# Integrate - Area Under the Curve

• November 12th 2008, 11:26 PM
jhomie
Integrate - Area Under the Curve
This is the question in my book and I can't seem to get an answer:

14 (a) For the function: http://thestudentroom.co.uk/latexren...e9f3f6cc14.png the area under the curve from x = 0 to 1 and the x axis, can be found with the equation http://thestudentroom.co.uk/latexren...b2b1b5acc4.png. What is the equation for the area under the curve from y = 0 to 1 and the y axis? I got http://thestudentroom.co.uk/latexren...2014b70b3c.png. Is this right?

It then says

(b) With the same function (http://thestudentroom.co.uk/latexren...e9f3f6cc14.png), the equation for x = 0 to 2 and the x axis can be found with: http://thestudentroom.co.uk/latexren...83d3901726.png. What is the equation for the area unde the curve from y = 0 to 2 and the y axis? I don't know what to do here

And finally.. (however, if we can get the right answer to (b) i'm sure (c) will be done too)

(c) For the same function (http://thestudentroom.co.uk/latexren...e9f3f6cc14.png) the equation for x = 2 to 3 and the x axis is http://thestudentroom.co.uk/latexren...d89489fecd.png. What would the equation be for the area under the curve for y = 2 to 3 and the y axis?
• November 13th 2008, 12:21 AM
mr fantastic
Quote:

Originally Posted by jhomie
This is the question in my book and I can't seem to get an answer:

14 (a) For the function: http://thestudentroom.co.uk/latexren...e9f3f6cc14.png the area under the curve from x = 0 to 1 and the x axis, can be found with the equation http://thestudentroom.co.uk/latexren...b2b1b5acc4.png. What is the equation for the area under the curve from y = 0 to 1 and the y axis? I got http://thestudentroom.co.uk/latexren...2014b70b3c.png. Is this right? Mr F says: Yes. And how you got this answer is probably the way you need to get the answers to the next two questions.

It then says

(b) With the same function (http://thestudentroom.co.uk/latexren...e9f3f6cc14.png), the equation for x = 0 to 2 and the x axis can be found with: http://thestudentroom.co.uk/latexren...83d3901726.png. What is the equation for the area unde the curve from y = 0 to 2 and the y axis? I don't know what to do here

Mr F says: The area of the rectangle is ${\color{red}2 \cdot 2^n = 2^{n+1}}$. Then the required area is ${\color{red}2^{n+1} - \frac{2^{n+1}}{n+1} = \, ....}$

And finally.. (however, if we can get the right answer to (b) i'm sure (c) will be done too)

(c) For the same function (http://thestudentroom.co.uk/latexren...e9f3f6cc14.png) the equation for x = 2 to 3 and the x axis is http://thestudentroom.co.uk/latexren...d89489fecd.png. What would the equation be for the area under the curve for y = 2 to 3 and the y axis?

(c) is left for you to then.
• November 13th 2008, 02:16 AM
jhomie
Quote:

Originally Posted by mr fantastic
(c) is left for you to then.

Thanks, I appreciate your help

The thing is however, that for (a) I made that equation based on numerical trials - i.e substituting numbers and subtracting from the area. I didn't get it in an algebraic way which is why I'm lost for (b).

For (b) however, $2^n+^1 -\frac {2^n+^1}{n+1}$ can be simplified to $\frac {2n\times 2^n}{n+1}$ ??

(c) I thought solving (b) would solve (c), but now when i have this extra info, i see that in (c) from 2 to 3 it makes a rectangle, but the y axis is different from the x axis, no?
• November 13th 2008, 03:48 AM
mr fantastic
Quote:

Originally Posted by jhomie
Thanks, I appreciate your help

The thing is however, that for (a) I made that equation based on numerical trials - i.e substituting numbers and subtracting from the area. I didn't get it in an algebraic way which is why I'm lost for (b).

For (b) however, $2^{n+1} -\frac {2^{n+1}}{n+1}$ can be simplified to $\frac {2n\times 2^n}{n+1}$ ??

