Let [t] denote the greatest integer leq t. Th | vedclass

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(B) Let $I = \int_{-1/2}^{1} ([2x] + |x|) \, dx$. We can split the integral as: $I = \int_{-1/2}^{1} [2x] \, dx + \int_{-1/2}^{1} |x| \, dx$. For the

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$3$

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(B) Let $I = \int_{-1/2}^{1} ([2x] + |x|) \, dx$. We can split the integral as: $I = \int_{-1/2}^{1} [2x] \, dx + \int_{-1/2}^{1} |x| \, dx$. For the first part,$\int_{-1/2}^{1} [2x] \, dx$: In the interval $[-1/2, 0)$,$2x \in [-1, 0)$,so $[2x] = -1$. In the interval $[0, 1/2)$,$2x \in [0, 1)$,so $[2x] = 0$. In the interval $[1/2, 1)$,$2x \in [1, 2)$,so $[2x] = 1$. Thus,$\int_{-1/2}^{1} [2x] \, dx = \int_{-1/2}^{0} (-1) \, dx + \int_{0}^{1/2} (0) \, dx + \int_{1/2}^{1} (1) \, dx = -[x]_{-1/2}^{0} + 0 + [x]_{1/2}^{1} = -(0 - (-1/2)) + (1 - 1/2) = -1/2 + 1/2 = 0$. For the second part,$\int_{-1/2}^{1} |x| \, dx$: Since $|x| = -x$ for $x $\int_{-1/2}^{0} (-x) \, dx + \int_{0}^{1} x \, dx = \left[-\frac{x^2}{2}\right]_{-1/2}^{0} + \left[\frac{x^2}{2}\right]_{0}^{1} = (0 - (-(-1/2)^2/2)) + (1/2 - 0) = -1/8 + 1/2 = 3/8$. Therefore,$I = 0 + 3/8 = 3/8$. The value requested is $8 \cdot I = 8 \cdot (3/8) = 3$.

Let $[t]$ denote the greatest integer $\leq t$. Then the value of $8 \cdot \int \limits_{-\frac{1}{2}}^{1}([2 x]+|x|) \,d x$ is .... .

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