13 Integrals of Symmetric Functions

If f(x) is continuous on [-a, a] and f is an even function, then

If f(x) is continuous on [-a, a] and f is an odd function, then

These properties of integrals of symmetric functions are very helpful when solving integrationproblems. Some of the more challenging problems can be solved quite simply by using thisproperty.

Module 3.14 Integration By Parts

Suppose that we have an integral such as

Similar to integrals solved using the substitution method, there are no general equations for thisindefinite integral. However there do not appear to be any clear substitutions that could be madeto simplify this integral. This brings us to an integration technique known as integration byparts, which will call upon our knowledge of the Product Rule for differentiation.

The Product Rule states: If f and g are differentiable functions, then

By taking the indefinite integral of both sides of the equation we have:

and we can rearrange this equation as

To make it easier to remember it is commonly written in the following notation. Let u=f(x) andv=g(x). Then the differentiables are du=f'(x)dx and dv=g'(x)dx, so by the substitution rule, theformula for integration by parts becomes:

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From our previous example, if we let u=x and dv=cosx, then du=dx and v=sinx. If we substitutethese values into the formula we have:

Module 3.15 Trigonometric Integrals

Suppose we have an integral such as

The easy mistake is to simply make the substitution u=sinx, but then du=cosxdx. So in order to

integrate powers of sine we need an extra cosx factor. Similarily, in order to integrate powers ofcosine we need an extra sinx factor. Thus for this example knowing we need an extra sinx factorto integrate powers of cosine we can separate one sine factor and convert the remaining sin4xto an expression involving cosine using the identity sin2x + cos2x = 1.

Now by using our knowledge of substitution we can evaluate the integral by letting u=cosx, thendu=-sinxdx and

Now consider the integral

If we were to use the method from the previous example and separate one cosine factor wewould be left with a factor of cosine of odd degree which isn't easily converted to sine. We mustnow consider the half angle formulas

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Using the half angle formula for cos2x, we have:

Strategy for Evaluating

(a)

If the power of sine is odd (m=2k+1), save one sine factor and use the identity sin

2

x +cos2x = 1 to convert the remaining factors in terms of cosine.

then substitute u=cosx.(b)

If the power of cosine is odd (n=2k+1), save one cosine factor and use the identity sin2x+ cos2x = 1 to convert the remaining factors in terms of sine.

then substitute u=sinx.(c)

If the powers of both sine and cosine are even then use the half angle identities.

In some cases it may be helpful to use the identity

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Now that we have learned strategies for solving integrals with factors of sine and cosine we canuse similar techniques to solve integrals with factors of tangent and secant. Using the identitysec2x = 1 + tan2x we are able to convert even powers of secant to tangent and vice versa. Nowwe will consider two examples to illustrate two common strategies used to solve integrals of theform

Suppose we have an integral such as

Observing that (d/dx)tanx=sec2x we can separate a factor of sec2x and still be left with an evenpower of secant. Using the identity sec2x = 1 + tan2x we can convert the remaining sec2x to anexpression involving tangent. Thus we have:

Then substitute u=tanx to obtain:

Note: Suppose we tried to use the substitution u=secx, then du=secxtanxdx. When we separateout a factor of secxtanx we are left with an odd power of tangent which is not easily converted tosecant.

Consider the integral

Since (d/dx)secx=secxtanx we can separate a factor of secxtanx and still be left with an evenpower of tangent which we can easily convert to an expression involving secant using theidentity sec2x = 1 + tan2x. Thus we have:

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Then substitute u=secx to obtain:

Note: Suppose we tried to use the substitution u=tanx, then du=sec2xdx. When we separate outa factor of sec2x we are left with an odd power of secant which is not easily converted totangent.

Strategy for Evaluating

(a)If the power of secant is even (n=2k, k>2) save a factor of sec2x and use the identitysec2x = 1 + tan2x to express the remaining factors in terms of tanx.

then substitute u=tanx.(b)

If the power of tangent is odd (m=2k+1), save a factor of secxtanx and use the identity

sec2

x = 1 + tan2

x to express the remaining factors in terms of secx.

then substitute u=secx.

Integrals of cotangent and cosecant are very similar to those with tangent and secant.

it is easy to see that integrals of the form can be solved by nearly identical

methods as are integrals of the form .

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Unlike integrals with factors of both tangent and secant, integrals that have factors of onlytangent, or only secant do not have a general strategy for solving. Use of trig identities,substitution and integration by parts are all commonly used to solve such integrals. Forexample,

If we make the substitution u=secx, then du=secxtanxdx, and we are left with the simple integral

Similarily we can use the same technique to solve

Another problem that may be encountered when solving trigonometric integrals are integrals ofthe form

Using the product formulas which are deduced from the addition/subtraction rules we havethe corresponding identities

Module 3.16 Trigonometric Substitution

Sometimes trigonometric substitutions are very effective even when at first it may not be soclear why such a substitution be made. For example, when finding the area of a circle or an

ellipse you may have to find an integral of the form where a>0.

