When we integrate function of acceleration, what do we get?

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Integrating the function of acceleration yields the velocity function when using an indefinite integral, which includes a constant of integration. A definite integral calculates the change in velocity over a specific time interval, represented as the difference between velocities at two points in time. Combining these concepts, the integral from a fixed time to a variable time provides the change in velocity since the initial time. However, this discussion highlights that while the integral gives the change in speed, it does not inherently convey the direction of that change. The distinction between vector quantities and their magnitudes is crucial in understanding this relationship.
AakashPandita
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Do we get function for velocity or do we get the function for the total change in velocity?
 
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What kind of integral do you mean? The "indefinite integral" is a function, the "definite integral" is a number. The "indefinite integral" of the acceleration function, \int a(t)dt is the velocity function, v(t), plus a "constant of integration". The definite integral, \int_{t_0}^{t_1} a(t)dt= v(t_1)- v(t_0), change in speed between times t_0 and t_1.

We can combine the two: writing \int_{t_0}^T a(t)dt where t_0 is a fixed time and T is a variable. It gives, for any T, the change in velocity since time t_0.
 
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okay. thanks again!
 
you said it tells us the change in speed.
that means we do not get direction of change?
 
Warning: All you wrote is correct for the vector quantities \vec{v} and \vec{a} but in general not for the magnitudes!
 
?i don't understand.
 

Thread 'Proving that convexity implies second order derivative being positive'

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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