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 ________________.

Solution

Maths sir · 4 steps

The smallest of all those brackets is just [x-10], so f(x) = [x] - 10 and the rest is counting.

1. Find the minimum. Since x - 10 < x - 9 < … < x - 1, and the greatest integer function never decreases, the smallest bracket is always [x - 10]. So f(x) = [x - 10] = [x] - 10.
2. Value on each piece. On [k, k+1) with k = 0, 1, …, 9, f(x) = k - 10. Each piece has length 1, so f takes the values -10, -9, …, -1, once each.
3. Three integrals. ∫₀¹⁰ f dx = -(10 + 9 + … + 1) = -55. Squaring removes the sign, so ∫₀¹⁰ f² dx = 1² + 2² + … + 10² = 385. Also ∫₀¹⁰ |f| dx = 55.

   -55 + 385 + 55 = 385

4. So. The first and third integrals cancel each other, and the sum is 385.

The common slip: 495. Treating ∫₀¹⁰ f dx as +55 instead of -55 gives 495. Every value of f here is negative.

The idea to keep. Over each unit interval, a greatest integer function is a constant, so its integral becomes a plain sum. Always check the sign: when f ≤ 0, ∫ f and ∫ |f| cancel.

Answer from the NTA final answer key. Solution by PrepWiser.