Finding Point c on f(x) = kx^n for Equal Area Between 0 and c

[Monocles]

Ok, so doing a physics problem the other day I had to find a point c on a graph such that there was an equal amount of area between 0 and c and c and a point b > c. I found that for the graph f(x) = kx, the point was b over the square root of 2. This intrigued me so I looked for a generalization. Does this look correct? I'm still a low level math student (only in Calculus 2) so I don't know anything about the correct way to solve this stuff. Is this a proof as well? If not, how would I prove it?

$$f(x) = kx^n$$


$$\int_{0}^{c} kx^n dx = \int_{c}^{b} kx^n dx$$


$$k\int_{0}^{c} x^n dx = k\int_{c}^{b} x^n dx$$


$$\left[ \frac{x^{n+1}}{n+1} \right]_{0}^{c} = \left[ \frac{x^{n+1}}{n+1} \right]_{c}^{b}$$


$$c^{n+1} = b^{n+1} - c^{n+1}$$


$$2c^{n+1} = b^{n+1}$$


$$c = \frac{b}{\sqrt[n+1]{2}}$$


thanks!

 

[Pere Callahan]

Seems correct to me. Explicit calculation is one way of proving things, so this can be considered a proof (if you supplement it with some text and a statement of what you have actually proven, assumptions ...) For example, must your "n" be a natural number, can it be negative, rational, complex..?

 

[Monocles]

OK, I decided to try and generalize it a bit more, but I hit a wall.

Maybe this should go in homework help section at this point? I don't know.

But anyways, I tried to generalize the same thing as the above, but with:


$$\sum_{i=1}^n k_i x^i$$


I won't bother writing out all my work, but here is what I come up with:


$$\sum_{i=1}^n c^{i+1} = \frac{1}{2} \sum_{i=1}^n b^{i+1}$$

Is there any way to simplify this further? It doesn't seem very useful in this form... (keep in mind I'm only in Calc 2 so if it would require Real Analysis or something like that its over my head at the moment)

 

[Pere Callahan]

Assuming this is correct ( I doubt it, what did you do with the factors $\frac{1}{i+1}$ coming from the integration of $x^i$...? How did you cancel the k's...?), have you ever heard the term "geometric series"?

 

[Monocles]

Oops yeah you're right, I messed up getting rid of k.

Here is what I had right before that:


$$2 \left( \sum_{i=1}^{n} \frac{k_i c^{i+1}}{i+1} \right) = \sum_{i=1}^{n} \frac{k_i b^{i+1}}{i+1}$$


Which I turned (incorrectly) into:


$$2 \left( \sum_{i=1}^{n} \frac{k_i}{i+1} + \sum_{i=1}^{n} c^{i+1} \right) = \sum_{i=1}^{n} \frac{k_i b^{i+1}}{i+1}$$


I don't think I would be allowed to do this, would I? If I can, I can still eliminate k, but expanding those series makes it look like that's wrong too.


$$2 \left( \sum_{i=1}^{n} \frac{k_i}{i+1} \times \sum_{i=1}^{n} c^{i+1} \right) = \sum_{i=1}^{n} \frac{k_i b^{i+1}}{i+1}$$


Edit: Yes I know about geometric series, I am having trouble seeing the connection to this though. I am getting pretty sleepy, though.

 

[Pere Callahan]

$$2 \left( \sum_{i=1}^{n} \frac{k_i c^{i+1}}{i+1} \right) = \sum_{i=1}^{n} \frac{k_i b^{i+1}}{i+1}$$


Which I turned (incorrectly) into:


$$2 \left( \sum_{i=1}^{n} \frac{k_i}{i+1} + \sum_{i=1}^{n} c^{i+1} \right) = \sum_{i=1}^{n} \frac{k_i b^{i+1}}{i+1}$$


I don't think I would be allowed to do this, would I?

Indeed, this is wrong.
Consider the case n=2:
$$\frac{k_1c^2}{2}+\frac{k_2c^3}{3}\neq\frac{k_1}{2}+\frac{k_2}{3}+c^2+c^3$$

Do you any reason to believe this...?

$$\sum_{i=1}^n{c^i}$$

by definition IS a geometric series. So if you're familar with the term and know in principle how to find a more explicit expression for them, where is your problem in applying it to this sum?