Integration Limits: Is Upper Limit Always Greater Than Lower?

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In integration, the upper limit is not always greater than the lower limit; it depends on the context of the problem. When calculating total work done in bringing a mass from infinity to a distance r from Earth, the limits should be set from negative infinity to r. This reflects the decrease in potential energy of the system, indicating that negative work is done. The integral can be expressed as ∫ from -∞ to r, emphasizing the work done against gravitational forces. Understanding these limits is crucial for accurate calculations in physics.
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is the upper limit always greater than the lower limit in integration?

what should be the limits if we need to calculate total work done in bringing a mass from infinity to distance r from earth.
 
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##\int_a^b f(x) \, dx = - \int_b^a f(x) \, dx##.
 
Miraj Kayastha said:
what should be the limits if we need to calculate total work done in bringing a mass from infinity to distance r from earth.

\intr
 
There is decrease in potential energy of system. So, negative work is done. -{infinity to r}
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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