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 be
treated as a positive (+) value, while the area of the region that lies below
the x-axis should be treated as a negative (-) value.
The image below illustrates this
concept. The positive area, above the x-axis, is shaded green and labelled
"+", while the negative area, below the x-axis, is shaded red and
labelled "-".
The integral of the function
f(x) from a to b is equal to the sum of the individual areas bounded by 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 of integral 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.
Examples
The Fundamental Theorem of Calculus
defines the relationship between the processes of differentiation and
integration. That relationship is that differentiation and integration are
inverse processes.
The Fundamental Theorem of Calculus : Part 1
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 to the the area bounded by the u-axis and the
function f(u) and the lines u=a and u=x will be continuous on [a,b] and
differentiable on (a,b). Most importantly, when we differentiate the function
g(x), we will find that it is equal to f(x). The graph to the right illustrates
the function f(u) and the area g(x).
The Fundamental Theorem of Calculus : Part 2
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 simply equal to the
value of the antiderivative
F(x) at b minus the value of F at a. This property allows us 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 simplified into one theorem.
The Fundamental Theorem of Calculus
Let f be a continuous function on
[a,b].
An indefinite integral has
the form
When evaluated, an indefinite
integral results in a function (or family of functions). An indefinite integral
of a function f(x) is also known as the antiderivative of f. A function
F is an antiderivative of f on an interval I, if F'(x) = f(x) for all x in I.
This is a strong indication that that the processes of integration and
differentiation are interconnected.
The following tables list the
formulas for antidifferentiation. These formulas allow us to determine the
function that results from an indefinite integral. Since the formulas are for
the most 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.
Note: After finding an indefinite integral, you can always check
to see if your answer is correct. Since integration and differentiation are
inverse processes, you can simply differentiate the function that results from
integration, and see if it is equal to the integrand.
Examples
2 | Find the general indefinite integrals
3 | Evaluate the definite integral
4 | Evaluate the definite integral of the absolute value of a function
3 | Evaluate the definite integral
4 | Evaluate the definite integral of the absolute value of a function
The total change theorem is an
adaptation of the second part of the Fundamental Theorem of Calculus. The Total
Change Theorem states: the integral of a rate of change is equal to the
total change.
If we know that the function f(x) is
the derivative of some function F(x), then the definite integral of f(x) from a
to b is equal to the change in the function F(x) from a to b.
Examples
Suppose that we have an integral
such as
With our current knowledge of
integration, we can't find the general equation of this indefinite integral.
There are no antidifferentiation formulas for this type of integral. However,
from our knowledge of differentiation, specifically the chain rule,
we know that 4x3 is the derivative of the function within the square
root, x4 + 7. We must also account for the chain rule when we are
performing 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 the range 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 equation for 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
values into 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 resulting
function, so that it is a function of x, rather than u.
The substitution rule also applies
to definite integrals. The Substitution Rule for Definite Integrals
states: If f is continuous on the range of u = g(x) and g'(x) is continuous on
[a,b], then
Examples
6 | Find the general indefinite integrals using
the substitution rule
7 | Evaluate the definite integral using the substitution rule
7 | Evaluate the definite integral using the substitution rule
These properties of integrals of
symmetric functions are very helpful when solving integration problems. Some of
the more challenging problems can be solved quite simply by using this
property.
Examples
Suppose that we have an integral
such as
Similar to integrals solved using
the substitution method, there are no general equations for this indefinite
integral. However there do not appear to be any clear substitutions that could
be made to simplify this integral. This brings us to an integration technique
known as integration by parts, which will call upon our knowledge of the
Product Rule for differentiation.
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) and v=g(x). Then the
differentiables are du=f'(x)dx and dv=g'(x)dx, so by the substitution rule, the
formula for integration by parts becomes:
From our previous example, if we let
u=x and dv=cosx, then du=dx and v=sinx. If we substitute these values into the
formula we have:
Note: By choosing u=x we obtain a simpler integral than we
started with. Had we chose u=cosx and dv=x then du=-sinx and v=(1/2)x2
so integration by parts gives:
This equation is correct, but the
integral is more difficult than the one we started with.
When choosing u and dv always try to
choose u=f(x) to be a function that becomes simpler when differentiated (or at
least not more complicated) and to choose dv=g'(x) to be a function that can be
easily integrated to give v.
Examples
9 | Find the general indefinite integral by
integration by parts
10 | Evaluate the definite integral by integration by parts
10 | Evaluate the definite integral by integration by parts
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 of
cosine we need an extra sinx factor. Thus for this example knowing we need an
extra sinx factor to integrate powers of cosine we can separate one sine factor
and convert the remaining sin4x to 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, then du=-sinxdx
and
Now consider the integral
If we were to use the method from
the previous example and separate one cosine factor we would be left with a
factor of cosine of odd degree which isn't easily converted to sine. We must
now consider the half angle formulas
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 sin2x + 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
11 | Find the indefinite trigonometric integral
12 | Using the half angle formulas solve the indefinite trigonometric integral
13 | Find the definite trigonometric integral
Now that we have learned strategies
for solving integrals with factors of sine and cosine we can use similar
techniques to solve integrals with factors of tangent and secant. Using the
identity sec2x = 1 + tan2x we are able to convert
even powers of secant to tangent and vice versa. Now we will consider two
examples to illustrate two common strategies used to solve integrals of the
form
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 even
power of secant. Using the identity sec2x = 1 + tan2x
we can convert the remaining sec2x to an expression 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 separate out a factor of secxtanx we are left with an
odd power of tangent which is not easily converted to secant.
