If ∫0¹(x²¹+x¹⁴+x⁷)(2 x¹⁴+3 x⁷+6)^(1 / 7) d x=1/l(11)^(m / n) where l, m, n ∈ ℕ, m and n are coprime then l+m+n is equal to ____________.

Solution

Maths sir · 4 steps

Push an x inside the seventh root, and the outside bracket becomes the derivative of what is inside.

1. Move x inside. Write x²¹+x¹⁴+x⁷ = (x²⁰+x¹³+x⁶) × x. For x ≥ 0, x = (x⁷)^(1/7), so this x can go inside the root as x⁷.

   I = ∫₀¹ (x²⁰+x¹³+x⁶)(2x²¹+3x¹⁴+6x⁷)^(1/7) dx

2. Substitute. Let t = 2x²¹+3x¹⁴+6x⁷. Then dt = 42(x²⁰+x¹³+x⁶) dx. The limits change from x = 0, 1 to t = 0, 11.
3. Integrate. I = 1/42∫₀¹¹ t^(1/7) dt = 1/42 × 7/8 t^(8/7)|₀¹¹ = 1/48(11)^(8/7).
4. So. l = 48, m = 8, n = 7, and 8 and 7 are coprime. So l+m+n = 48+8+7 = 63.

The common slip: 57. Forgetting the 7/8 from integrating t^(1/7) leaves 1/42, which gives l = 42 and a sum of 57, not 63.

The idea to keep. When a fractional power sits on a polynomial, multiply inside the root by a power of x so that its derivative matches the factor outside. Then one substitution finishes the integral.

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