How do you solve the definite integral ∫|x^3-x|dx with limits from -1 to 2?

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Discussion Overview

The discussion revolves around solving the definite integral ∫|x^3-x|dx from -1 to 2, focusing on the challenges posed by the absolute value in the integrand. Participants explore how to determine the intervals where the expression inside the absolute value is positive or negative, and how to approach the integration process step by step.

Discussion Character

  • Homework-related
  • Exploratory
  • Technical explanation

Main Points Raised

  • One participant requests a stepwise solution for the integral, indicating a need for guidance.
  • Another participant suggests making a drawing to better understand the function, highlighting the importance of visualizing the problem.
  • A participant expresses difficulty with the modulus sign and determining the correct limits for positive and negative values.
  • Discussion includes identifying the zero points of the function x^3 - x, with participants noting these points as -1, 0, and 1.
  • There is mention of antisymmetry in the function, which some participants find relevant for understanding the integral's behavior.
  • A later reply emphasizes the need to split the integral into sections based on where the function is positive or negative, providing a general rule for handling absolute values.

Areas of Agreement / Disagreement

Participants generally agree on the zero points of the function and the need to analyze the function's behavior across different intervals. However, there is no consensus on the best approach to proceed with the integration, as some participants express confusion and seek clarification.

Contextual Notes

Participants mention the need for visual aids and the exploration of symmetry, indicating that the discussion is still developing with respect to the integration process and the handling of absolute values. Some assumptions about the function's behavior may not be fully explored.

Prasun-rick
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Member warned that homework questions must be posted in the HW & CW sections
∫|x^3-x|dx with limits from -1 to 2.Can anyone kindly show the step wise solution or suggest how to proceed .Thanks in advance​
 
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Can you tell us first what makes it troubling for you?
 
First step: make a drawing !
 
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Krylov said:
Can you tell us first what makes it troubling for you?
The modulus sign! I In which limit I have to take positive and negative in the modulus I could not understand!
 
BvU said:
First step: make a drawing !
Well I am not able to draw precisely and that is where the problem lies !
 
Doesn't have to be precise. Simply investigate the function a little.
Where it comes from on the left, where it goes through zero and where it goes to on the right.
Perhaps there is even a little symmetry to explore and make good use of ...:rolleyes:
 
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BvU said:
Doesn't have to be precise. Simply investigate the function a little.
Where it comes from on the left, where it goes through zero and where it goes to on the right.
Perhaps there is even a little symmetry to explore and make good use of ...:rolleyes:
well kindly please tell me the terminal points ! I found it as -1,0 and 1 am I correct??
 
I don't know what terminal points are o_O but the zero points of ##\ \ x^3-x = x(x+1)(x-1) \ \ ## are indeed -1, 0, 1 .

Between -1 and 0 two out of three factors are < 0 and the other is > 0, so the expression is positive.
Between 0 and 1 ...
There is antisymmetry : ##\ \ (-x)^3 -(-x) = -(x^3 - x) ##, so ...
 
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BvU said:
I don't know what terminal points are o_O but the zero points of ##\ \ x^3-x = x(x+1)(x-1) \ \ ## are indeed -1, 0, 1 .

Between -1 and 0 two out of three factors are < 0 and the other is > 0, so the expression is positive.
Between 0 and 1 ...
There is antisymmetry : ##\ \ (-x)^3 -(-x) = -(x^3 - x) ##, so ...
Thanks I finally got that !Thank you very much ! :smile:
 
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Hey Prasun-rick.

You will have to split up the function within the absolute value function so that it is positive and negative and create separate integrals for each section.

Just remember that you define |x| = x if x > 0 and -x if x < 0 (and 0 if x = 0).
 
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