Question

The value of $\underset{0}{\overset{0.5}{\int }}\left(\right\{2x\}-1)\left(\right\{3x\}-1)dx$ is

$($Where $\{.\}$ denotes fractional part of $x)$

• A. $\frac{19}{36}$
• B. $\frac{19}{72}$
• C. $\frac{11}{72}$
• D. $\frac{19}{11}$

Answer 1

The correct option is C $\frac{11}{72}$

Let $I=\underset{0}{\overset{0.5}{\int }}\left(\right\{2x\}-1)\left(\right\{3x\}-1)dx$
$\Rightarrow I=\underset{0}{\overset{0.5}{\int }}(2x-[2x]-1)(3x-[3x]-1)dx$
$I=\underset{0}{\overset{1/3}{\int }}(2x-1)(3x-1)dx+\underset{1/3}{\overset{1/2}{\int }}(2x-1)(3x-2)dx$ $I=\underset{0}{\overset{1/3}{\int }}(6x^2-5x+1)dx+\underset{1/3}{\overset{1/2}{\int }}(6x^2-7x+2)dx$
$\Rightarrow I={(2x^3-\frac{5x^2}{2}+x)|}_0^{1/3}+{(2x^3-\frac{7x^2}{2}+2x)|}_{1/3}^{1/2}$
$\therefore I=\frac{11}{72}$