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I am really at a loss for what Stewart is trying to say. The question is:

Use Definition 2 to find an expression for the area under the graph of $f$ as a limit. Do not evaluate the limit.

$$f(x) = \frac {2x}{x^2+1},\quad 1 \le x \le 3$$

Definition 2 is:

The area $A$ of the region $S$ that lies under the graph of a continuous function $f$ is the limit of the sum of the areas of approximating rectangles. $$A = \lim_{x\to\infty}R_n=\lim_{x\to\infty}[f(x_1)\Delta x+f(x_2)\Delta x+\dots+f(x_n)\Delta x]$$

He then somehow reworks this into some insane sigma expression that has $n$s and numbers and I have no idea what is going on and neither does he really because he doesn't explain anything.

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1 Answer 1

up vote 8 down vote accepted

He doesn’t explain anything there in the question because he’s already explained it in the text, probably with examples.

Let’s forget the summation notation for now and concentrate on the ideas. You need to divide the interval $[1,3]$ into $n$ equal subintervals. Since $[1,3]$ is $3-1=2$ units long, each of the $n$ subintervals must be $\frac2n$ units long. For instance, if $n$ were $5$, you’d divide the interval into subintervals of length $\frac25=0.4$: $[1.0,1.4],[1.4,1.8],[1.8,2.2],[2.2,2.6]$, and $[2.6,3.0]$. $\Delta x$ denotes the length of the subintervals, so in this problem $\Delta x=\frac2n$. (In my numerical example $\Delta x=\frac25$.)

The numbers $x_0,x_1,x_2,\dots,x_n$ are the endpoints of the subintervals. The first subinterval is $[x_0,x_1]$; the second is $[x_1,x_2]$; the $n$-th and last is $[x_{n-1},x_n]$; and in general the $k$-th subinterval is $[x_{k-1},x_k]$. Of course this means that $x_0=1$ and $x_n=3$, since the first interval must begin at $1$, and the last must end at $3$. So how do we find the other numbers $x_k$?

Remember that each interval $[x_{k-1},x_k]$ is $\Delta x=\frac2n$ units wide. Thus, $x_1$ is $\frac2n$ units to the right of $1$, $x_2$ is $\frac2n$ units to the right of $x_1$, and so on. This means that

$$\begin{align*} x_0&=1\\ x_1&=1+\frac2n\\ x_2&=1+\frac2n+\frac2n=1+2\left(\frac2n\right)\\ x_3&=1+2\left(\frac2n\right)+\frac2n=1+3\left(\frac2n\right)\\ x_4&=1+3\left(\frac2n\right)+\frac2n=1+4\left(\frac2n\right) \end{align*}$$

and so on. In general $$x_k=1+k\left(\frac2n\right)=1+\frac{2k}n\;.$$

Now we can start to plug some specifics into the expression

$$f(x_1)\Delta x+f(x_2)\Delta x+\dots+f(x_n)\Delta x\;:$$

it’s $$f\left(1+\frac2n\right)\left(\frac2n\right)+f\left(1+\frac4n\right)\left(\frac2n\right)+f\left(1+\frac6n\right)\left(\frac2n\right)+\dots+f\left(1+\frac{2n}n\right)\left(\frac2n\right)\;,$$ where I’ve simply plugged in the values that we found above for $x_1,x_2,\dots,x_n$ and $\Delta x$. There’s one term for each of the $n$ subintervals, and the term for the $k$-th subinterval is $$f(x_k)\Delta x=f\left(1+\frac{2k}n\right)\left(\frac2n\right)\;.$$ Since $k$ runs from $1$ through $n$ (first interval through $n$-th and last interval), that rather large and unwieldy sum can be abbreviated to $$\sum_{k=1}^nf\left(1+\frac{2k}n\right)\left(\frac2n\right)\;.$$ Every term has a factor of $\frac2n$, so I can factor it out, if I wish, to get $$\frac2n\sum_{k=1}^nf\left(1+\frac{2k}n\right)\;.\tag{1}$$

We’re not done, though, because we actually know what the function $f$ is: $$f(x)=\frac{2x}{x^2+1}\;.$$ Inserting that into $(1)$ gives us $$\frac2n\sum_{k=1}^n\frac{2\left(1+\frac{2k}n\right)}{\left(1+\frac{2k}n\right)^2+1}\;.$$

Now $$\frac{2\left(1+\frac{2k}n\right)}{\left(1+\frac{2k}n\right)^2+1}$$ looks pretty horrendous, but it can be simplified quite a bit:

$$\begin{align*} \frac{2\left(1+\frac{2k}n\right)}{\left(1+\frac{2k}n\right)^2+1}&=\frac{2+\frac{4k}n}{1+\frac{4k}n+\frac{4k^2}{n^2}+1}\\ &=\frac{\frac{2n+4k}n}{\frac{2n^2+4kn+4k^2}{n^2}}\\ &=\frac{n(2n+4k)}{2n^2+4kn+4k^2}\\ &=\frac{n(n+2k)}{n^2+2kn+2k^2}\;, \end{align*}$$

so $$\frac2n\sum_{k=1}^n\frac{2\left(1+\frac{2k}n\right)}{\left(1+\frac{2k}n\right)^2+1}=\frac2n\sum_{k=1}^n\frac{n(n+2k)}{n^2+2kn+2k^2}=2\sum_{k=1}^n\frac{n+2k}{n^2+2kn+2k^2}\;.$$ Finally(!), we want the limit of this sum as $n\to\infty$:

$$A=\lim_{n\to\infty}2\sum_{k=1}^n\frac{n+2k}{n^2+2kn+2k^2}=2\lim_{n\to\infty}\sum_{k=1}^n\frac{n+2k}{n^2+2kn+2k^2}\;.$$

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I am confused where the 2 came from for the 2/n for the width of the rectangle I think. Also when you are factoring the summation after plugging in the function but not just get ride of the plugged in function on the top and bottom? –  user138246 Apr 12 '12 at 13:47
    
I think I figured out where the 2 came from, it is the entire of length of the interval so any calculation for finding the area will take that entire interval and divide it into n lengths. –  user138246 Apr 12 '12 at 14:25
    
@Jordan: Yes, the $2$ is $3-1$, the length of the interval. –  Brian M. Scott Apr 12 '12 at 19:14
    
You used 1+ 2k/2 is that because this particular function starts at one? If it was from 0-3 it would start at 0? –  user138246 Apr 13 '12 at 13:48
    
@Jordan: That’s right: started at the lefthand end of the $[1,3]$ interval. –  Brian M. Scott Apr 13 '12 at 15:11

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