The idea is good, the calculations are quite messy :)
$y=(\{2 x\}-1) (\{3 x\}-1)$ on $[0,1]$
becomes
$y=1-\{2 x\}-\{3 x\}+\{2 x\}\{3 x\}$
where
$\left\{ {2x} \right\} = \left\{ {\begin{array}{*{20}{l}}
{2x,\;0 \leqslant x \leqslant \frac{1}{2}} \\
{2x - 1,\frac{1}{2} < x \leqslant 1\;}
\end{array}} \right.
$
$\left\{ {3x} \right\} = \left\{ {\begin{array}{*{20}{l}}
{3x,\;0 \leqslant x \leqslant \frac{1}{3}} \\
{3x - 1,\;\frac{1}{3} < x \leqslant \frac{2}{3}} \\
{3x - 2,\;\frac{2}{3} < x \leqslant 1}
\end{array}} \right.
$
$
\left\{ {2x} \right\}\left\{ {3x} \right\} = \left\{ {\begin{array}{*{20}{l}}
{6{x^2}}&{0 \leqslant x \leqslant \frac{1}{3}} \\
{2x\left( {3x - 1} \right)}&{\frac{1}{3} < x \leqslant \frac{1}{2}} \\
{\left( {2x - 1} \right)\left( {3x - 1} \right)}&{\frac{1}{2} < x \leqslant \frac{2}{3}} \\
{\left( {2x - 1} \right)\left( {3x - 2} \right)}&{\frac{2}{3} < x \leqslant 1}
\end{array}} \right.
$
So the integral must be calculated on each interval for each function
$$\int_0^1 1 \, dx- \left(\int_0^{\frac{1}{2}} 2 x \, dx+\int_{\frac{1}{2}}^1 (2 x-1) \, dx\right)-\left(\int_0^{\frac{1}{3}} 3 x \, dx+\int_{\frac{1}{3}}^{\frac{2}{3}} (3 x-1) \, dx+\int_{\frac{2}{3}}^1 (3 x-2) \, dx\right)\\ +\left(\int_0^{\frac{1}{3}} 6 x^2 \, dx+\int_{\frac{1}{3}}^{\frac{1}{2}} \left(6 x^2-2 x\right) \, dx+\int_{\frac{1}{2}}^{\frac{2}{3}} \left(6 x^2-5 x+1\right) \, dx+\int_{\frac{2}{3}}^1 \left(6 x^2-7 x+2\right) \, dx\right)=\frac{19}{72}$$
Hope this helps