2. Integrals
As we mentioned earlier, calculus revolves around two main concepts, derivatives and
integrals.
1.
Derivatives:
We have seen that derivatives correspond physically to an
instaneous rate of change of one variable y which is related to
another variable x by a formula of the form y = f(x). This is the
y y - y0
limit of average rates of change
=
as the changes approach
x x - x0
zero. Thus we divide the change in y by the change in x and take
the limit as the changes approach zero.
2.
Integrals:
There are two important aspects to integrals. These are called
indefinite integrals and definite integrals.
Indefinite integrals (or antiderivatives): As we have seen, this is the process of
recovering a formula from its derivative. For example, finding an object's
position when one is given the velocity v as a function of time t, i.e.
v = f(t).
Definite integrals: It turns out that one way find how far you have traveled in
certain period of time is to divide the time period into small segments and
sum up the car's velocity during each segment times the length of the time
segment. In the limit as the length of the segments approach zero, you get
the distance traveled. This illustrates the second aspect of integrals,
namely as a limit of sums. This type of integral is called a definite integral.
We turn to this now.
2.1 Sums
Before looking in detail at definite integrals, it will be helpful to first spend some time
discussing sums. In particular, we look at summation notation which is a common way
of indicating certain sums compactly.
Example 1. Consider the sum of the squares of consecutive integers starting at 12 and
ending with 52. Thus we are talking about
(1)
S = 12 + 22 + 32 + 42 + 52
If we want to know the actual value of the sum, then we can compute it out. It is
S = 1 + 4 + 9 + 16 + 25 = 55. However, at the moment we are interested in an alternative
2.1 - 1
way of specifying a sum instead of writing it out as in (1). This is called summation
notation. Using summation notation, the formula (1) would be expressed as
5
S =  j2
(2)
j=1
5
Thus  j2 is simply another way to express the sum 12 + 22 + 32 + 42 + 52. Summation
j=1
notation uses the summation symbol  which is a capital Greek sigma. With the sigma
one needs to specify four things:
(i)
a summation variable
(ii)
a starting value, which must be an integer
(iii)
an ending value, which also must be an integer
(iv)
a formula expressing the general term in the sum in terms of the
summation variable
Their positions relative to the sigma are as follows
ending value

formula
summation variable = starting value
Note that there is always an equal sign between the summation variable and the starting
5
value. In the sum  j2
j=1
(i)
the summation variable = j
(ii)
the starting value = 1
(iii)
the ending value = 5
(iv)
the formula = j2
2.1 - 2
5
Any letter can be used for the summation variable. The sum  j2 could equally well be
j=1
5
written as  k2. So when converting a sum from the familiar form as in (1) to
k=1
summation notation as in (2), you can use any variable as the summation variable.
n
In general, the sum  f(j) denotes
j=m
n
(3)
 f(j) = f(m) + f(m+1) +  + f(n)
j=m
i.e. we evaluate the formula f(j) with every value of j beginning with the starting value
and ending with the ending value and then we add up the values we get after we evaluate
f(j) with these values of j. Thus, going from summation notation to the usual notation for
a sum is straight forward.
7
Example 2. What does the sum
 1 +i i represent?
i=3
One has
7
 1 +i i
3
4
5
6
7
= 1+3 + 1+4 + 1+5 + 1+6 + 1+7
i=3
3
4
5
6
7
= 4 + 5 + 6 + 7 + 8
Another element of choice that one has is whether the value of the summation variable
corresponds to the position of the corresponding term in the sum.
Example 3. Write the sum S = 24 + 25 + 26 + 27 using sigma notation.
7
Comparing with (3) we see that one possibility is S =  2k. In this case the value of k
k=4
does not correspond to the position of the corresponding term in the sum. For example,
the first term is 24, but this corresponds to k = 4. Sometimes we want to write the sum so
that the value of the summation variable corresponds to the position of the term in the
n
sum. In this case the sum will have the form  f(j) for some ending value n and formula
j=1
f(j). The ending value n will be the number of terms in the sum. For example, with the
sum S = 24 + 25 + 26 + 27 one has n = 4 since there are four terms in the sum. There are
two ways to come up with the formula f(j).
