Integrals

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Chapter 5
Integrals
5.2 Area
5.3 The Definite Integral
5.4 The Fundamental Theorem of Calculus
5.5 The Substitution Rule
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4.2 Area
Area Problem Find the area of the region S that
lies under the curve yf(x) from a to b.(see
Figure 1)
Figure 1
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Idea for problem solving First approximate the
region S by polygons, and then take the limit of
the area of these polygons.(see the following
example)
Example 1 Find the area under the parabola yx2
from 0 to 1.
Solution We start by dividing the interval 0,
1 into n-subintervals with equal length, and
consider the rectangles whose bases are these
subintervals and whose heights are the values of
the function at the right-hand endpoints.
y
Figure 2
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Then the sum of the area of these rectangles
is
As n increases, Sn becomes a better and better
approximation to the area of the parabolic
segment. Therefore we define the area A to be the
limit of the sums of the areas of these
rectangles, that is,
Applying the idea of Example 1 to the more
general region S of F.1, we introduce the
definition of the area as following

Step 1 Partition--Divide the interval a, b
into n smaller subintervals by choosing partition
points x0 , x1 , x2 ,., xn so that
ax0 lt x1ltx2 ltlt
xnb
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This subdivision is called a partition of
a,b and we denote it by P . Let
denote the length of ith subinterval
xi-1 , xi , and P (the norm of P) denotes
the length of longest subinterval. Thus
Step 2 ApproximationBy the partition above,
the area of S can be approximated by the sum of
areas of n rectangles .
Using the partition P one can divide the
region S into n strips (see F.3). Now, we choose
a number in each subinterval, then each
strip Si can be approximated by a rectangle Ri
(see F.4).
The sum of areas of these rectangles as an
approximation is
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approximated by
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Step 3 Taking limitNotice that the
approximation appears to become better and better
as the strips become thinner and thinner. So we
define the area of the region as the limit value
(if it exists) of the sum of areas of the
approximating rectangles, that is
(1)
Remark 1 It can be shown that if f is
continuous, then the limit (1) does exist.
Remark 2 In step 1, we have no need to divided
the interval a,b into n subintervals with equal
length. But for purposes of calculation, it is
often convenient to take a partition that divides
the interval into n subintervals with equal
length.(This is called a regular partition)
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Example 2 Find the area under the parabola
yx21 from 0 to 2.
Solution Since yx21 is continuous, the limit
(1) must exist for all possible partition P of
the interval a, b as long as P 0. To
simplify things let us take a regular partition.
Then the partition points are x00,
x12/n, x24/n, , xi2i/n, , xn2n/n2 So the
norm of P is
P2/n Let us choose the point to be
the right-hand endpoint
xi2i/n By definition, the area is


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Example 3 Find the area under the cosine curve
from 0 to b, where
Solution We choose a regular partition P so that
Pb/n and we
choose to be the right-hand endpoint of
the ith sub-interval
xiib/n Since P 0 as n ,
the area under the cosine curve from 0 to b is
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/section 5.2 end
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5.3 The Definite Integral
In Chapters 6 and 8 we will see that limit of
form
occurs in a wide variety of situations not only
in mathematics but also in physics, Chemistry,
Biology and Economics. So it is necessary to give
this type of limit a special name and notation.
1. Definition of a Definite Integral
If f is a function defined on a closed
interval a, b, let P be a partition of a, b
with partition points x0 , x1 , x2 ,., xn ,
where ax0 lt x1ltx2 ltlt xnb
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upper limit
integrand
Note 1
integral sign
lower limit
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Note 4 Geometric interpretations
For the special case where f(x)gt0,
the area under the graph of f
from a to b.
In general, a definite integral can be
interpreted as a difference of areas
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Note 5 In the case of agtb and ab, we
extend the definition of as
follows
If agtb, then
If ab, then
Example 1 Express
as an integral on the interval 0, .
Example 2 Evaluate the integral
by interpreting in terms of areas.
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Solution We compute the integral as the
difference of the areas of the two triangles
y
yx-1
A1
o
3
A2
1
x
-1
2. Existence Theorem
Theorem If f is either continuous or monotonic
on a, b, then f is integrable on a, b that
is , the definite integral exists.
Remark 1 If f is discontinuous at some points,
then might exist or it might
not exist. But if f is piecewise continuous, then
f is integrable.
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Remark 2 It can be shown that if f is integrable
on a, b, then f must be a bounded function on
a, b.
3. Integral Formulas under Regular Partition
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Theorem If f is integrable on a, b, then
Example 3 Express as
an integral on the interval 1, 2.

