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How can be

$\frac{1}{0.5}\int_0^1 t\sin{(2\pi t)}\ dt = \frac{-1}{\pi}$

and inside this interval sin signal is defined, i.e. both its $+ive$ part and $-ive$ part of the wave is present.

Shouldn't be the above mentioned expression be $0$ because integration of sinusoid over complete interval is $= 0$.

Regards

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    correct but one thing is wrong i-e means That is2012-11-01

2 Answers 2

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As $\,1/0.5=2\,$ , we have by parts

$u=t\,\,,\,u'=1\;\;\;,\;\;v'=\sin 2\pi t\,\,,\,v=-\frac{1}{2\pi}\cos 2\pi t\Longrightarrow$

$2\int_0^1t\sin 2\pi t\,dt=\left.-\frac{1}{\pi}t\cos 2\pi t\right|_0^1+\frac{1}{\pi}\int_0^1\cos 2\pi t\,dt=-\frac{1}{\pi}$

as the RHS integral's clearly zero.

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Check out this plot of the integrand $t \sin(2 \pi t)$ from $t = 0$ to $=1$. You can see that the "negative part" outweighs the "positive part".

Now, if you didn't have that factor of $t$ in the integrand, then the "positive part" and the "negative part" would cancel perfectly, as you see here.