Integrals: Method of Exhaustion

[courtrigrad]

Find the area of $$y = 2x^{2}$$ using the method of exhaustion. I know the area to be $$\frac{2x^{3}}{3}$$ from the rules. However, in Apostol's book, it shows you that the area under of $$y = x^{2}$$ is $$\frac{x^{3}}{3}$$. So I will consider that a given. That means, intuitively, the area of $$y = 2x^{2}$$ is $$\frac{2x^{3}}{3}$$.

I drew the graph, and found that the area of the rectangle was: $$(\frac{b}{n})(\frac{2k^{2}b^{2}}{n^{2}})$$ = $$\frac{b^{3}}{n^{3}} 2k^{2}$$. So, $$S_{n} = \frac{2b^{3}}{n^{3}}(1^{2}+2^{2} + .. + n^{2})$$ and $$s_{n} \frac{2b^{3}}{n^{3}}(1^{2}+ 2^{2} + ... + (n-1)^{2})$$

We know that $$1^{2} + 2^{2} + ... + n^{2} = \frac{n^{3}}{3} + \frac{n^{2}}{2} + \frac{n}{6}$$.

So $$1^{2} + 2^{2} + ... + (n-1)^{2}} < \frac{2n^{3}}{3}< 1^{2}+2^{2}+...+n^{2}$$. Multiplying both sides by $$\frac{2b^{3}}{n^{3}}$$ we get $$s_{n} < \frac{4b^{3}}{3} < S_{n}$$. But I know the area to be $$\frac{2b^{3}}{3}$$. Where did I make my mistake?

Thanks

 

[AKG]

The following is wrong:

$$1^{2} + 2^{2} + ... + n^{2} = \frac{n^{3}}{3} + \frac{n^{2}}{2} + \frac{n}{6}$$

It should be:

$$1^{2} + 2^{2} + ... + n^{2} = \frac{n^{3}}{3} + \frac{n^{2}}{2} + \frac{n}{6} + \frac{1}{6}$$

but that won't make a big difference. The real problem is in:

$$1^{2} + 2^{2} + ... + (n-1)^{2}} < \frac{2n^{3}}{3}< 1^{2}+2^{2}+...+n^{2}$$

it should be:

$$1^{2} + 2^{2} + ... + (n-1)^{2}} < \frac{n^{3}}{3}< 1^{2}+2^{2}+...+n^{2}$$

I have to say that it's very frustrating to help someone like you:

I drew the graph, and found that the area of the rectangle was: $(\frac{b}{n})(\frac{2k^{2}b^{2}}{n^{2}})$

You don't even bother to say which rectangle. We can't read your mind. And what are b, k, and n? Also, it doesn't make sense to say, "the area of y = x²". After taking far more effort than I should have to take in order to help you I actually managed to figure out what all this stuff you meant was. Next time, if you want help, make the effort to be clear; that way, it isn't difficult for others to help you just because they can't understand you. I mean, it's not as though you just have a problem with English and don't know you to express yourself. You just started using letters and variables with no explanation of what they denoted. You may as well have defined these letters in your native language, it would have been better than not defining them at all, as you did.

 

[HallsofIvy]

Isn't it obvious that
$$1^{2} + 2^{2} + ... + (n-1)^{2}} < \frac{2n^{3}}{3}< 1^{2}+2^{2}+...+n^{2}$$
is wrong?

If n= 2 then the two sums are 1²= 1 and 1²+ 2²= 5 but $\frac{2(2^3)}{3}= \frac{16}{3}=5\frac{1}{3}$ which is not between 1 and 5!

Did you lose the "6" denominator somewhere?

 

[courtrigrad]

Sorry for the ambiguity. It's just that this is my first time really studying the subject. But I learned that $$1^{2} + 2^{2} + ... + (n-1)^{2}} < \frac{n^{3}}{3}< 1^{2}+2^{2}+...+n^{2}$$ should be assumed.

 

[courtrigrad]

Also for $$y = ax^{2} + c$$ what would the upper and lower sums be? I know that the area bounded by the curve and the x-axis is $$\frac{ax^{3}}{3} + cx$$. Let's say that the length of the interval is b . We divide the interval into equal subintervals of length $$\frac{b}{n}$$. The area of a rectangle k is $$(\frac{b}{n})(a(\frac{kb}{n})^{2}+c)$$. So the lower sum $$s_{n} = 1^{2} + 2^{2} + ... + (n-1)^{2}( \frac{ab^{3}}{n^{3}}) + \frac{b^{3}}{n^{3}}(1+2+... + n)$$. So do I just use the fact that $$1+2 +... + n = \frac{1}{2}n(n-1)$$?

Thanks

 

[AKG]

Again, you're not clear. There are infinitely many intervals of length b. Since I already figured out what you had done in your first post, I know what you mean now, but you aren't communicating in such a way that makes it easy for people to help you. Draw a picture and upload it. Clearly define what you're talking about, and show your work. I can tell you that your s_n is wrong, and I can even tell you what the answer should be, but I won't help you as it will encourage you to continue to ask questions the way you have.

 

[courtrigrad]

I already got it.

Thanks ...