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Step-by-Step Solution
Step 1: Express the integrand in terms of the fractional part
We are asked to compute
\int_{1}^{2} \lvert 2x - [3x] \rvert \, dx.
Recall that the greatest integer function [3x] can be written as
[3x] = 3x - \{3x\} ,
where \{3x\} denotes the fractional part of 3x . Hence,
2x - [3x] = 2x - (3x - \{3x\}) = \{3x\} - x.
Thus, the integral becomes
\displaystyle \int_{1}^{2} \lvert \{3x\} - x \rvert \, dx.
Step 2: Determine the sign of \{3x\} - x on [1, 2]
For x in [1, 2], observe that 1 \le x \le 2 and 0 \le \{3x\} < 1. Since x \ge 1,
\{3x\} - x \le 1 - x \le 0.
Hence, \{3x\} - x is non-positive on the entire interval [1, 2] . Therefore,
\lvert \{3x\} - x \rvert = x - \{3x\}.
So our integral simplifies to
\displaystyle \int_{1}^{2} \bigl(x - \{3x\}\bigr) \, dx.
Step 3: Break the integral into two separate integrals
We can write
\displaystyle \int_{1}^{2} \bigl(x - \{3x\}\bigr)\,dx \;=\; \int_{1}^{2} x \, dx \;-\; \int_{1}^{2} \{3x\}\,dx.
Step 4: Evaluate \displaystyle \int_{1}^{2} x \, dx
This is a straightforward integral:
\displaystyle \int_{1}^{2} x \, dx = \left[ \frac{x^2}{2} \right]_{1}^{2} = \frac{(2)^2}{2} - \frac{(1)^2}{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2}.
Step 5: Evaluate \displaystyle \int_{1}^{2} \{3x\}\, dx
Let us transform the variable: set t = 3x. Then dt = 3\,dx, or dx = \frac{dt}{3}.
When x=1, t=3. When x=2, t=6. Hence,
\displaystyle \int_{1}^{2} \{3x\}\,dx \;=\; \int_{t=3}^{t=6} \{t\}\,\frac{dt}{3} \;=\; \frac{1}{3}\,\int_{3}^{6} \{t\}\,dt.
Over each integer interval of t, from n to n+1, the fractional part \{t\} goes from 0 to 1. Between t=3 and t=6, we have three integer intervals: [3,4], [4,5], and [5,6]. Thus,
\displaystyle \int_{3}^{6} \{t\}\,dt
= \int_{3}^{4} \{t\}\,dt
+ \int_{4}^{5} \{t\}\,dt
+ \int_{5}^{6} \{t\}\,dt.
But \{t\} on each such interval runs from 0 to 1 as t goes from an integer to the next. Therefore, each part equals
\displaystyle \int_{0}^{1} u\,du = \frac{1}{2}.
Hence,
\displaystyle \int_{3}^{6} \{t\}\,dt = 3 \times \frac{1}{2} = \frac{3}{2}.
Multiplying by \frac{1}{3},
\displaystyle \int_{1}^{2} \{3x\}\,dx = \frac{1}{3} \times \frac{3}{2} = \frac{1}{2}.
Step 6: Combine the two results
Putting these parts together:
\displaystyle \int_{1}^{2} \bigl(x - \{3x\}\bigr)\,dx
= \biggl(\frac{3}{2}\biggr) - \biggl(\frac{1}{2}\biggr)
= 1.
Thus, the value of the given definite integral is
\boxed{1}.
