If I₁ = ∫₀¹ ( 1 - x⁵⁰ )¹⁰⁰ dx and
I₂ = ∫₀¹ ( 1 - x⁵⁰ )¹⁰¹ dx such
that I₂ = αI₁ then α equals to :

Solution

Maths sir · 4 steps

Integrate I₂ by parts with 1 as the second function, and I₂ comes back on the right side.

1. By parts. Write I₂ = ∫₀¹ 1 · (1 - x⁵⁰)¹⁰¹ dx. Differentiate (1 - x⁵⁰)¹⁰¹ and integrate 1 to get x. The boundary term [x(1 - x⁵⁰)¹⁰¹]₀¹ is 0 at both limits.

   I₂ = 0 - ∫₀¹ x · 101(1 - x⁵⁰)¹⁰⁰(-50x⁴⁹) dx = 5050∫₀¹ x⁵⁰(1 - x⁵⁰)¹⁰⁰ dx

2. Rebuild the integrals. Write x⁵⁰ = 1 - (1 - x⁵⁰). Then ∫₀¹ x⁵⁰(1 - x⁵⁰)¹⁰⁰ dx = I₁ - I₂, so I₂ = 5050(I₁ - I₂).
3. Solve. I₂ + 5050I₂ = 5050I₁, so 5051I₂ = 5050I₁ and I₂ = 5050/5051I₁.
4. So. α = 5050/5051, which is option (b). This value is less than 1, as it should be: on [0, 1], (1 - x⁵⁰)¹⁰¹ ≤ (1 - x⁵⁰)¹⁰⁰, so I₂ < I₁.

The idea to keep. For ∫ (1 - xᵏ)ⁿ dx, integrate by parts with 1 as the second function. Then split xᵏ = 1 - (1 - xᵏ) so the next power's integral reappears and you can solve for it.

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