Fractional Integration

Note that since $x\ge 0,$ the integrand is the same as $(1-\{2x\})(1-\{3x\}).$ Thinking of this as $1-u$ vs. $u,$ it's all the same since it doesn't matter if we're working "up" or "down" from the fractional parts going from 0 to 1. (This is admittedly handwave-y, and it would be great for someone to post a rigorous proof of this.) Thus, our final answer will be $\int_0^1 \{2x\}\{3x\}\,dx.$

There are a few ways to handle this integral. The most straightforward is to just break it up into the intervals where "things change"; e.g., where $2x$ or $3x$ become $\ge 1, \ge 2.$ This gives $\frac{19}{72},$ so the answer is $A+B = 19 + 72 = 91.$

$\begin{aligned} I & = \int_0^1 (\{2x\}-1)(\{3x\}-1)\ dx \\ & = \int_0^1 (2x-\lfloor 2x \rfloor -1)(3x-\lfloor 3x \rfloor -1)\ dx \\ & = \int_0^\frac 13 (2x-1)(3x-1)\ dx + \int_\frac 13^\frac 12 (2x-1)(3x-2)\ dx + \int_\frac 12^\frac 23 (2x-2)(3x-2)\ dx + \int_\frac 23^1 (2x-2)(3x-3)\ dx \\ & = \int_0^\frac 13 \left(6x^2-5x+1\right)\ dx + \int_\frac 13^\frac 12 \left(6x^2-7x+2\right)\ dx + \int_\frac 12^\frac 23 \left(6x^2-10x+4\right)\ dx + \int_\frac 23^1 \left(6x^2-12x+6\right)\ dx \\ & = 2x^3-\frac 52x^2+x \ \bigg|_0^\frac 13 + 2x^3-\frac 72x^2+2x \ \bigg|_\frac 13^\frac 12 + 2x^3-5x^2+4x \ \bigg|_\frac 12^\frac 23 + 2x^3-6x^2+6x \ \bigg|_\frac 23^1 \\ & = \small 2x^3 \ \bigg|_0^1 - 6 (1) + (6-5)\left(\frac 23\right)^2 + \left(5-\frac 72\right)\left(\frac 12\right)^2 + \left(\frac 72-\frac 52\right)\left(\frac 13\right)^2 + 6(1) - (6-4)\left(\frac 23\right) - \left(4-2\right)\left(\frac 12\right) - \left(2-1\right)\left(\frac 13\right) \\ & = 2 - 6 + \frac 49 + \frac 38 + \frac 19 + 6 - \frac 43 - 1 - \frac 13 \\ & = \frac {19}{72} \end{aligned}$

$\implies A+B = 19+72 = \boxed{91}$