Let Max_(0 ≤ x ≤ 2) { (9 - x²)/(5 - x) } = α and Min_(0 ≤ x ≤ 2) { (9 - x²)/(5 - x) } = β .
If ∫_(β - 8/3)^(2α - 1) Max{ (9 - x²)/(5 - x),x }dx = α ₁ + α ₂log ₑ( 8/15 ) then α ₁ + α ₂ is equal to _____________.

Solution

Maths sir · 4 steps

Use t = 5 - x to find α, β, then split the integral where the curve crosses y = x.

1. Find alpha and beta. Put t = 5 - x, so t ∈ [3, 5] and (9 - x²)/(5 - x) = 10 - (t + 16/t). Since t + 16/t is smallest at t = 4, the maximum is 10 - 8 = 2 at x = 1. At the ends, t = 3 gives 5/3 and t = 5 gives 9/5, so α = 2, β = 5/3, and the limits are -1 to 3.
2. Which one is larger. (9 - x²)/(5 - x) - x = (9 - 5x)/(5 - x), and 5 - x > 0 on [-1, 3]. So the fraction is the larger one for x ≤ 9/5, and x is the larger one for x ≥ 9/5.

   ∫₋₁^(9/5)(x + 5 - 16/(5 - x))dx + ∫_(9/5)³ x dx

3. Evaluate each part. First part: [x²/2 + 5x]₋₁^(9/5) = 378/25 and [16 ln(5 - x)]₋₁^(9/5) = 16 ln(16/5)/6 = 16 ln 8/15. Second part: 1/2(9 - 81/25) = 72/25.
4. So. The integral is 450/25 + 16 ln 8/15 = 18 + 16 ln 8/15. So α₁ = 18, α₂ = 16 and α₁ + α₂ = 34.

The common slip: 2. Getting the sign of the log term wrong makes α₂ = -16 and gives 2. The antiderivative of -16/(5 - x) is +16 ln(5 - x), which is what produces +16 ln 8/15.

The idea to keep. For an integral of max or min of two functions, first find where they cross, then integrate the larger one on each piece. Divide out the polynomial before integrating a rational function.

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