# Math Help - Integrate - Area Under the Curve

1. ## 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: the area under the curve from x = 0 to 1 and the x axis, can be found with the equation . What is the equation for the area under the curve from y = 0 to 1 and the y axis? I got . Is this right?

It then says

(b) With the same function (), the equation for x = 0 to 2 and the x axis can be found with: . 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 () the equation for x = 2 to 3 and the x axis is . What would the equation be for the area under the curve for y = 2 to 3 and the y axis?

2. 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: the area under the curve from x = 0 to 1 and the x axis, can be found with the equation . What is the equation for the area under the curve from y = 0 to 1 and the y axis? I got . 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 (), the equation for x = 0 to 2 and the x axis can be found with: . 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 () the equation for x = 2 to 3 and the x axis is . 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.

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

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?

4. Originally Posted by jhomie

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.

5. 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)

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?

6. 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)

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.

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

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.

8. 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