Area Under Curve: Calculate Integral from 0 to 1

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Discussion Overview

The discussion revolves around calculating the area under the curve defined by the function \(f(x) = x^3\) within specific bounds. Participants explore the integral that corresponds to the area between the graphs of \(y = -1\) and \(y = f(x)\) for \(x\) in the interval \([-1, 0]\), and between \(y = 1\) and \(y = f(x)\) for \(x\) in the interval \([0, 1]\). The conversation includes aspects of symmetry, signed areas, and the interpretation of integrals.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants propose that the area can be calculated using the integral \(A = 2\int_0^1 (1 - x^3) \, dx\), suggesting symmetry in the regions.
  • Others question the necessity of splitting the interval and discuss the implications of signed areas, noting that integrating \(f(x) = x^3\) from \(-1\) to \(1\) results in zero.
  • A participant suggests that the integral should account for the areas above and below the x-axis separately, leading to a formulation of the area as \(A = \int_{-1}^0 (0 - x^3) \, dx + \int_{0}^1 (x^3 - 0) \, dx\).
  • There is a request for clarification on the original question, particularly regarding the definition of \(f(x)\) and the reasoning behind the split intervals.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of the area calculation, with some agreeing on the symmetry argument while others emphasize the importance of signed areas and the implications of integrating over the entire interval. The discussion remains unresolved regarding the best approach to calculate the area.

Contextual Notes

Participants highlight limitations in understanding the original question and the assumptions made about the function \(f(x)\). There is also a discussion about the implications of signed versus unsigned areas in the context of definite integrals.

karush
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A region is bounded between the graphs of y = -1 and y = f(x) for x between -1 and 0, and between the graphs of y = 1 and y = f(x) for x between 0 and 1.
Give an integral that corresponds to the area of this region.

however the answer to this is $\int_0^1 2(1-x^3)dx$

if I graph this i get a just an interval of 0 to 1
 
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karush said:
A region is bounded between the graphs of y = -1 and y = f(x) for x between -1 and 0, and between the graphs of y = 1 and y = f(x) for x between 0 and 1.
Give an integral that corresponds to the area of this region.

however the answer to this is $\int_0^1 2(1-x^3)dx$

if I graph this i get a just an interval of 0 to 1

Please post the entire question, preferably as asked.

In particular what is \(f(x)\)? Presumably \(f(x)=x^3\), but I am only guessing.

(If it is as I guess consider just one part of the graph and use symmetry to get the area for the other)

CB
 
Let $f(x)= x^3$

A region is bounded between the graphs of $y = -1$ and $y = f(x)$ for $x$ between $-1$ and $0$, and between the graphs of $y = 1$ and $y = f(x)$ for x between $0$ and $1$. Give an integral that corresponds to the area of this region.

OK, sorry, I left out the $f(x)$, this is exactly the way it was given, however not sure why they split the interval
 
Hello, karush!

I don't suppose you made a sketch . . .

A region is bounded between the graphs of: $y = \text{-}1,\;y = x^3$ on $[\text{-}1,0]$
. . and between the graphs of: $y = 1,\;y = x^3$ on $[0,1]$.
Give an integral that corresponds to the area of this region.
Code: 
                |
                |          *
                |
               1+ - - - - *
                |::::::::*|
                |::::::*  |
     -1         |:::*     |
  ----+---------+---------+----
      |     *:::|         1
      |  *::::::|
      |*::::::::|
      * - - - - +
              -1|
     *          |
                |
There are two sections to consider.
Since the two sections are symmetric (have equal areas)
. . we have find the area of one section and double.

A \;=\;2\int^1_0(1-x^3)\,dxWhy did they split the region?

Evaluate: .\int^1_{\text{-}1}(1-x^3)\,dx. . and tell me what you get.
 
ok so if we take the full interval of -1 to 1 they cancel to zero but since the areas are = at the given intervals then the area of the -1 to 1 is just double or 2 times.

https://www.physicsforums.com/attachments/511

thanks and I will more careful with the OP
 
karush said:
ok so if we take the full interval of -1 to 1 they cancel to zero but since the areas are = at the given intervals then the area of the -1 to 1 is just double or 2 times.

https://www.physicsforums.com/attachments/511

thanks and I will more careful with the OP

The integral of \(f(x)=x^3\) from \(-1\) to \(1\) is indeed zero, but that is not what you are asked to find. You are asked to find the area of a region, and if you look at soroban's post you will see a sketch of that region.

Also the way the question is worded you are not dealing with signed areas, so if it had specified the area in your diagram it would have been asking for:

\[A=\int_{-1}^0 (0-x^3)\; dx + \int_{0}^1 (x^3-0)\; dx=2 \int_{0}^1 (x^3-0)\; dx\]

CB
 

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