Navigating Maths: Finding Answers to Limit Questions

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The discussion revolves around solving a limit problem involving the sum of powers of integers and its relation to calculus concepts. Participants share their methods and insights, with one user noting that the limit converges to a simple expression of 1/(p+1) for positive real values of p. The conversation also touches on the confusion surrounding Riemann sums and the teaching order of definite and indefinite integrals in calculus. Users express varying levels of understanding and seek clarification on mathematical concepts, highlighting the complexity of learning advanced mathematics. Overall, the thread illustrates the challenges and collaborative nature of tackling mathematical problems.
  • #31
I think I'll open a thread about this, I really don't think that's a very good way to have it taught...unless of course integration is defined from the outset to be the inverse operation of differentiation, and then later on you find out it just so happens to also under the area under the curve.

EDIT: CompuChip has beaten me =]

EDIT: Sigh actually, forgive me please, It is definitely possible to teach things in a different order with different definitions, and though some ways may be harder and be more deceiving to the student about the development of calculus, its still the same. Sorry guys, just my personal taste here.
 
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  • #32
area under the curve is definite integration , indefinite integration in not area under the curve it is area + value of function at initial point...however i just believe that there is no confusion in taking indefinite integration as an inverse function or operation as you see in cases of inverse operations we need not to define the operation in terms of other simpler operations

we can just define it as the inverse of well defined function.

like you doing division when multiple is known is just like doing integration (indefinite) when its derivative is known

and definite integrals can just use the concept of indefinite integration, and thus it is possible to reverse the relation and define the other in terms of first ...but in fundamental theorem of calculus regarding definite integration ,we cannot proceed unless we know the anti derivative of the function f(b)-f(a)

which itself uses the idea of something being like inverse operation...leaving the method of Riemann...
 
  • #33
Could you perhaps reword the second half of your post? It's a bit confusing.

If we define the definite integral just to be the indefinite integral evaluated at certain points, and the indefinite integral to be the inverse of differentiation, how do we reunite this with the Fundamental theorem of Calculus, and the fact that it gives the area under a curve? Even if we can, this just seems so un-natural.
 
  • #34
you see that both are inter related and can be defined in terms of each other

but what important is which one is easier to whom
many authors prefer to introduce the concept of indefinite first and definite later, but reverse can not be wrongif function be f(x)

its anti derivative will be \int f(x)*dx=F(x)+C

how do we know F(x), clearly by considering its inverse nature with differentiation

now if we want to integrate it between b,a

=F(b)-F(a)this can be represented as
f(b)+c-(F(a)+c)

\int f(x)*dx (x=b) - \int f(x)*dx (x=a)
however the entire relation can be expressed ,reversing the role of definite with indefinite , still it will be alright
but what is common between both the definitions , it is the reversibility with differentiation as even in definite integrals we need to find F(x) i,e anti derivativewhich not done by any specific process ...

this what i mean to say that both definitions of integrals i.e the relation , is correct , and is just the matter of personal interpretation .......like log(1+x)=the log series

but this does not mean that log is defined in that form , the definition of log is different and it is just the relation....
 
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