Can someone please tell me how you solve this intergal?

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The discussion centers on solving the integral for gravitational potential energy, specifically E = ∫ (Gm₁m₂/r²) dr = - (Gm₁m₂/r). Participants clarify that finding the anti-derivative of the integrand, particularly with α = -2, leads to the result -1/x. The relationship between force and energy is established, emphasizing that the integral of force over distance yields work, measured in Joules, which is equivalent to energy. The conversation highlights the importance of limits of integration in determining the work done by gravitational force between two points.

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E = \int \frac{Gm_1m_2}{r^2} dr = - \frac{Gm_1m_2}{r}

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.
 
\int x^\alpha dx=\frac{x^{\alpha+1}}{\alpha+1}

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

try this with \alpha=-2.

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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