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

Correct Answer: 385

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