Consider the integral
Since (d/dx)secx=secxtanx we can
separate a factor of secxtanx and still be left with an even power of tangent
which we can easily convert to an expression involving secant using the
identity sec2x = 1 + tan2x. Thus we have:
Then substitute u=secx to obtain:
Note: Suppose we tried to use the substitution u=tanx, then
du=sec2xdx. When we separate out a factor of sec2x we are
left with an odd power of secant which is not easily converted to tangent.
Strategy for Evaluating
(a)
If the power of secant is even (n=2k, k>2) save a
factor of sec2x and use the identity sec2x = 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 sec2x = 1 + tan2x to
express the remaining factors in terms of secx.
then substitute u=secx.
Note: If the power of secant is even and the power of tangent is
odd then either method will suffice, although there may be less work involved
to use method (a) if the power of secant is smaller, and method (b) if the
power of tangent is smaller.
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 .
can
be solved by nearly identical methods as are integrals of the form .
Unlike integrals with factors of
both tangent and secant, integrals that have factors of only tangent, 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. For example,
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
17 | Find the definite trigonometric integral
18 | Find the definite trigonometric integral
19 | Find the indefinite trigonometric integral
Another problem that may be
encountered when solving trigonometric integrals are integrals of the form
Using the product formulas
which are deduced from the addition/subtraction rules we have the corresponding identities
Sometimes trigonometric
substitutions are very effective even when at first it may not be so clear 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.
where
a>0.
It is difficult to make a
substitution where the new variable is a function of the old one, (for example,
had we made the substitution u = a2 - x2, then du= -2xdx,
and we are unable to cancel out the -2x.) So we must consider a change in
variables where the old variable is a function of the new one. This is where
trigonometric identities are put to use. Suppose we change the variable from x
to 
by
making the substitution x = a sin θ. Then using the trig
identity 
we
can simplify the integral by eliminating the root sign.
by
making the substitution x = a sin θ. 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 The Substitution Rule in reverse. If x=g(t) then by restricting the boundaries on
g we can assure that 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.
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".
Integration of rational functions by
partial fractions is a fairly simple integrating technique used to simplify one
rational function into two or more rational functions which are more easily
integrated.
Think back to the steps taken when
adding or subtracting fractions that do not have the same denominator. First
you find the lowest common multiple of the two denominators and then cross
multiply with the numerators accordingly. eg.
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.
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 less than the degree
of the denominator. If the function is improper; that is the degree of the numerator
is greater than or equal to the degree of the denominator, then we must first
use long division to divide the denominator into the numerator until we obtain
a remainder, such that it's degree is less than the denominator. Then if
possible the above process is used to simplify the proper 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 more partial fractions.
Case 1
The denominator is a product of distinct linear factors (no factor is repeated or a constant mulptiple of another).
The denominator is a product of distinct linear factors (no factor is repeated or a constant mulptiple of another).
For example, 
Since the degree of the numerator is
less than the degree of the denominator we don't need to divide. The denominator
can be factored as follows:
Since the denominator has distinct
linear factors we can write the rational fraction as the sum of two or more
partial fractions as follows:
By multiplying both sides by 
we
have:
we
have:
From this equation we can match
terms of the same degree to determine the coefficients by solving the following
system of equations:
Case 2
The denominator is a product of
linear functions, some of which are repeated.
For example, 
Since the degree of the numerator is
greater than the degree of the denominator we must factorize 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:
From this equation we can match
terms of the same degree to determine the coefficients by solving the following
system of equations:
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 the form Ax + B.
For example, 
Since the degree of the numerator is
less than the degree of the denominator we do not have to divide first.
Since x3 + 4x = x(x2
+ 4) can't be factored any further we have:
multiplying both sides by x(x2
+ 4), we have:
From this equation we can match
terms of the same degree to determine the coefficients by solving the following
system of equations:
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 denominator that is repeated or is
a constant multiple of another.
For example, 
If we were to expand the denominator
we would see that its degree is greater than the the degree 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:
From this equation we can match
terms of the same degree to determine the coefficients by solving the following
system of equations:
In all of the previous tutorials we
have dealt with integrals with a continous function f on a finite
interval [a,b]. In this section we will consider two types of integrals
known as improper integrals. The first type of improper integral are those
defined on an infinite interval, and the second are those where the function f
has an infinite discontinuity in [a,b].
Type 1: Infinite Intervals
Type 2: Discontinous Integrands
For more practice with the concepts
covered in this tutorial, visit the Integral Problems page at the link below.
The solutions to the problems will be posted after these chapters are covered
in your calculus course.
To test your knowledge of integration problems, try taking the general integrals test on the iLrn website or the advanced integrals test at the link below.
To test your knowledge of integration problems, try taking the general integrals test on the iLrn website or the advanced integrals test at the link below.
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