Is This Proof Correct for the Area Under a Sinusoidal Graph from 0 to 2π?

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Discussion Overview

The discussion revolves around the correctness of a proof related to calculating the area under a sinusoidal graph from 0 to 2π, particularly addressing the integration of a function that is sinusoidal from 0 to π and zero from π to 2π. The scope includes mathematical reasoning and conceptual clarification regarding integration limits and the representation of piecewise functions.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents a graph of a sinusoidal wave from 0 to π and a zero function from π to 2π, questioning the method used by others to calculate the area under the curve.
  • Another participant asks for clarification on why the initial group believes their method is correct, suggesting there may be a misunderstanding.
  • A participant insists that the integral should only be calculated from 0 to π, arguing against splitting the limits of integration from 0 to 2π due to the zero function in the second interval.
  • One participant agrees with the objection raised, stating that the equality involving the integral from 0 to 2π is not correct unless the function is properly defined as piecewise.
  • A later reply supports the piecewise definition of the function, indicating that the integration can be correctly expressed in terms of the defined intervals.
  • Areas of Agreement / Disagreement

    Participants express disagreement regarding the validity of splitting the integral limits and the correctness of the initial method proposed by the group. There is no consensus on the resolution of the mathematical claims presented.

    Contextual Notes

    Participants discuss the implications of defining the function piecewise and the mathematical laws governing integration, but the discussion does not resolve the underlying assumptions or definitions that may affect the conclusions drawn.

mjerrar
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I ve discussed with a lot of people at my university including a teacher but they equate the above equation inorder to get the area under a graph which is sinusoidal for 0 to pi and a zero funtion for pi to 2pi...

So please give me a proof so that I can show them...i.e. their method is mathematically wrong...
 
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Errr, why would they say that? Did they give any explanation for it?? Maybe you misunderstood what they said??
 
The graph was actually of a sinusoidal wave from 0 to pi...and from pi to 2pi it remained zero...making the periodic function for the ac wave...
Now the problem was that they said to find the equivalent dc current...i.e. to find the average value u use this formula to find the area...

integrate Asinxdx from o to 2pi...
whereas I insisted it will be
integrate Asinxdx from 0 to pi...as the sinusoidal graph is only for this interval

and then divide the area by 2pi...

and then they said u can split the limits...i.e.
integrate sinxdx from o to 2pi = integrate sinxdx from o to pi + integrate 0dx from pi to 2pi...

this is where I objected that is equality sign is not correct ...it doesn't satisfy mathematical laws...
am I correct..??
and please if I am...help me ...prove it to them...
 
You're right. Writing

\int_0^{2\pi}\sin(x)dx=\int_0^\pi\sin(x)dx+ \int_\pi^{2\pi}0dx

is obviously not correct.

However, if they wrote

f(x)=\left\{\begin{array}{l} \sin(x)~\text{if}~0\leq x\leq \pi\\ 0~\text{if}~\pi\leq x\leq 2\pi \end{array}\right.

then writing

\int_0^{2\pi}f(x)dx=\int_0^\pi\sin(x)dx+\int_\pi^{2\pi}0dx

is correct.
 
thanks alott...micromass...hope this satisfies them...
 

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