Method of Exhaustion - Archimedes's Method

**RedBarchetta** · January 1st 2009, 02:23 PM

Find the area under the curve using Archimedes's method for 2x+1 from zero to some number b.

So we divide the area under the parabolic segment into rectangles. Where we divide the segment b into n parts where we have (b/n) sections and k=1,2,3,...,n.

So the area of the rectangles are:

.....but the answer should be +b not +b/n. Any help is appreciated. Thank you!

**ThePerfectHacker** · January 1st 2009, 04:40 PM

We will compute the areas two ways. One way through the left endpoints one way through the right endpoints. The left endpoints for $f(x)=2x^2+1$ are smaller than the true area of the parabola, is given by, $\sum_{k=0}^{n-1} f\left( \frac{bk}{n} \right) \cdot \frac{b}{n}$ . And the right endponts are larger than the true area of the parabola, is given by, $\sum_{k=1}^n f\left( \frac{bk}{n} \right) \cdot \frac{b}{n}$ .

Thus, $\sum_{k=0}^{n-1} \frac{2b^3k^2}{n^3} + \frac{b}{n} \leq A \leq \sum_{k=1}^n \frac{2b^3k^2}{n^3} + \frac{b}{n}$ .

Now use the formula $\sum_{k=0 \text{ or } 1}^m k^2 = \frac{m(m+1)(2m+1)}{6}$ .

**RedBarchetta** · January 1st 2009, 05:02 PM

Shouldn't the first term contain b^3 and not b^2?

**ThePerfectHacker** · January 1st 2009, 05:32 PM

Originally Posted by RedBarchetta

Shouldn't the first term contain b^3 and not b^2?

It is fixed now.

**RedBarchetta** · January 1st 2009, 05:52 PM

....

So do you evaluate that last summation by replacing m with n-1 and n respectively? I'm still not sure how this leads to the formula for the area.

**ThePerfectHacker** · January 1st 2009, 07:48 PM

Careful! $\sum_{k=1}^n \frac{b}{n} = \frac{b}{n}+...+\frac{b}{n} = n\cdot \frac{b}{n} = b$ .

After you do that take the limit $n\to \infty$

**RedBarchetta** · January 1st 2009, 10:13 PM

I'm not sure if i'm following this....

**ThePerfectHacker** · January 2nd 2009, 08:45 AM

Originally Posted by RedBarchetta

I'm not sure if i'm following this....

When you have $\sum_{k=1}^n a_k$ it means $a_1+a_2+...+a_n$ .

Now if $a_k = \tfrac{b}{n}$ then $\sum_{k=1}^n a_k = \underbrace{\tfrac{b}{n}+...+\tfrac{b}{n}}_{n \text{ times }} = b$

EDIT: Fixed

**Danny** · January 2nd 2009, 08:48 AM

Typo - n times. Fixed ;-)

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