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(A) We need to evaluate the integral $I = \int_1^2 |2x - [3x]| dx$.The function $[3x]$ changes its value at $x = \frac{n}{3}$ for integers $n$. In the

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(A) We need to evaluate the integral $I = \int_1^2 |2x - [3x]| dx$.The function $[3x]$ changes its value at $x = \frac{n}{3}$ for integers $n$. In the interval $[1, 2]$,the points of discontinuity are $x = \frac{4}{3}, \frac{5}{3}$.We split the integral as follows:$I = \int_1^{4/3} |2x - 3| dx + \int_{4/3}^{5/3} |2x - 4| dx + \int_{5/3}^2 |2x - 5| dx$Since $2x $I = \int_1^{4/3} (3 - 2x) dx + \int_{4/3}^{5/3} (4 - 2x) dx + \int_{5/3}^2 (5 - 2x) dx$$I = [3x - x^2]_1^{4/3} + [4x - x^2]_{4/3}^{5/3} + [5x - x^2]_{5/3}^2$$I = (4 - 16/9) - (3 - 1) + (20/3 - 25/9) - (16/3 - 16/9) + (10 - 4) - (25/3 - 25/9)$$I = (20/9 - 2) + (35/9 - 32/9) + (6 - 50/9) = 20/9 - 18/9 + 3/9 + 54/9 - 50/9 = 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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