Can someone please tell me how you solve this intergal?

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Discussion Overview

The discussion revolves around the process of solving an integral related to gravitational force and its connection to energy. Participants explore the mathematical steps involved in integrating the force function and the conceptual understanding of how this relates to potential energy.

Discussion Character

  • Mathematical reasoning
  • Conceptual clarification
  • Homework-related

Main Points Raised

  • One participant presents the integral of gravitational force and questions the derivation of the negative sign in the resulting expression for energy.
  • Another participant suggests finding an anti-derivative of the integrand as a solution approach.
  • A participant provides a specific integral formula for power functions and applies it to the case where α = -2, seeking to understand its implications.
  • There is a discussion about the relationship between potential energy and distance, with a participant noting that potential energy is proportional to height.
  • Another participant explains that the integral of force represents energy by summing the product of force and an infinitesimal distance, linking this to the concept of work and energy units.

Areas of Agreement / Disagreement

Participants express varying levels of understanding regarding the integral and its implications for energy, with some clarifying concepts while others seek further explanation. No consensus is reached on the overall understanding of the integral's significance.

Contextual Notes

Some participants' contributions rely on specific mathematical assumptions and definitions, which may not be universally accepted or understood. The discussion does not resolve the underlying conceptual connections between force, work, and energy.

zeromodz
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[tex]E = \int \frac{Gm_1m_2}{r^2} dr = - \frac{Gm_1m_2}{r}[/tex]

The answer is there, but how do you get that (especially how do you get the negative sign)? Also, why exactly does taking the integral with respect to the distance of force give you energy?
 
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What do you mean by "solving"??

Find an anti-derivative of the integrand, then you're through.
 
[tex]\int x^\alpha dx=\frac{x^{\alpha+1}}{\alpha+1}[/tex]

try this with [tex]\alpha=-2[/tex].
 
Petr Mugver said:
[tex]\int x^\alpha dx=\frac{x^{\alpha+1}}{\alpha+1}[/tex]

try this with [tex]\alpha=-2[/tex].

Does it equal?

-1/x ? okay I understand now. But, how does this equal energy?
 
here it is
 

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Potential Energy, mgh [h=height] , is proportional to distance,
is it not ? Same idea.
Note that the derivative of Energy with respect to distance is Force
 
The integral of the force equals energy because, intuitively, the integral is just summing up the product of the gravitational force, and an infinitesimal distance dr. Since Force.Distance = (Newtons)(Meters) =Newton-Meters = Joules = Work, and since joules are also units of energy, you get energy. If you include limits of integration, say from a to b, this gives you the amount of work done by the gravitational force from point a to b, and thus the difference in potential energy from a to b.
 

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