Calc 1
Fri 11/10/08
6.4 - Defn of Area; Sigma Notation
When I explained section 6.1, I breie
y talked about approximating area under curves using
rectangles. The problem is, if we want a lot of rectangles to get the area a pretty good
approximation, we need to add up a whole bunch of stu▯ and it got really tedious to write + over
and over. Well, rather than writing something plus something plus something a whole bunch of
times, mathematicians came up with a shortcut. This is capital sigma: ▯. It means sum. ie, if i
were to say
1 + 2 + 3 + 4 + 5 + ▯▯▯ + 10
this is the same as expressing it in terms of ▯:
X10
k
k=1
Have you ever taken a computer science or a programming class before? This works the same way
as a loop does. It says start at k = 1 and keep incrementing k by one until you get to 10. Oh, and
by the way, add up all the k’s while you’re doing that. Notice this does not actually solve the
adding problem. It merely rewrites it so that it’s less to write (imagine if i were to ask you to add
up all the numbers from 1 to 100! Writing it down in this ▯ notation is much easier than writing
1 + 2 + ▯▯▯ + 100!). We shall get to tricks involving what to do about actually adding the numbers
in a bit. For now, we’re just interested in the actual notation.
We’re also not limited to just adding the numbers between 1 and 10 either. We can do other
things. Say i wanted to add all the squares from 1 to 100. This formula would change slightly to
X0
k2
k=1
It expands out to 1 + 4 + 9 + 16 + ▯▯▯ + 81 + 100.
▯ by itself just means add stu▯ up. The k = 1 on the bottom says we’re dealing with the variable
k here, and it’s starting at 1. i just as well could have started at 3 if i wanted to. Finally, the 10
on top means when i have to stop adding. When i reach k = 10, stop. That’s the last one. to the
right of ▯ there can be any function of variable k (say, f(k)). You don’t have to have k in that
function, of course.
5
X
1 = 1 + 1 + 1 + 1 + 1 = 5
k=1
That just sums up the number 1 ▯ve times.
other examples:
X6
2k = 2(3) + 2(4) + 2(5) + 2(6)
k=3
X0
k = (▯4) + (▯3) + (▯2) + (▯1) + (0) 3
k=▯4
1 You can also change the limits of these sums. To do this, make sure that when you plug in that
▯rst number you get what you’re supposed to get.
Example: we have 1 + 4 + 9 + 16 + 25. We can express this as
5
X
k2
k=1
or perhaps
4
X 2
(k + 1)
k=0
Notice that when we plug in k = 0, we get 1 ; k = 1 gives 2 ; up to k = 4 which gives 5 = 25.
Now, sums have a few nice properties that i think i should mention. Namely, if you’re adding a
bunch of stu▯ together that has a common factor, you can take out that factor. ie,
n n
X X
cak= c ak
k=1 k=1
where k is our function (i might switch it o▯ betweek f(k) and a , but they’re essentially the
same thing) and c is just a constant.
n n n
X X X
(k + k ) = ak+ bk
k=1 k=1 k=1
n n n
X X X
(k ▯ k ) = ak▯ bk
k=1 k=1 k=1
Now that we got some of those properties out of the way, here are three summations that you
should become familiar with. In fact, you should become so familiar with them that you should
memorize them!
n
X n(n + 1)
k = 1 + 2 + ▯▯▯ + n =
k=1 2
Xn
k = 1 + 2 + 3 + ▯▯▯ + n =n(n + 1)(2n + 1)
6
k=1
▯ ▯
Xn n(n + 1)2
k = 1 + 2 + ▯▯▯ + n =
2
k=1
These play a crucial role in ▯nding areas because you don’t actually have to add a whole bunch of
stu▯ up! You can quickly multiply a couple numbers together and voila, you’re already done!
2 Okay, so now let’s use this new knowledge and apply it to the area stu▯. As you may recall, when
an interval [a;b] is split into n equal subintervals, we found that the width of each subinterval was
b▯a. The b ▯ a part is the total length of the interval, and we want to divide that entire length up
n
n times. Well, to make things a little more convenient (and more confusing, because what would
calculus be without the confusion?), we’re just going to call that ▯x.
We can of course evaluate the rectangle’s height at any point within each interval we want. These
points that will determine each rectangle’s height are going t1 be2x ;x n:::;x . This means that
▯ ▯
the heights themselves are f1x );:::;fnx ).
▯
So, the ▯rst rectangle would be of width ▯x and have height 1(x ), while the second rectangle has
width ▯x and height f(x ), and so on. This gives us the total area of all the rectangles to be
▯ ▯ 2 ▯
▯xf(x 1 + ▯xf(x )2+ ▯▯▯ + ▯xf(x )n Remember, we’re adding the areas of all the rectangles
together. This shortens to
n
X ▯
A ▯ f(xk)▯x
k=1
As I mentioned before, the more rectangles you do, the better approximation you have. So, take
the limit of the number of subintervals (ie, the number of rectangles!), and you will get not an
approximation, but the exact area.
X n
A = lim f(xk)▯x
n!1
k=1
Also, in your homework, you’ll be approximating areas using right endpoints, left endpoints, and
midpoints. I am just going to do right endpoints, because it’s basically all the same thing, just a
▯
di▯erent xkis used, that’s all. But, here they all are:
▯
Left: k = a + (k ▯ 1)▯x
▯
Right: xk= a + k▯x
▯ 1▯
Mid: x▯ = a + k ▯ ▯x
k 2
3
Example: use right endpoints to ▯nd the area under f(x) = x over [0;1].
Solution: ▯rst, let’s see what ▯x is.
▯x = b ▯ a= 1 ▯ 0= 1
n n n
and since we’re doing right end points, our x ’s are going to be
k
▯ k k
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