# Thread: How would you tackle this question? [Area]

1. ## How would you tackle this question? [Area]

I'm not asking you to do it for me (unless you want to ), but i just dont know what to do, so what would you do..?

Heres the Q:

For the function , 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:

Is this ratio true for the general case from x = a to b such that a < b, and for the regions defined below?

Area A: , , and the y axis
Area B: , , 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

2. Hello, jhomie!

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

Area A

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

Area B

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

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

3. Originally Posted by Soroban
Hello, jhomie!

Area A

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

Area B

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

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

yea, i understand that. $\frac{n}{1} = n:1$ 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 $y=x^\frac{4}{3}$ i cant seem to get the same ratio. I might be doing something wrong, but dunno