2.1 - 3
Method 1. Suppose one already has the sum written in summation notation as with
7
S =  2k, then one can make the change of variables j = k –3, so that k = 4 corresponds to
k=4
7
4
k=4
j=1
j = 1 and k = 7 corresponds to j = 4. Also, k = j + 3. Then S =  2k =  2j+3.
Method 2. Another way to achieve this is to proceed from scratch writing whatever is
changing from term to term in terms of the position of the term. For example, with
S = 24 + 25 + 26 + 27 one has
4
S = 24 + 25 + 26 + 27 = 21+3 + 22+3 + 23+3 + 24+3 =  2j+3
j=1
Usually converting a sum from the usual notation to summation notation is fairly straight
forward since a formula f(j) for the general term will either be supplied or there will at
least be some suggestion of it.
Example 4. Convert the sum S =
sin(x)
1
+
sin(2x)
3
+
sin(3x)
5
sin(nx)
+  + 2n-1 to summation
notation.
This sum illustrates a feature that is present in many sums, namely the number of terms in
the sum is a variable, in this case n. The formula for the general term is suggested by the
n
th
th
formula for the n term, i.e the j term
sin(jx)
is 2j-1 .
Thus S =
.
 sin(jx)
2j-1
j=1
Occasionally one has to come up with a formula that describes all the terms. Furthermore,
the formula must have the property that as you go from one term to the next, you increase
the variable by 1.
Example 5. Convert the sum S = 7 + 11 + 15 +
+ 107 to summation notation.
Note that each term is four more than the previous, so that
S = 7 + (7+4) + (7+8) + + (7+100)
= [7 + (0)(4)] + [7 + (1)(4)] + [7 + (1)(4)] +
25
=  (7 + 4j)
j=0
2.1 - 4
+ [7 + (25)(4)]
Summation formulas. Certain sums have a nice formula for their value. One of the
most famous is the sum of the first n positive integers, i.e.
n
S = 1+2+3+ + n = j
j=1
There is a clever method for finding the value of this sum. We just write the sum in
reverse order and add it to the original sum.
S = 1 + 2
+ 3 + + (n-2) + (n-1) + n
S = n + (n-1) + (n-2) + + 3 + 2 + 1
We add the two lines adding vertically. Each term j in the first line is added to the
corresponding term n-1-j in the second giving
2S = (n+1) + (n+1) + (n+1) +
+ (n+1) + (n+1) + (n+1) = n(n+1)
So
n
(4)
1+2+3+
+ n = j =
j=1
n(n+1)
2
Another well known formula is for the sum of the squares of the first n positive integers.
It is
n
(5)
12 + 22 + 32 + + n2 =  j2 =
j=1
n(n+1)(2n+1)
6
This can be proved in a manner similar to (4), but the details are a bit more intricate.
Yet another formula is for the sum of the powers of some fixed numbers, i.e.
xn+1 - 1
= x = x-1
j=0
n
(6)
2
3
S = 1+x+x +x +
n
+ x
j
To prove this we multiple both sides of S = 1 + x + x2 + x3 + + xn by x and subtract
from S. We get
S – xS = (1 + x + x2 + x3 + + xn) - (x + x2 + x3 + + xn + xn+1)
So
(1-x)S = 1 - xn+1
from which (6) follows.
2.1 - 5
Three other general formulas for sums are
(7)
(8)
(9)
n
n
n
j=n
j=n
j=n
 (aj + bj) =  aj +  bj
n
n
n
j=n
j=n
j=n
 (aj - bj) =  aj -  bj
n
n
j=n
j=n
 (caj) = c  aj
n
(10)
 c = nc
j=n
The last formula just says that if we add c to itself n times we get nc.
n
Example 6. Find S =  3j2 - 4j + 7. Using (7), (8) and (9) we get
j=1
n
n
n
j=1
j=1
j=1
S = 3  j2 - 4  j +  7
Using (4), (5) and (10) we get
n(n+1)(2n+1)
- 4n(n+1)
2
n(n2 - 5n + 8)
=
2
S =
2.1 - 6
+ 7n =
n[(n+1)(2n+1) - 8n - 8 + 14
2