# Newton's Method

1. Oct 10, 2007

### becca4

1. The problem statement, all variables and given/known data
Problem:
A (area) = $$\int^{e}_{1}$$ ln(x) dx = 1

Now we let k be such that 0 $$\leq$$ k $$\leq$$1

Consider the line y = k.

Find k so that area computed by A is exactly one half.

2. Relevant equations

So, first, I found point of intersection:

k = ln (x)

e$$^{k}$$ = x​

Now I have:

1/2 = $$\int^{e}_{e^{k}}$$ ln (x) dx​

3. The attempt at a solution

I'm having a hard time grasping the big picture of this, that's pretty much why I'm stuck. I know that Newton's Method is used to find the roots of a function, but this area twist is really giving me a hard time. I'm supposed to use N.M. to solve for k after setting up integral to compute A/2. THEN I have to experiment to find 2A/3 and A/1000. Can anyone enlighten me? Where is this zero happening?? Am I supposed to use $$\int$$ ln (x) -k dx??

HELP!!

2. Oct 10, 2007

### HallsofIvy

Staff Emeritus
The problem, as stated, doesn't make a whole lot of sense. The "area computed by A" does not depend on k. Do you mean to say that y= k divides the area computed by A in half? That's different from saying that $\int_{e^k}^e ln(x)dx= 1/2$ which has nothing to do with A.

In any case, you will have to integrate $\int ln(x) dx$. Can you do that?

3. Oct 10, 2007

### becca4

I copied the problem verbatim from the handout I was given, so I'm sorry that it doesn't make a whole lot of sense to you...

Yes, I mean to say that the line y = k splits the area in half. And yes, I know how to integrate ln(x), but it doesn't do me a whole lot if I don't know what I'm doing. (

So I have:

$$\int^{e}_{e^{k}}$$ ln (x) dx = x ln x - x ]$$^{e}_{e^{k}}$$

= ( e ln(e) - e ) - (e$$^{k}$$ ln(e$$^{k}$$ )

= (e*1 -e) - ( e$$^{k}$$*k)

= - ( e$$^{k}$$*k)
Since that integral is equal to 1/2 (since the line cuts it in half)

1/2 = -k * e$$^{k}$$

0 = -k * e$$^{k}$$ - 1/2 <--- Is this supposed to be the equation I use N.M. on?

4. Oct 10, 2007

### HallsofIvy

Staff Emeritus

You forgot the "-x". the integral is
$$-ke^k+ e^k$$
and so this is $e^k(1-k)$
No, because you neglected the "-x" your equation is
$$e^k (1-k)= 1/2$$

Last edited: Oct 10, 2007
5. Oct 10, 2007

### becca4

Ok, yeah, that makes sense. So now,

$$k - ( e^k (1-k) ) / d/dk ( e^k (1-k) )$$ ​

should give me what k converges to, meaning the value of k that when 1/2 is subtracted, equals to zero, right??

So for 2/3, I'd use this equation to solve for k using Newton's method,
$$e^k (1-k)= 2/3$$ ​

and for 1/1000,
$$e^k (1-k)= 1/1000$$​

Am I on the right track?

6. Oct 10, 2007

### Gib Z

Are you 100% sure that your original integral wasn't $$\int^e_1 \frac{1}{x} dx = 1$$ instead? Because $$\int^e_1 \log x dx = -1$$ in case you didn't check with the anti derivative... Also the integral of 1/x from t to 1, wrt t, is a common definition of the natural logarithm, so you might have just typed down something you were thinking about a few lines ahead. Just check up on the question.

7. Oct 11, 2007

### HallsofIvy

Staff Emeritus
Since log(x) (ln(x)) is positive for x> 1, how do you get -1 as its integral? I get 1 for the integral!

8. Oct 11, 2007

### Gib Z

That's a good point...

$$\int^e_1 \log x dx = x(\log x -1) \right|^e_1 = e(1-1) - 1(\log 1 -1)= 0 - (\log 1 -1) = 1$$..

Great, just great, i forgot the brackets on the log 1 -1 when I was doing it in my head, now i've made an ass of myself =]. Please ignore any thing i have said.

9. Oct 11, 2007

### becca4

So I am on the right track?

10. Oct 11, 2007

### Gib Z

Yes probably, seeing as Halls is helping you