Evaluate Integral: sinπ(k/2) with k = 1 to 6

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Discussion Overview

The discussion revolves around evaluating the sum of the function \( f(x) = \sin(\pi x) \) at specific points defined by \( x_k = k/2 \) for \( k \) ranging from 1 to 6. Participants explore the formulation of this sum and its interpretation in the context of calculus.

Discussion Character

  • Homework-related, Mathematical reasoning

Main Points Raised

  • One participant poses a question about forming a Riemann sum with the function \( f(x) = \sin(\pi x) \) and expresses confusion about the process.
  • Another participant asserts that the problem is straightforward, indicating that the sum evaluates to 1 without further elaboration.
  • A third participant expresses gratitude for the clarification, noting a tendency to overcomplicate calculus problems.
  • A fourth participant acknowledges the commonality of such confusion in calculus.

Areas of Agreement / Disagreement

The discussion does not reach a consensus on the complexity of the problem, with differing perceptions of its difficulty among participants.

Contextual Notes

Participants do not clarify the assumptions behind their interpretations of the sum or the implications of the Riemann sum formulation.

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Question,
Evaluate:
[tex]\sum_{k=1}^{\ 6}\ f(xk)[/tex]
where
[tex]\ xk\ = \ k/2[/tex]
and
[tex]\ f(x)=sin\pi\ x[/tex]

OK,
does this mean that that I should form a RiemannSum with;
[tex]sin\pi\ (k/2)* delt(k/2)= \sum_{k=1}^{\ 6}f(k/2)delt(k/2)[/tex]

Im confused.
 
Last edited:
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I cannot see your problem, it is a trivial substitution

[tex]\sum_{k=1}^{\ 6}\ f(xk) = \sum_{k=1}^{\ 6}\ \sin(k\frac{\pi}{2})=1[/tex]
 
Thankyou,
sometimes I get muddled when working with calculus, and often percieve things as being a lot harder than they are.

Thanks again.
 
Ok, quite an usual occurrence!
 

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