[ . ] represents greatest integer function, then $\int_{-1}^1(x[1+\sin \pi x]+1) d x=$

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Aicharya · Accepted Answer

$\int_{-1}^1(x[1+\sin \pi x]+1) d x$ $I=\int_{-1}^1(x[1+\sin \pi x]+1) d x$ We know that, $-1 \leq \sin \pi x \leq 1$ $I=\int_{-1}^0(x[1+\sin \pi x]+1) d x+\int_0^1(x(1+\sin \pi x)+1) d x$ $\begin{aligned} & \Rightarrow \quad I=\int_{-1}^0(x \cdot 0+1) d x+\int_0^1(x+1) d x \ & \Rightarrow \quad I=[x]_{-1}^0+\left[\frac{x^2}{2}+xight]_0^1 \ & \Rightarrow \quad I=0-(-1)+\left[\frac{1}{2}+1ight] \ & \Rightarrow \quad I=2+\frac{1}{2}=\frac{5}{2}\end{aligned}$

[ . ] represents greatest integer function, then $\int_{-1}^1(x[1+\sin \pi x]+1) d x=$

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