A bee with velocity vector r'(t) starts out at the origin at t=0 and flies around for T seconds. Where is the bee located at time T if ∫(0-T) r'(u)du = 0? Also, what does the quantity ∫(0-T) ||r'(u)||du represent?

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A bee with velocity vector r'(t) starts out at the origin at t=0 and flies around for T seconds. Where is the bee located at time T if ∫(0-T) r'(u)du = 0? Also, what does the quantity ∫(0-T) ||r'(u)||du represent?

Mathematics
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what does (0-T) represent?
integral from 0 to T

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sorry it's the only way I know how to type it
so those are the limits, ok
do you recall the definition of: \[\int_{a}^{b}f(x)~dx=??\]
the fundamental thrm of calculus is also what it is called
It's F(b) - F(a)
correct, so if F(b)-F(a) = 0, then either a=b, or F has the same value at multiple input levels similar to a periodic function.
for example, cos(pi) - cos(0) = 0 since cos(pi)=cos(0)
but the question is, what does taking the integral of a function tell us?
it can tell you what the area under a curve is
and my example is spose to read cos(2pi) = cos(0), but the site is lagging for me
it's ok
the area under a curve, yes ... but that reflects displacement
ever wonder why in some cases we get a negative area? and in other cases we dont?
Yes because the integral of velocity is displacement
right?
negative displacement ?
when we want to find the area between 2 curves, area is a positive value always ... its an absolute value. And in those cases we have to concern ourselves with the places that one curve crosses over the other.
displacement on the other hand reflects is a vector, it has magnitude and direction.
F(b)-F(a) = 0 assumes that at a and b, we are in the same place. Either a = b, or we have simply ended were we started.
Then the location of the bee is still the origin at time T?
yes, at time T, the bee is simply back in the same place it was at when the time was equal to 0
But why substitute u in for t for the integration? That's the most confusing part. Does that have any effect on the result?
0 and T are values that the variable u can take on. the function itself is dependant on the value of u ...
What is the difference between ∫(0-T) r'(u)du = 0 and ∫(0-T) r'(t)dt = 0 Why did they substitute ?
hmm, im not absolutely sure, but i think it has something to do with what is called a 'dummy' variable - meant to avoid confusion between the overuse of the letter T
oh ok .. well thank you !
good luck

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