Show that an area function is constant with fund. thm of calc

Click For Summary

Homework Help Overview

The discussion centers around demonstrating that the area function for a parabola in the first quadrant is constant, specifically using the Fundamental Theorem of Calculus (FTC). The original poster seeks to understand the validity of applying the FTC in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster computes an integral to suggest that the area is constant but questions the applicability of the FTC due to the integrand's division by a variable. Other participants propose showing that the derivative of the area function is zero as an alternative approach.

Discussion Status

Participants are exploring different methods to demonstrate the constancy of the area function. There is an ongoing examination of the implications of using the FTC and the conditions under which it applies. The conversation reflects a mix of interpretations and approaches without reaching a definitive conclusion.

Contextual Notes

Participants note that the area function should be considered only for positive values of a, as the focus is on the first quadrant. There is also a mention of a correction to a previous typo in the area function presented by the original poster.

dustbin
Messages
239
Reaction score
6

Homework Statement



I need to show that the area function for a parabola in the first quadrant is constant.

Homework Equations



A(a) = \int^a_0 \frac{1}{a}-\frac{x^2}{a^3}\,dx

The Attempt at a Solution



Computing this integral gives an area of 2/3. Since the area will always be 2/3 and does not depend on the value of a, then the area function is constant. However, my question is about showing that the area function is constant using the Fundamental Theorem of Calculus.

Can I use the FTC, legitimately? Or is the hypothesis of the FTC not satisfied because of the division by a in the integrand?
 
Last edited:
Physics news on Phys.org
Another way to show ##A(a)## is constant would be to show ##A'(a)=0##.
 
Right, but wouldn't that be using the FTC? Taking the derivative would result in:

= \frac{1}{a}-\frac{a^2}{a^3} = \frac{1}{a}-\frac{1}{a} = 0


NOTE: I made a mistake in my original area function above. I fixed the typo.
 
If a = 0, then the area under the parabola between 0 and 0 is clearly zero, so it's reasonable to assume that a > 0. You don't have to consider a < 0, since you're concerned with the area in the first quadrant.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
5
Views
2K
  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K