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Question:

Find the value of the integral: $$\int^{3\pi\over 2}_{\pi \over 2}[2\sin x] dx$$ Where $[y]$ represents the greatest integer less than or equal to $y$.

I know that I will have to spit the integral wherever the value of $2\sin x$ approaches an integer. However, shouldn't the answer be zero? The graph will be as negative as it is positive. Is there a flaw here somewhere?

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    $\begingroup$ You could use the formula $\int_{a}^{b} f(x)dx = \int_{0}^{b} f(a+b-x)dx$ and then take the case of $[\sin x] + [-\sin x] = -1$ as the other case integral would turnout to be equal to zero . $\endgroup$ – Sujith Sizon Oct 10 '15 at 4:24
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It should not be zero. As you define it, $[y]$ is the greatest integer less or equal to the number. This is better known as the integer part, [2.01] = 2, [3.94] = 3, etc. With that in mind, for the domain you selected and keeping in mind that $\sin(x) \in [-1,1]$, for $x\in (\pi /2, \pi]$ you will have $[\sin (x)] = 0$, and then for $x\in (\pi, 3\pi /2]$, $[\sin (x)] = -1$, so then your integral would be

$$ \int_{\pi/2}^{3\pi /2}[\sin (x)]\,dx = 0\int_{\pi/2}^{\pi}\,dx - \int_{\pi}^{3\pi /2}\,dx = -\pi/2 $$

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Since we are getting a strong smell of

$$U.L. + L.L. = \pi$$

and we know the basic formula of $[.]$ GIF as

$$[x]+[-x]=-1$$ (actually a two valued result but we are ignoring the latter case)

We could use corollary to eliminate GIF in the beginning itself

$$\int_{a}^{b} f(x)dx = \int_{0}^{b} f(a+b-x)dx$$

NOTE: Here we will only take the case of $[x]+[-x]=-1$ as the other case gives $=0$ and integration will yield $=0$ itself ie: for x= integer cases

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