Question

$\int\limits^2_1 |2x-[3x]|dx =?$

Answer 1

Answer:

REFER TO THE ATTACHMENT

Answer 2

We're asked to integrate the function,

$\displaystyle\longrightarrow f(x)=|2x-[3x]|$

in the interval $[1,\ 2].$

Let,

$\longrightarrow [3x]=a$

$\longrightarrow3x\in[a,\ a+1)$

$\longrightarrow x\in\left[\dfrac{a}{3},\ \dfrac{a+1}{3}\right)$

Thus,

$\longrightarrow[3x]=3\quad\forall x\in\left[1,\ \dfrac{4}{3}\right)$

$\longrightarrow[3x]=4\quad\forall x\in\left[\dfrac{4}{3},\ \dfrac{5}{3}\right)$

$\longrightarrow[3x]=5\quad\forall x\in\left[\dfrac{5}{3},\ 2\right)$

Value of $[3x]$ is 6 for $x=2$ but this can be ignored.

Therefore,

$\displaystyle\longrightarrow\int\limits_1^2|2x-[3x]|\ dx=\int\limits_1^{\frac{4}{3}}|2x-3|\ dx+\int\limits_{\frac{4}{3}}^{\frac{5}{3}}|2x-4|\ dx+\int\limits_{\frac{5}{3}}^2|2x-5|\ dx$

But,

$\longrightarrow x\in\left[1,\ \dfrac{4}{3}\right)$

$\longrightarrow 2x\in\left[2,\ \dfrac{8}{3}\right)$

$\longrightarrow 2x-3\in\left[-1,\ -\dfrac{1}{3}\right)$

$\Longrightarrow |2x-3|=-(2x-3)$

And,

$\longrightarrow x\in\left[\dfrac{4}{3},\ \dfrac{5}{3}\right)$

$\longrightarrow 2x\in\left[\dfrac{8}{3},\ \dfrac{10}{3}\right)$

$\longrightarrow 2x-4\in\left[-\dfrac{4}{3},\ -\dfrac{2}{3}\right)$

$\Longrightarrow |2x-4|=-(2x-4)$

And,

$\longrightarrow x\in\left[\dfrac{5}{3},\ 2\right)$

$\longrightarrow 2x\in\left[\dfrac{10}{3},\ 4\right)$

$\longrightarrow 2x-5\in\left[-\dfrac{5}{3},\ -1\right)$

$\Longrightarrow |2x-5|=-(2x-5)$

Therefore,

$\displaystyle\longrightarrow\int\limits_1^2|2x-[3x]|\ dx=-\int\limits_1^{\frac{4}{3}}(2x-3)\ dx-\int\limits_{\frac{4}{3}}^{\frac{5}{3}}(2x-4)\ dx-\int\limits_{\frac{5}{3}}^2(2x-5)\ dx$

$\displaystyle\longrightarrow\int\limits_1^2|2x-[3x]|\ dx=-\left[x^2-3x\right]_1^{\frac{4}{3}}-\left[x^2-4x\right]_{\frac{4}{3}}^{\frac{5}{3}}-\left[x^2-5x\right]_{\frac{5}{3}}^2$

$\displaystyle\longrightarrow\int\limits_1^2|2x-[3x]|\ dx=-\big[x(x-3)\big]_1^{\frac{4}{3}}-\big[x(x-4)\big]_{\frac{4}{3}}^{\frac{5}{3}}-\big[x(x-5)\big]_{\frac{5}{3}}^2$

$\displaystyle\longrightarrow\int\limits_1^2|2x-[3x]|\ dx=-\left[-\dfrac{20}{9}+2\right]-\left[-\dfrac{35}{9}+\dfrac{32}{9}\right]-\left[-6+\dfrac{50}{9}\right]$

$\displaystyle\longrightarrow\underline{\underline{\int\limits_1^2|2x-[3x]|\ dx=1}}$

Hence 1 is the answer.