How to Find the Area Bounded by a Curve Using Integrals

Click For Summary

Homework Help Overview

The discussion revolves around finding the area bounded by the curve defined by the polynomial function y = x^3 - 2x^2 - 5x + 6, the x-axis, and the vertical lines x = -1 and x = 2. Participants are exploring the setup of integrals to calculate this area.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the correct setup of integrals for the specified intervals and question the original poster's approach to integrating over the curve. There is a focus on understanding the behavior of the curve in relation to the x-axis.

Discussion Status

Some participants have offered guidance on the integral setup, pointing out potential errors in the original poster's work. The conversation indicates a productive exploration of the problem, with participants engaging in clarifying the integral's formulation.

Contextual Notes

There is mention of the curve intersecting the x-axis at specific points, and the original poster notes that the graph goes below the x-axis, which may affect the area calculation. Additionally, there are references to homework template requirements that are not being adhered to.

SherlockOhms
Messages
309
Reaction score
0
Find the area bound by the curve y = x^3 - 2x^2 - 5x + 6, the x-axis and the lines x = -1 and x = 2. The answer is 157/12.

The curve cuts the x-axis at x = -2, 1 and 3. I've shown my general idea on the attachment. I didn't end up with the correct answer so could somebody explain to me where I've gone wrong?
 
Physics news on Phys.org
ImageUploadedByPhysics Forums1365170546.687284.jpg
 
DAPOS said:
Find the area bound by the curve y = x^3 - 2x^2 - 5x + 6, the x-axis and the lines x = -1 and x = 2. The answer is 157/12.

The curve cuts the x-axis at x = -2, 1 and 3. I've shown my general idea on the attachment. I didn't end up with the correct answer so could somebody explain to me where I've gone wrong?
Your integral for the interval [1, 2] is not set up correctly. It should look like this:
$$\int_1^2 (0 - y)dx $$
Do you see why?

BTW, do not discard the three parts of the homework template. They are there for a reason.
 
Mark44 said:
Your integral for the interval [1, 2] is not set up correctly. It should look like this:
$$\int_1^2 (0 - y)dx $$
Do you see why?

BTW, do not discard the three parts of the homework template. They are there for a reason.

Thanks for that. The graph goes below the x-axis.
I sent that from my phone and the templates don't actually show for you to use.
Thanks again.
 

Similar threads

Replies
5
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
4
Views
2K
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K