If [x] is the greatest integer leq x,then pi^ | vedclass

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(B) Let $I = \pi^{2} \int_{0}^{2} \sin \frac{\pi x}{2} (x-[x])^{[x]} dx$. Since $[x] = 0$ for $x \in [0, 1)$ and $[x] = 1$ for $x \in [1, 2)$,we split

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$4(\pi-1)$

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(B) Let $I = \pi^{2} \int_{0}^{2} \sin \frac{\pi x}{2} (x-[x])^{[x]} dx$. Since $[x] = 0$ for $x \in [0, 1)$ and $[x] = 1$ for $x \in [1, 2)$,we split the integral: $I = \pi^{2} \left[ \int_{0}^{1} \sin \frac{\pi x}{2} (x-0)^0 dx + \int_{1}^{2} \sin \frac{\pi x}{2} (x-1)^1 dx \right]$ $I = \pi^{2} \left[ \int_{0}^{1} \sin \frac{\pi x}{2} dx + \int_{1}^{2} (x-1) \sin \frac{\pi x}{2} dx \right]$ For the first part: $\int_{0}^{1} \sin \frac{\pi x}{2} dx = [-\frac{2}{\pi} \cos \frac{\pi x}{2}]_0^1 = 0 - (-\frac{2}{\pi}) = \frac{2}{\pi}$. For the second part,use integration by parts: $\int (x-1) \sin \frac{\pi x}{2} dx = (x-1)(-\frac{2}{\pi} \cos \frac{\pi x}{2}) - \int 1 \cdot (-\frac{2}{\pi} \cos \frac{\pi x}{2}) dx = -\frac{2(x-1)}{\pi} \cos \frac{\pi x}{2} + \frac{4}{\pi^2} \sin \frac{\pi x}{2}$. Evaluating from $1$ to $2$: $[-\frac{2(2-1)}{\pi} \cos \pi + \frac{4}{\pi^2} \sin \pi] - [-\frac{2(1-1)}{\pi} \cos \frac{\pi}{2} + \frac{4}{\pi^2} \sin \frac{\pi}{2}] = [\frac{2}{\pi} + 0] - [0 + \frac{4}{\pi^2}] = \frac{2}{\pi} - \frac{4}{\pi^2}$. Summing the parts: $I = \pi^2 [\frac{2}{\pi} + \frac{2}{\pi} - \frac{4}{\pi^2}] = \pi^2 [\frac{4}{\pi} - \frac{4}{\pi^2}] = 4\pi - 4 = 4(\pi-1)$.

If $[x]$ is the greatest integer $\leq x$,then $\pi^{2} \int_{0}^{2}\left(\sin \frac{\pi x}{2}\right)(x-[x])^{[x]} d x$ is equal to :

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Assertion $(A)$: $\int_{\frac{\pi}{2}}^{\frac{3 \pi}{2}} [2 \sin x] dx = 0$,where $[.]$ denotes the greatest integer function.
Reason $(R)$: $2 \sin x$ is a decreasing function in $\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right]$.

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