(c) I thought solving (b) would solve (c), but now when i have this extra info, i see that in (c) from 2 to 3 it makes a rectangle, but the y axis is different from the x axis, no?

$2^{n+1} -\frac {2^{n+1}}{n+1} = \frac{(n+1)2^{n+1}}{n+1} -\frac {2^{n+1}}{n+1} = \frac{n \cdot 2^{n+1}}{n+1}$.

(c) Get the length and width of the rectangle. Hence get the area of the rectangle. Subtract the area between the curve and the x-axis from the area of the rectangle.
• November 13th 2008, 04:08 AM
jhomie
Quote:

Originally Posted by mr fantastic
$2^{n+1} -\frac {2^{n+1}}{n+1} = \frac{(n+1)2^{n+1}}{n+1} -\frac {2^{n+1}}{n+1} = \frac{n \cdot 2^{n+1}}{n+1}$.

(c) Get the length and width of the rectangle. Hence get the area of the rectangle. Subtract the area between the curve and the x-axis from the area of the rectangle.

Thanks, for your help I have now done (a) (b) and (c) (Rofl)

2 more Qs though: The first is:

Doesn't $\frac{n \cdot 2^{n+1}}{n+1}$ the same/simplify to $\frac {2n\times 2^n}{n+1}$?

_____

The 2nd Q is somewhat related to the (a/b/c), but it is for integration 3 which i learnt last year and my paper 2 exam in 2 weeks is on this.

Area A is the area under the curve for x = a to b and the x axis
Area B is the area under the curve for y = a to b and the y axis

It says: " For a function $y=x^n$, the ratio of Area A: Area B is $n:1$. Given the function $y=x^n$ from x = a to x = b such that a < b, does this ratio hold true for the regions defined below? Prove and explain why or why not.

Area A: $y=x^n$, $y=a^n$, $y=b^n$ and the y axis
Area B: $y=x^n$, x = a, x = b, and the x axis

I'm not asking you to do it for me; you've already helped me a lot, but I don't know what to do or where to start, so just guide me a little?

• November 13th 2008, 04:12 AM
mr fantastic
Quote:

Originally Posted by jhomie and edited by Mr F (fixed the latex error)
Thanks, for your help I have now done (a) (b) and (c) (Rofl)

2 more Qs though: The first is:

Doesn't $\frac{n \cdot 2^{n+1}}{n+1}$ the same/simplify to $\frac{2n\times 2^n}{n+1}$?
[snip]

Yes it does. Sorry I didn't look carefully enough.
• November 13th 2008, 04:14 AM
mr fantastic
Quote:

Originally Posted by jhomie
Thanks, for your help I have now done (a) (b) and (c) (Rofl)

2 more Qs though: The first is:

Doesn't $\frac{n \cdot 2^{n+1}}{n+1}$ the same/simplify to $\frac {2n\times 2^n}{n+1}$?

_____

The 2nd Q is somewhat related to the (a/b/c), but it is for integration 3 which i learnt last year and my paper 2 exam in 2 weeks is on this.

Area A is the area under the curve for x = a to b and the x axis
Area B is the area under the curve for y = a to b and the y axis

It says: " For a function $y=x^n$, the ratio of Area A: Area B is $n:1$. Given the function $y=x^n$ from x = a to x = b such that a < b, does this ratio hold true for the regions defined below? Prove and explain why or why not.

Area A: $y=x^n$, $y=a^n$, $y=b^n$ and the y axis
Area B: $y=x^n$, x = a, x = b, and the x axis

I'm not asking you to do it for me; you've already helped me a lot, but I don't know what to do or where to start, so just guide me a little?

A ratio of n : 1 means the areas are a fraction $\frac{1}{n+1}$ and $\frac{n}{n+1}$ of the rectangular area.
• November 13th 2008, 01:05 PM
jhomie
Quote:

Originally Posted by mr fantastic
A ratio of n : 1 means the areas are a fraction $\frac{1}{n+1}$ and $\frac{n}{n+1}$ of the rectangular area.

Yup, I understand that, but I don't understand the rest of it. I don't know what is wants me and how to do it with the regions specified