Question

Let
$\begin{array}{l}f\left(x\right) = min \\{\left[x – 1\right], \left[x – 2\right], …., \left[x – 10\right]\\}\end{array}$
where [t] denotes the greatest integer ≤ t. Then
$\begin{array}{l} \displaystyle\int\limits_{0}^{10}f\left(x\right)dx+\displaystyle\int\limits_{0}^{10}\left(f\left(x\right)\right)^2dx+\displaystyle\int\limits_{0}^{10}\left|f\left(x\right)\right|dx\end{array}$
is equal to ________.

Answer 1

$\begin{array}{l}\because f\left(x\right) = min \\{\left[x – 1\right], \left[x – 2\right], ….., [x – 10]\\} = \left[x – 10\right] \end{array}$
Also,\begin{array}{l}\left|f\left(x\right)\right|=\left\\{\begin{matrix}-f\left(x\right), & \text{if } x\leq 10 \\\f\left(x\right), & \text{if } x \geq 10 \\\\\end{matrix}\right.\end{array}
$\begin{array}{l} \therefore\ \displaystyle\int\limits_{0}^{10}f\left(x\right)dx+\displaystyle\int\limits_{0}^{10}\left(f\left(x\right)\right)^2dx+\displaystyle\int\limits_{0}^{10}\left(-f\left(x\right)\right)dx\end{array}$
$\begin{array}{l} =\displaystyle\int\limits_{0}^{10}\left(f\left(x\right)\right)^2dx\end{array}$
\begin{array}{l}= 10^2 + 9^2 + 8^2 + …… + 1^2\\\ =\frac{10\times11\times21}{6} \\\= 385\end{array}