Expert: Paul Klarreich - 1/9/2010

Question
The value of the limit
lim       n
n->INF E(sigma)((pi/6n)tan(ipi/18n))
        i=1

is equal to the area below the graph of a function f(x)on an interval [A,B]. Find f, A & B. Do no evaluate the limit.


Can you please HELP? I was never taught sigma notation and how to understand it for the calculus class I took before the one I am in now. Could you please help explain in detail how to do this problem and set it up to get the answer. thanks.

Answer
Questioner: brooke
Country: United States
Category: Calculus
Private: No
Subject: math help
Question: The value of the limit
lim       n
n->INF E(sigma)((pi/6n)tan(ipi/18n))
       i=1

is equal to the area below the graph of a function f(x)on an interval [A,B]. Find f, A & B. Do no evaluate the limit.


Can you please HELP? I was never taught sigma notation and how to understand it for the calculus class I took before the one I am in now. Could you please help explain in detail how to do this problem and set it up to get the answer. thanks.
........................................
This is your old friend the 'match game' -- I am sure you have seen it on late-night TV.

In your text you have found the

DEFINITION OF THE DEFINITE INTEGRAL, which says:

{x=b
|     f(x) dx =
}x=a

limit(as n -> infinity) of SUM(for k=1 to n) (f(x[k]) delta-x

[From now on, I will write dx there.  Sloppy, but shorter.]
......................................
Now, then, we can match that up to what you wrote:

lim       n
n->INF E(sigma)((pi/6n)tan(ipi/18n))
       i=1
       
First, I will change your 'i's to 'k's.  (I love using k's -- I don't like i's.)


lim       n
n->INF E(sigma)((pi/6n)tan(k pi/18n))
       k=1

Then I shall reformat it. (Not changing anything here, just retyping it.)


limit(as n -> INFINITY) of SUM(for k=1 to n) ((pi/6n)tan(k pi/18n))
Now we can try to match it with the definition:

limit(as n -> infinity) of SUM(for k=1 to n) (f(x[k]) dx

This looks like it will be something like:

{x=b
|   C tan(x) dx =
}x=a

limit(as n -> infinity) of SUM(for k=1 to n) (tan(x[k]) dx

Now we have  x[k] = k pi/18n, and  x[n], the last one, would be pi/18.

Also, the 'zeroth' would be x[0] = 0 pi/18n = 0.  So we now look like:

{x=pi/18
|    C tan(x) dx =
}x=0

limit(as n -> infinity) of SUM(for k=1 to n) (tan(x[k]) dx

and the 'dx', the spacing, should be  pi/18n.

But we have a  pi/6n factor, which is 3 * pi/18n.  So the dx = pi/18n, and we should have the final answer:

{x=pi/18
|    3 tan(x) dx =
}x=0

...................................
Uncomfortable with that?  Let's work it backwards:

{x=pi/18
|    3 tan(x) dx = a limit of a sum, which we will work up. (not work out, up)
}x=0

The interval from 0 to pi/18 shall be divided into 'n' pieces, each = pi/18n, which is our dx.

The 'typical' value of x[k] is  k times that (plus the left endpoint, which happens to be zero here.)

x[k] = k pi/18n

So we get a typical element of 3 tan(x[k]) dx =

3 tan(k pi/18n) pi/18n =

tan(k pi/18n) pi/6n

Now we SUMMIFY that (sorry about making up silly words, but...)

SUM(for k = 1 to n) tan(k pi/18n) pi/6n

and then LIMITIZE (are you getting sick, yet?)

limit(as n -> infinity) SUM(for k = 1 to n) tan(k pi/18n) pi/6n

and that is what you started with.

=========================================================

Are you getting the idea?  Don't focus on many things at once.  Just look at one -- the typical element and work that out.

BTW, note the new subject line.  That is really what you are studying here.