JEE Main 2020 · 7 January · morning · Single correct · +4 / −1

If ƒ(a + b + 1 - x) = ƒ(x), for all x, where a and b are fixed positive real numbers, then
1/(a + b)∫ₐᵇ x( f(x) + f(x + 1) ) dx is equal to:

  1. (a) ∫ₐ₋₁ᵇ⁻¹ f(x+1)dx
  2. (b) ∫ₐ₊₁ᵇ⁺¹ f(x + 1)dx
  3. (c) ∫ₐ₋₁ᵇ⁻¹ f(x)dx
  4. (d) ∫ₐ₊₁ᵇ⁺¹ f(x)dx
Answer and solution

Answer (a) ∫ₐ₋₁ᵇ⁻¹ f(x+1)dx

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