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I need to estimate the area between the functions

$f(x) = \log(x+1)/x \, , f(x-1), \, y=0$, and $y=a$. where $a>1$. Now I have tried quite a few ways to do this, but nothing comes to mind.

I tried writing out the taylor series, I tried changing this around. Nothing really gave a decent approximation.

A decent in my mind would be anything greater than one decimal. Eg an error less than $E<10^2$.

Cheers =)

EDIT: After making a nice drawing it seems that if $a>>0$ then $A \approx a + \frac{1}{12}\pi^2$

Image

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1. where did you find this problem? 2. Estimate the area restricted you mean find the differences of the area of the two curves? I am asking because I don't know what you mean –  yiyi May 13 '12 at 17:37
    
If you poured water into the space between the two functions, how much water would there be betwee two verticals rods, placed at y = 0 and y = a. Where a is some positive number. Will put up an image. –  N3buchadnezzar May 13 '12 at 17:40
    
Then you can't you $\int_0^a f(x)-f(x-1) dx$, I did that in Maple and got: $ -1/6\,{\pi }^{2}-{\it dilog} \left( a+1 \right) +{\it dilog} \left( a \right)$ –  yiyi May 13 '12 at 17:46
1  
The function are bound by vertical lines, not horizontal lines... –  N3buchadnezzar May 13 '12 at 17:55

2 Answers 2

up vote 4 down vote accepted

Short answer:

Any horizontal line $y=c$ with $0< c\leq a$ intersects the set $B$ between the two curves in a segment of length $1$. Therefore by Cavalieri's principle the area $\mu(B)$ is given by $\mu(B)=a$.

More info:

Cavalieri's principle (maybe known to Archimedes) is a $17^{\rm th}$ century version of the Fubini theorem. It says the following: When two three-dimensional bodies are cut by every horizontal plane in two equal areas then the two bodies have the same volume. (You need this principle to prove that the volume of any pyamid is given by $V={1\over3}G\cdot h$, where $G$ is the area of its base and $h$ denotes its height.)

For a formal proof of $\mu(B)=a$ in your example let $$g:\quad y\mapsto x=g(y)\qquad(0<y< \infty)$$

be the inverse of the function $$f:\quad x\mapsto {\log(1+x)\over x}\qquad(-1<x<\infty)\ .$$ (This means: Look at your figure considering $y$ as independent vaiable.) Then the domain $B$ in question is defined by $$B:=\{(x,y)\ |\ 0<y\leq a,\ g(y)\leq x\leq g(y)+1\}\ .$$ It follows that the area $\mu(B)$ can be computed as $$\mu(B)=\int_B{\rm d}(x,y)=\lim_{\epsilon\to 0+}\int_\epsilon^a\bigl(\int_{g(y)}^{g(y)+1} \ dx\bigr)\ dy=\lim_{\epsilon\to 0+}\int_\epsilon^a 1\ dy=\lim_{\epsilon\to 0+}(a-\epsilon)=a\ ,$$ where the inner integration was with respect to $x$.

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Oh, is that because the area has just been distorted, but its still the same amount? Never heard of Cavalieri's principle before, and just asking to make sure I understand it correctly. and the area of the space is just Length * width and the width is one, because f(x-1) is just a horizontal shift of one? thus the length is a, thus that is the reason A=a*1=a??? –  yiyi May 14 '12 at 2:17

You can also use the Cavalieri integral directly. Assume that $g(y)$ is defined as in the above answer (inverse of $f(x)$). Then the cavalieri integral of the region is equal to $\int_{g(y)}^{g(y)+1}a dx=\lim_{c\rightarrow\infty}\int_{-1}^{0}a\cdot h'(x)dx=a\lim_{c\rightarrow\infty}\int_{-1}^0 1dx=a$. Where $h(x)=x-g(a)+c$ is the inverse transformation function and $c$ is the $x$-intercept of $g(y)$. Alternatively let $p(x) = x+g(a)-c$ be the transformation function then $\int_{g(y)}^{g(y)+1}a dx =\int_{g(y)}^{g(y)+1}a \cdot 1(x) dx =\lim_{c\rightarrow\infty}\int_c^{c+1}a.1(p(x))dx=\lim_{c\rightarrow\infty}\int_c^{c+1}adx=a$. See Cavalieri integration, QM 35(3) 2012 pp. 265-296 for more details.

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