Verified Solution • 2025–26 Session moderate

QJEEM22Ma3288: JEE Main 2022 Mathematics - Class 12 JEEM22Ma3288 Solution: Let where [t] denotes the greatest integer . Then is equal to…

Let f(x)=min⁡{[x−1],[x−2],…,[x−10]}f(x)=\min \{[x-1],[x-2], \ldots,[x-10]\} where [t] denotes the greatest integer ≤t\leq \mathrm{t} . Then ∫010f(x)dx+∫010(f(x))2 dx+∫010∣f(x)∣dx\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 ________________.

Step-by-Step Verified Solution

∵\because f(x)=min⁡ {[x−1],[x−2], ......, [x−10]}=[x−10]f(x) = \min \,\{ [x - 1],[x - 2],\,......,\,[x - 10]\} = [x - 10]

Also |f(x)|={−f(x),ifx≤10f(x),ifx≥10

∴\therefore ∫010f(x)dx+∫010(f(x))2dx+∫010(−f(x))dx\int\limits_0^{10} {f(x)dx + \int\limits_0^{10} {{{(f(x))}^2}dx + \int\limits_0^{10} {( - f(x))dx} } }

=∫010(f(x))2dx= \int\limits_0^{10} {{{(f(x))}^2}dx}

=102+92+82 + ..... + 12= {10^2} + {9^2} + {8^2}\, + \,.....\, + \,{1^2}

=10×11×216=385= {{10 \times 11 \times 21} \over 6} = 385

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