If [x] denotes the greatest integer function | vedclass

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(C) Let $I = \int_{-\frac{3}{2}}^{\frac{3}{2}}[2x-3] dx$. Let $t = 2x-3$, then $dt = 2dx$, so $dx = \frac{dt}{2}$.When $x = -\frac{3}{2}$, $t = -6$. W

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$10$

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(C) Let $I = \int_{-\frac{3}{2}}^{\frac{3}{2}}[2x-3] dx$. Let $t = 2x-3$, then $dt = 2dx$, so $dx = \frac{dt}{2}$.When $x = -\frac{3}{2}$, $t = -6$. When $x = \frac{3}{2}$, $t = 0$.$I = \frac{1}{2} \int_{-6}^{0} [t] dt = \frac{1}{2} \sum_{n=-6}^{-1} \int_{n}^{n+1} n dt = \frac{1}{2} \sum_{n=-6}^{-1} n = \frac{1}{2} (-6-5-4-3-2-1) = \frac{1}{2} (-21) = -\frac{21}{2}$.Thus, $k = -\frac{21}{2}$.Finally, $\left|k+\frac{1}{2}\right| = \left|-\frac{21}{2} + \frac{1}{2}\right| = \left|-\frac{20}{2}\right| = |-10| = 10$.

If $[x]$ denotes the greatest integer function of $x$ and $\int_{-\frac{3}{2}}^{\frac{3}{2}}[2x-3] dx = k$, then $\left|k+\frac{1}{2}\right| = $

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