Let [ t ] denote the greatest integer less th | vedclass

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(A) Let $I = \int_{1}^{2} |2x - [3x]| dx$. Since $1 \le x \le 2$,we have $3 \le 3x \le 6$. We divide the interval $[1, 2]$ based on the values of $[3x

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(A) Let $I = \int_{1}^{2} |2x - [3x]| dx$. Since $1 \le x \le 2$,we have $3 \le 3x \le 6$. We divide the interval $[1, 2]$ based on the values of $[3x]$: Case $1$: $1 \le x Case $2$: $4/3 \le x Case $3$: $5/3 \le x \le 2$,then $5 \le 3x \le 6$,so $[3x] = 5$. The integrand is $|2x - 5| = 5 - 2x$. Now,$I = \int_{1}^{4/3} (3 - 2x) dx + \int_{4/3}^{5/3} (4 - 2x) dx + \int_{5/3}^{2} (5 - 2x) dx$. Evaluating each integral: $\int_{1}^{4/3} (3 - 2x) dx = [3x - x^2]_{1}^{4/3} = (4 - 16/9) - (3 - 1) = 20/9 - 2 = 2/9$. $\int_{4/3}^{5/3} (4 - 2x) dx = [4x - x^2]_{4/3}^{5/3} = (20/3 - 25/9) - (16/3 - 16/9) = 35/9 - 32/9 = 3/9 = 1/3$. $\int_{5/3}^{2} (5 - 2x) dx = [5x - x^2]_{5/3}^{2} = (10 - 4) - (25/3 - 25/9) = 6 - 50/9 = 4/9$. Summing these: $I = 2/9 + 3/9 + 4/9 = 9/9 = 1$.

Let $[ t ]$ denote the greatest integer less than or equal to $t$. Then the value of $\int_{1}^{2} |2x - [3x]| dx$ is

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