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

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Solution

$\int_{1}^{2} | 2 x - [3 x] | d x$
Substituting $3 x = t$

$= \frac{1}{3} \int_{3}^{6} ∣ ∣ ∣ \frac{2 t}{3} - [t] ∣ ∣ ∣ d t$
$= \frac{1}{9} [\int_{3}^{6} | 2 t - 3 [t] |] d t$
$= \frac{1}{9} [\int_{3}^{4} | 2 t - 9 | + \int_{4}^{5} | 2 t - 12 | + \int_{5}^{6} | 2 t - 15 |] d t$
$= \frac{1}{9} [\int_{3}^{4} (9 - 2 t) + \int_{4}^{5} (12 - 2 t) + \int_{5}^{6} (15 - 2 t)] d t$
$= \frac{1}{9} [9.1 + 12.1 + 15.1 - [4^{2} - 3^{2}] - [5^{2} - 4^{2}] - [6^{2} - 5^{2}]]$
$= \frac{1}{9} [36 - [4^{2} - 3^{2} + 5^{2} - 4^{2} + 6^{2} - 5^{2}]]$
$= \frac{1}{9} [36 - 36 + 9] = 1$

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