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It is difficult to make a substitution where the new variable is a function of the old one, (forexample, had we made the substitution u = a2 - x2, then du= -2xdx, and we are unable to cancelout the -2x.) So we must consider a change in variables where the old variable is a function ofthe new one. This is where trigonometric identities are put to use. Suppose we change the

variable from x to by making the substitutionx=a sin. Then using the trig identity

we can simplify the integral by eliminating the root sign.

By changing x to a function with a different variable we are essentially using the TheSubstitution Rule in reverse. If x=g(t) then by restricting the boundaries on g we can assurethat g has an inverse function; that is, g is one-to-one. In the example above we would require

to assure has an inverse function.

If we look at the Substitution Rule and replace u with x and x with t, we obtain

This is known as the "inverse substitution".

Module 3.17 Integration of Rational Functions By Partial Fractions

Integration of rational functions by partial fractions is a fairly simple integrating technique usedto simplify one rational function into two or more rational functions which are more easilyintegrated.

Think back to the steps taken when adding or subtracting fractions that do not have the samedenominator. First you find the lowest common multiple of the two denominators and then crossmultiply with the numerators accordingly. eg.

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Well the same process applies when dealing with polynomial fractions. eg.

Now by reversing this process we can simplify a function such as into two fractions

which are more easily integrated.

This process is possible when the function is proper; that is the degree of the numerator is lessthan the degree of the denominator. If the function is improper; that is the degree of thenumerator is greater than or equal to the degree of the denominator, then we must first use longdivision to divide the denominator into the numerator until we obtain a remainder, such that it'sdegree is less than the denominator. Then if possible the above process is used to simplify theproper function.

To complete some of the problems in this section it will be useful to know the table integral

In general there are 4 cases to consider to express a rational function as the sum of two or morepartial fractions.

Case 1The denominator is a product of distinct linear factors (no factor is repeated or a constantmulptiple of another).

For example,

Since the degree of the numerator is less than the degree of the denominator we don't need todivide. The denominator can be factored as follows:

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Since the denominator has distinct linear factors we can write the rational fraction as the sum oftwo or more partial fractions as follows:

By multiplying both sides by we have:

From this equation we can match terms of the same degree to determine the coefficients bysolving the following system of equations:

Case 2

The denominator is a product of linear functions, some of which are repeated.

For example,

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Since the degree of the numerator is greater than the degree of the denominator we mustfactorize by long division.

So we can now factor the denominator to obtain:

Since the linear factor (x-2) occurs twice, the partial fraction decomposition is:

When we multiply both sides by the least common denominator we get:


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Case 3

The denominator contains irreducible quadratic factors, none of which are repeated.

When reducing such functions to partial fractions if there is a term in the denominator of the

form ax2 + bx + c, where b2 - 4ac < 0, then the numerator for that partial fraction will be of theform Ax + B.

For example,

Since the degree of the numerator is less than the degree of the denominator we do not have todivide first.

Since x3 + 4x = x(x2 + 4) can't be factored any further we have:

multiplying both sides by x(x2 + 4), we have:


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Case 4

The denominator contains a repeated irreducible quadtratic factor.

Functions of this form are the same as those in case 3 only there is a term in the denominatorthat is repeated or is a constant multiple of another.

For example,

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If we were to expand the denominator we would see that its degree is greater than the thedegree of the numerator so we do not have to divide first.

Since the function cannot be factored any further we have:

multiplying both sides by (x + 1)(x2 + 4)2, we have:


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Module 3.18 Comparison between Differentiation and integration:

1. Both are operations on functions.

3. We have already seen that all functions are not differentiable. Similarly, all functions are not

integrable. We will learn more about non-differentiable functions and non-integrable functions inhigher classes.

4. The derivative of a function, when it exists, is a unique function. The integral of a function isnot so. However, they are unique up to an additive constant, i.e., any two integrals of a functiondiffer by a constant.

5. When a polynomial function P is differentiated, the result is a polynomial whose degree is 1less than the degree of P. When a polynomial function P is integrated, the result is a polynomialwhose degree is 1 more than that of P.

6. We can speak of the derivative at a point. We never speak of the integral at a point, we speak

of the integral of a function over an interval on which the integral is defined.

7. The derivative of a function has a geometrical meaning, namely, the slope of the tangent tothe corresponding curve at a point. Similarly, the indefinite integral of a function representsgeometrically, a family of curves placed parallel to each other having parallel tangents at thepoints of intersection of the curves of the family with the lines orthogonal (perpendicular) to theaxis representing the variable of integration.

8. The derivative is used for finding some physical quantities like the velocity of a movingparticle, when the distance traversed at any time t is known. Similarly, the integral is used incalculating the distance traversed when the velocity at timet is known.

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Example 2: Find the following intergrals;

3 Integrals

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