Let [t] denote the largest integer less than or equal to t. If ∫0³([x²]+[x²/2]) d x=a+b √2-√3-√5+c √6-√7, where a, b, c ∈ Z, then a+b+c is equal to __________.

Solution

Maths sir · 5 steps

Each integer the inside function crosses adds a strip of height 1 from that point to 3; add those lengths.

1. Count the jumps. For 0 ≤ x < 3, [x²] equals the number of integers k from 1 to 8 with √k ≤ x. So each k adds 1 to the integrand on the stretch from √k to 3, which contributes 3 - √k to the integral.

   ∫₀³ [x²] dx = Σₖ₌₁⁸(3-√k) = 24 - (1+√2+√3+2+√5+√6+√7+2√2)

2. First integral. Collecting terms, ∫₀³ [x²] dx = 21 - 3√2 - √3 - √5 - √6 - √7.
3. Second integral. On [0, 3], x²/2 runs up to 4.5, so it crosses the integers 1, 2, 3, 4 at x = √2, 2, √6, 2√2. The same counting gives 12 - (√2 + 2 + √6 + 2√2) = 10 - 3√2 - √6.
4. Add and match. The total is 31 - 6√2 - √3 - √5 - 2√6 - √7. Comparing with the given form, a = 31, b = -6, c = -2.
5. So. a + b + c = 31 - 6 - 2 = 23.

The common slip: 39. Dropping the minus signs and taking b = 6, c = 2 gives 31 + 6 + 2 = 39. Both coefficients are negative.

The idea to keep. For ∫₀ᵇ [f(x)] dx with f increasing from 0, find the points xₖ where f reaches each integer. The integral is Σ (b - xₖ), with no piece-by-piece splitting.

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