Answer
If the purpose is to find an approximation to
an integral, it is usually better to choose
to be midpoint of the subinterval, which we
denote by . Any Riemann sum is an
approximation to an integral, but if we use
midpoints and a regular partition we get the
following approximation
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Midpoint Rule where and
Using the Midpoint Rule with n5 we can get an
approximation of integral (see
page 277).
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4. Properties of the Integral
Suppose all of the following integrals exist. Then
Example 4 Using the properties above and the
results
to evaluate
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Order properties of the integral Suppose the
following integrals exist and altb. 5. If f(x)gt0
and altxltb, then 6. If f(x)gtg(x) for altxltb, then
7. If mltf(x)ltM for altxltb, then 8.
Proof of Property 7 Since mltf(x)ltM , Property 6
gives
Using Property 1, we obtain
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Example 5 Show that
Solution Notice that
for 1ltxlt4.
Since xx for xgt0, we have
Thus, by Property 7,
Example 6 Proof that
/section 4.3 end
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4.4 The Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus in this
section gives the precise inverse relationship
between the derivative and the integral. It
enables us to compute areas and integrals very
easily without having to compute them as limits
of sums as we did in sections 4.2 and 4.3.
1. Fundamental Theorem
For a continuous function f on a, b, we
define a new function g by
Computing the derivative of g(x) we obtain
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For now let us assume that hgt0. Since f is
continuous on x, xh, the Extreme Value Theorem
says that there are number u and v in x, xh
such that f(u)m and f(v)M, where m and M are
the absolute minimum and maximum values of f on
x, xh.
By Property 7, we have
Since hgt0, we can divide this inequality by h
Combining it with (2) gives
(3)
Inequality (3) can be proved in a similar manner
for the case where hlt0.
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Since f is continuous at x, and u, v lie
between x and xh, we have
We conclude, from (3) and the Squeeze Theorem,
that
(4)
If xa and b, then Equation (4) can be
interpreted as one- side limit. Then Theorem
2.1.8 shows that g is continuous on a, b.
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Note This Theorem can be written in Leibniz
notation as
(5)
Roughly speaking, equation (5) says that we first
integrate f and then differentiate the result,
we get back to the original function f.
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Example 1 Find the derivative of the function
g(x)
Example 2 Find
Example 3 Find
Example 4 Find
The second part of the Fundamental Theorem of
Calculus provides us with a much simpler method
for the evaluation of integrals.
The Fundamental Theorem of Calculus, Part 2 If f
is continuous on a, b, then
where F is any antiderivative of f, that is
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Proof We know from part 1 that g is an
antidervitive of f. If F is any other
antidervitive of f on a, b, then f and g differ
only by a constant (6)
F(x)g(x)C for altxltb. But both f and g are
continuous on a, b and so, by taking limits of
Equation (6)(as x a- and x b ), we see
that it also holds when xa and xb. Noticing
g(a)0 and using Eq.(8) with xa and xb, we
have F(b)-F(a)g(b)C-g(a)Cg(b)-g
(a)
Notation Set F(x) F(b)-F(a). So we have
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Example 5 Evaluate the integral
Example 6 Find the area under the cosine curve
from 0 to b, where
2. The Indefinite Integral
Because of the relation given by the
Fundamental Theorem between antidervitive and
integral, the notation is
traditionally used for an
antidervative of f and called an indefinite
integral. Thus
(7)
Attention A definite integral is a number,
whereas an indefinite integral is a function.
Then the connection between them is given by Part
2 of the Fundamental Theorem, i.e.,
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(7)
3. Table of Indefinite Integrals
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Example 7 Find the general indefinite integral

Example 8 Evaluate
Example 9 Evaluate
Example 10 What is wrong with the following
calculation?
End /section 4.4
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4.5 The Substitution Rule
Because of the Fundamental Theorem of
Calculus , we can integrate a function if we know
an antiderivative. But the anti-differentiation
formulas in Section 4.4 do not suffice to
evaluate integrals of more complicated functions.
It is necessary for us to develop some
techniques to transform a given integral into one
of the forms in the table.
The substitution rule drives from the Chain
Rule
Integrating it, we get
(1)
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If we make the substitution ug(x), then
Eq.(1) becomes
or, writing
, we get
Thus we have proved the following rule
The Substitution Rule for Indefinite Integral
If ug(x) is a differentiable function whose
range is an interval I and f is continuous on I,
then
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Example 6 Evaluate
Example 7 Evaluate
Example 8 Evaluate
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The next theorem uses the Substitution Rule
for Definite Integral to simplify the calculation
of integral of functions that possess symmetry
properties.
Example 9 Evaluate
Example 10 Show that