# How would you tackle this question? [Area]

• Nov 15th 2008, 08:48 PM
jhomie
How would you tackle this question? [Area]
I'm not asking you to do it for me (unless you want to(Wink) ), but i just dont know what to do, so what would you do..?

Heres the Q:

For the function http://thestudentroom.co.uk/latexren...e9f3f6cc14.png, the ratio of area A:Area B is n:1, where A is the area from y = 0 to 1 and the y axis, and B is the area from x = 0 to 1 and the x axis. As shown:

http://img89.imageshack.us/img89/5300/gramb1.png
Is this ratio true for the general case http://thestudentroom.co.uk/latexren...e9f3f6cc14.png from x = a to b such that a < b, and for the regions defined below?

Area A: http://thestudentroom.co.uk/latexren...e9f3f6cc14.png, http://thestudentroom.co.uk/latexren...8891219205.png, http://thestudentroom.co.uk/latexren...e82284722e.png and the y axis
Area B: http://thestudentroom.co.uk/latexren...e9f3f6cc14.png, http://thestudentroom.co.uk/latexren...deb8e35f9a.png, http://thestudentroom.co.uk/latexren...16a6df9829.png and the x axis

If so prove it; if not why not?

___

Once again, I'm not asking you to do it for me, I just need help as to what to do

I've proven that the ratio holds for like x = 0 to 5, or 2 to 7, but what else do i need to do?/ What is the question really asking for
• Nov 15th 2008, 10:12 PM
Soroban
Hello, jhomie!

Quote:

For the function: $\displaystyle y = x^n$, the ratio of area $\displaystyle A$ to area $\displaystyle B$ is $\displaystyle n:1$
where $\displaystyle A$ is the area from $\displaystyle y = 0\text{ to }1$ and the y-axis,
and $\displaystyle B$ is the area from $\displaystyle x = 0\text{ to }1$ and the x-axis.
Code:

|
1+ - - - - -*
|::::::::::|
|::: A :::*|
|::::::::* |
|::::::*  |
|:::*  B  |
--*----------+--
|          1

Area A

$\displaystyle A \;=\;\int^1_0\left(1 - x^n\right)\,dx \;=\;x - \frac{x^{n+1}}{n+1}\,\bigg]^1_0 \;=\;1 - \frac{1}{n+1} \;=\;\frac{n}{n+1}$

Area B

$\displaystyle B \;=\;\int^1_0 x^n\,dx \;=\;\frac{x^{n+1}}{n+1}\,\bigg]^1_0 \;=\;\frac{1}{n+1}$

Therefore: .$\displaystyle \frac{A}{B} \;=\;\frac{\frac{n}{n+1}}{\frac{1}{n+1}} \;=\;\frac{n}{1}$

• Nov 15th 2008, 10:21 PM
jhomie
Quote:

Originally Posted by Soroban
Hello, jhomie!

Area A

$\displaystyle A \;=\;\int^1_0\left(1 - x^n\right)\,dx \;=\;x - \frac{x^{n+1}}{n+1}\,\bigg]^1_0 \;=\;1 - \frac{1}{n+1} \;=\;\frac{n}{n+1}$

Area B

$\displaystyle B \;=\;\int^1_0 x^n\,dx \;=\;\frac{x^{n+1}}{n+1}\,\bigg]^1_0 \;=\;\frac{1}{n+1}$

Therefore: .$\displaystyle \frac{A}{B} \;=\;\frac{\frac{n}{n+1}}{\frac{1}{n+1}} \;=\;\frac{n}{1}$

yea, i understand that. $\displaystyle \frac{n}{1} = n:1$ right?

But that is just for 0 to 1, theres still a and b both -ve, a -ve b positive, then changing n so that it is positive integer, negative integers, fraction and irrational. Is that what i'm supposed to do?

For $\displaystyle y=x^\frac{4}{3}$ i cant seem to get the same ratio. I might be doing something wrong, but dunno