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

[mjerrar]

http://imageshack.us/photo/my-images/17/4andfinal.jpg/

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...

 

[micromass]

Errr, why would they say that? Did they give any explanation for it?? Maybe you misunderstood what they said??

 

[mjerrar]

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...

 

[micromass]

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.

 

[mjerrar]

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