JEE MAIN - Mathematics (2022 - 27th July Evening Shift - No. 17)
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
385
Explanation
$$\because$$ $$f(x) = \min \,\{ [x - 1],[x - 2],\,......,\,[x - 10]\} = [x - 10]$$
Also $$|f(x)| = \left\{ {\matrix{ { - f(x),} & {if\,x \le 10} \cr {f(x),} & {if\,x \ge 10} \cr } } \right.$$
$$\therefore$$ $$\int\limits_0^{10} {f(x)dx + \int\limits_0^{10} {{{(f(x))}^2}dx + \int\limits_0^{10} {( - f(x))dx} } } $$
$$ = \int\limits_0^{10} {{{(f(x))}^2}dx} $$
$$ = {10^2} + {9^2} + {8^2}\, + \,.....\, + \,{1^2}$$
$$ = {{10 \times 11 \times 21} \over 6} = 385$$
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