Understanding the Relationship Between Work and Kinetic Energy in Calculus

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The discussion clarifies the relationship between work and kinetic energy in calculus, specifically addressing the integral of dx. It explains that while the integral of dx is x, in the context of definite integrals, it must be evaluated from initial to final positions, resulting in F[Xf - Xo]. This evaluation leads to the expression F*delta(x), where delta(x) represents the change in position. The conversation emphasizes understanding the logic behind these calculations. Overall, the thread effectively breaks down the integral's application in physics.
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The question is, why? In general calc, the integral of dx is just x, right? So why does that integral become delta x? Whats the logic behind that?
 
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Yes. The integral of dx is x. Since it is a definite integral, you must evaluate x from Xo to Xf. Therefore, you will have F[Xf - Xo] otherwise known as F*delta(x).
 
Ahhh gotcha, thanks!
 
For simple comparison, I think the same thought process can be followed as a block slides down a hill, - for block down hill, simple starting PE of mgh to final max KE 0.5mv^2 - comparing PE1 to max KE2 would result in finding the work friction did through the process. efficiency is just 100*KE2/PE1. If a mousetrap car travels along a flat surface, a starting PE of 0.5 k th^2 can be measured and maximum velocity of the car can also be measured. If energy efficiency is defined by...

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