Finding the Area Under a Parabola Using Archimedes' Method

Click For Summary
SUMMARY

This discussion focuses on using Archimedes' method to calculate the area under a parabola defined by the equation y = a*b^2 + c over the interval 0 < x < b. The user employs the summation of upper and lower rectangles, with specific areas calculated as a(k^2)(b/n)^3 + (bc)/n. The main issue arises when summing the rectangles, where the user mistakenly concludes with (bc)/n instead of the correct total of bc. The resolution highlights the importance of recognizing the arithmetic summation of identical terms.

PREREQUISITES
  • Understanding of calculus concepts, specifically integration and area under curves.
  • Familiarity with Archimedes' method for approximating areas.
  • Knowledge of summation notation and arithmetic series.
  • Basic algebra skills for manipulating equations and terms.
NEXT STEPS
  • Study the principles of Archimedes' method in detail.
  • Learn about Riemann sums and their application in calculus.
  • Explore the concept of definite integrals and their geometric interpretations.
  • Practice problems involving area calculations under various curves.
USEFUL FOR

Students of calculus, educators teaching integration techniques, and anyone interested in historical mathematical methods for area approximation.

Sculptured
Messages
23
Reaction score
0
I've started with Apostol's first volume of calculus and I'm already having a problem with the first exercise in the book. Weak. It asks to use Archimedes' method to find the area under a parabola from 0<x<b for a parabola where y = a*b^2 + c. Using the summation of upper/lower rectangles of width b/n and height a(k*b/n)^2 +c. Now each rectangle has an area of a(k^2)(b/n)^3 + (bc)/n. k goes from 0 to n-1 for the lower rectangles and from 1 to n for the upper rectangles. My problem comes when summing up all of the rectangles using the arithmetic summation. I get the first portion correct, but end up with (bc)/n instead of bc as the second term in the area formula. Any way that I'm approaching this incorrectly or should approach this?
 
Physics news on Phys.org
The sum bc/n+bc/n+...+bc/n (n times) is bc.
 
Ah, right. Stupid of me to miss that result.
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 22 ·
Replies
22
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 24 ·
Replies
24
Views
4K