# Math Help - Method of Exhaustion - Archimedes's Method

1. ## Method of Exhaustion - Archimedes's Method

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!

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

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

4. Originally Posted by RedBarchetta
Shouldn't the first term contain b^3 and not b^2?
It is fixed now.

5. ....

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.

6. 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 $n\to \infty$

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

8. 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

9. Typo - n times. Fixed ;-)