Let β(m, n)=∫0¹ xᵐ⁻¹(1-x)ⁿ⁻¹ d x, m, n>0. If ∫0¹(1-x¹⁰)²⁰ d x=a × β(b, c), then 100(a+b+c) equals _________.

Solution

Maths sir · 3 steps

Put t = x¹⁰ so the bracket becomes (1-t)²⁰ and the integral turns into a Beta function.

1. Substitute. Let x¹⁰ = t, so x = t^(1/10) and dx = 1/10 t^(-9/10) dt. The limits stay 0 to 1, because t = 0 at x = 0 and t = 1 at x = 1.

   ∫₀¹ (1-x¹⁰)²⁰ dx = 1/10∫₀¹ t^(1/10-1)(1-t)²¹⁻¹ dt = 1/10 β(1/10, 21)

2. Match the form. Compare with tᵐ⁻¹(1-t)ⁿ⁻¹. Here m - 1 = -9/10 gives m = 1/10, and n - 1 = 20 gives n = 21. So a = 1/10, b = 1/10 and c = 21.
3. So. a + b + c = 0.1 + 0.1 + 21 = 21.2, and 100 × 21.2 = 2120. The answer is (d).

The idea to keep. If you see (1 - xᵏ)ⁿ on [0, 1], substitute t = xᵏ. The dx adds a power of t, and that power gives the first Beta argument.

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