If [ cdot ] represents the greatest integer f | vedclass

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(C) Let $I = \int_{-1}^1 (x[1+\sin(\pi x)]+1) dx$. Using the property $\int_{-a}^a f(x) dx = \int_{-a}^a f(-x) dx$,we can split the integral. However,

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$\frac{5}{2}$

Vedclass · Answer

(C) Let $I = \int_{-1}^1 (x[1+\sin(\pi x)]+1) dx$. Using the property $\int_{-a}^a f(x) dx = \int_{-a}^a f(-x) dx$,we can split the integral. However,let us evaluate the function $f(x) = x[1+\sin(\pi x)]+1$. For $x \in [-1, 0]$,$\sin(\pi x) \in [-1, 0]$,so $1+\sin(\pi x) \in [0, 1]$. Thus,$[1+\sin(\pi x)] = 0$ for $x \in (-1, 0)$. For $x \in [0, 1]$,$\sin(\pi x) \in [0, 1]$,so $1+\sin(\pi x) \in [1, 2]$. Thus,$[1+\sin(\pi x)] = 1$ for $x \in (0, 1)$. Therefore,$I = \int_{-1}^0 (x \cdot 0 + 1) dx + \int_0^1 (x \cdot 1 + 1) dx$. $I = \int_{-1}^0 1 dx + \int_0^1 (x+1) dx$. $I = [x]_{-1}^0 + [\frac{x^2}{2} + x]_0^1$. $I = (0 - (-1)) + ((\frac{1}{2} + 1) - 0)$. $I = 1 + \frac{3}{2} = \frac{5}{2}$.

If $[ \cdot ]$ represents the greatest integer function,then $\int_{-1}^1 (x[1+\sin(\pi x)]+1) dx = $

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