Let f (x) = min\ [x - 1], [x - 2], , [x - 10]\ where [t] denotes...

Given, $f\left(x\right) = min\left\{\left[x - 1\right], \left[x - 2\right], \ldots , \left[x - 10\right]\right\}$

$⇒ f\left(x\right) = \left[x\right] - 10$

Now finding ${∫}_{0}^{10}f\left(x\right) \cdot dx = - 10 - 9 - 8 - \ldots . . - 1$

$= - \dfrac{10 \cdot 11}{2} = - 55$

Finding${∫}_{0}^{10}\left(f{\left(x\right)}^{2}\right)dx = {10}^{2} + {9}^{2} + {8}^{2} + \ldots + {1}^{2}$

$= \dfrac{10 \cdot 11 \cdot 21}{6} = 385$

And ${∫}_{0}^{10}\left|f\left(x\right)\right| = 10 + 9 + 8 + \ldots . + 1$

$= \dfrac{10 \cdot 11}{2} = 55$

So, ${∫}_{0}^{10}f\left(x\right)dx + {∫}_{0}^{10}{\left(f\left(x\right)\right)}^{2}dx + {∫}_{0}^{10}\left|f\left(x\right)\right|dx$

$= - 55 + 385 + 55 = 385$