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← JEE Main 2022 · 27 Jul · Shift 2
MathematicsInteger+4 / -1Question 27 of 90

JEE Main 2022 · 27 Jul · Shift 2, Mathematics question 27

Let f(x)=min⁡{[x−1],[x−2],…,[x−10]}f(x)=\min\{[x-1],[x-2],\ldots,[x-10]\} where [t][t] denotes the greatest integer ≤t\le t. Then ∫010f(x) dx+∫010(f(x))2dx+∫010∣f(x)∣ dx\displaystyle\int_0^{10}f(x)\,dx+\int_0^{10}(f(x))^2dx+\int_0^{10}|f(x)|\,dx is equal to ______.

Answer and worked solution

Official answer: 385

Minimum. f(x)=[x]−10f(x)=[x]-10, which takes the values −10,…,−1-10,\dots,-1 on the unit intervals of [0,10)[0,10).

Integrals. −55+385+55=385-55+385+55=385.

Answer: 385