Why do you calculate the area UNDER a curve with integration?

  • Context: Undergrad 
  • Thread starter Thread starter user111_23
  • Start date Start date
  • Tags Tags
    Area Curve Integration
Click For Summary

Discussion Overview

The discussion revolves around the reasoning behind calculating the area under a curve using integration. Participants explore the practicality and definitions related to finding areas, particularly in relation to closed figures and bounded functions.

Discussion Character

  • Exploratory, Technical explanation, Conceptual clarification

Main Points Raised

  • Some participants suggest that calculating the area outside a closed figure is impractical, leading to the focus on areas under curves.
  • Others argue that the area outside of a closed figure is infinite unless under specific conditions, which supports the use of integration.
  • It is noted that similar reasoning applies to finding areas outside of shapes like squares or circles.
  • A participant describes the process of defining integration through the division of an interval into segments and constructing rectangles, leading to the conclusion that the integral represents the area under the curve.

Areas of Agreement / Disagreement

Participants express varying views on the practicality and definitions surrounding area calculations, indicating that multiple competing perspectives remain without a clear consensus.

Contextual Notes

Some assumptions about the definitions of area and integration are not explicitly stated, and the discussion does not resolve the implications of these definitions.

user111_23
Messages
84
Reaction score
0
I'm think it's because it's impractical to find the area outside a closed figure. But I'm still not sure.
 
Physics news on Phys.org
Because that's how it's defined? The area outside of a closed figure is going to be infinite unless you have a very contrived scenario
 
For the same reason it doesn't make sense to find the area outside of a square, circle, or anything else.
 
One of the ways that integration of a bounded function f on the segment [a,b] defined is by dividing [a,b] into an increasingly large number of segments. Then for each segment construct a rectangle whose width is the length of the segment and whose height is the value of the function at some point in the segment. The sum of the area of these rectangles approaches the integral as the segments get smaller.
It's easy to see that by this definition the integral is equal to the area between the function and the x-axis, or, the area "under" the line.
 

Similar threads

Undergrad The net area and total area involving definite integrals
  • · Replies 2 ·
Replies
2
Views
2K
High School Question about units for "area under curve"
  • · Replies 2 ·
Replies
2
Views
3K
High School Integration from "Area Under Curve" Perspective: Explained
  • · Replies 11 ·
Replies
11
Views
2K
Undergrad Question about Integrals to Determine Volume vs Surface Area
  • · Replies 5 ·
Replies
5
Views
4K
High School Calculating shaded area: I'm getting discrepancies between methods
  • · Replies 3 ·
Replies
3
Views
3K
Undergrad How to evaluate the enclosed area of this implicit curve?
  • · Replies 2 ·
Replies
2
Views
3K
Undergrad Negative area above x-axis from integrating x^2?
  • · Replies 4 ·
Replies
4
Views
3K
High School Explain Integration to me, please
  • · Replies 5 ·
Replies
5
Views
3K
High School Why is this definite integral a single number?
  • · Replies 20 ·
Replies
20
Views
4K
Undergrad Double integrals - do areas cancel?
  • · Replies 24 ·
Replies
24
Views
4K