Can you minimize?

The given integral is:- $f(x) = \int_{0}^{1} t|t-x|dt$ This can be broken into two parts. (Here we assume that $x$ lies between $0$ and $1$ ):- $f(x) = \int_{0}^{x} t(x-t) dt + \int_{x}^{1} t(t-x)dt$ Solving the above integral we get:- $f(x) = \dfrac{x^{3}}{3} -\dfrac{x}{2} + \dfrac{1}{3}$ Differentiating w.r.t. $x$ :- $f'(x) = x^{2}- \dfrac{1}{2}$ For minima, $f'(x)=0$ $x^{2}=\dfrac{1}{2}$ $x=\dfrac{1}{\sqrt{2}}$ Hence, $f(x) = \dfrac{1}{6\sqrt{2}} -\dfrac{1}{2\sqrt{2}} + \dfrac{1}{3}$ Hence, $f(x)= \dfrac{2-\sqrt{2}}{6}$ Therefore:- $\omega = 0.097631$

Note that $f(x) = \displaystyle \int_0^1 t|t-x|\ dt = \begin{cases} \displaystyle \int_0^1 t (t-x) \ dt & \text{for }x < t \\ \displaystyle \int_0^1 t (x-t) \ dt & \text{for }x \ge t \end{cases}$

For $x \le 0$ , $f(x) = \displaystyle \int_0^1 t(t-x) \ dt = \frac 13 - \frac x2$ $\implies \min (f(x)) = f(0) = \dfrac 13$ .

For $x \ge 0$ , $f(x) = \displaystyle \int_0^1 t(x-t) \ dt = \frac x2 - \frac 13$ $\implies \min (f(x)) = f(1) = \dfrac 16$ .

For $0 < x < 1$ ,

$\begin{aligned} f(x) & = \int_0^1 t |t-x| \ dt \\ & = \underbrace{\int_0^x t (x-t) \ dt}_{t < x} + \underbrace{\int_x^1 t (t-x) \ dt}_{t> x} \\ & = \frac {x^3}2 - \frac {x^3}3 + \frac 13 - \frac x2 - \frac {x^3}3 + \frac {x^3}2 \\ & = \frac {x^3}3 - \frac x2 + \frac 13 \\ \implies f'(x) & = x^2 - \frac 12 & \small \color{#3D99F6} \text{Equating to }0 \\ x & = \frac 1{\sqrt 2} & \small \color{#3D99F6} \text{Since }0 < x < 1 \\ \implies \min(f(x)) & = f \left(\frac 1{\sqrt 2}\right) = \frac {2-\sqrt 2}6 \approx 0.097631 & \small \color{#3D99F6} \text{Since }f'' \left(\frac 1{\sqrt 2}\right) > 0 \end{aligned}$

Therefore, $\lfloor 1000\omega \rfloor = \boxed{97}$ .