Let $f(x)=\min \{[x-1],[x-2], \ldots,[x-10]\}$ where [t] denotes the greatest integer $\leq \mathrm{t}$. Then $$\int\limits_{0}^{10} f(x) \mathrm{d} x+\int\limits_{0}^{10}(f(x))^{2} \mathrm{~d} x+\int\limits_{0}^{10}|f(x)| \mathrm{d} x$$ is equal to ________________.
Answer (integer)
385
Solution
<p>$\because$ $f(x) = \min \,\{ [x - 1],[x - 2],\,......,\,[x - 10]\} = [x - 10]$</p>
<p>Also $$|f(x)| = \left\{ {\matrix{
{ - f(x),} & {if\,x \le 10} \cr
{f(x),} & {if\,x \ge 10} \cr
} } \right.$$</p>
<p>$\therefore$ $$\int\limits_0^{10} {f(x)dx + \int\limits_0^{10} {{{(f(x))}^2}dx + \int\limits_0^{10} {( - f(x))dx} } } $$</p>
<p>$= \int\limits_0^{10} {{{(f(x))}^2}dx}$</p>
<p>$= {10^2} + {9^2} + {8^2}\, + \,.....\, + \,{1^2}$</p>
<p>$= {{10 \times 11 \times 21} \over 6} = 385$</p>
About this question
Subject: Mathematics · Chapter: Definite Integration · Topic: Properties of Definite Integrals
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