Let f(x)= min \[x-1],[x-2],,[x-10] where [t] denotes the greatest integer ≤ t. Then ∫₀¹⁰ f(x) d x+∫₀¹⁰(f(x))² ~d x+∫₀¹⁰|f(x)| d x is equal to _ _ _ _ _ _ _ _.

Integral Calculus: JEE Main 2022 Mathematics question

JEE Main 2022 · 27 Jul, Evening shift · Q33

JEE Main 202227 Jul, Evening shiftMathematicsMediumIntegral Calculus369× since 2002

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 ________________.

Your answerType your answer. Integers and decimals are accepted.

JEE Main 2022 Integral Calculus Question with Answer (27 Jul Evening shift, Q33) | JEE Arc