The value of ₀^5 (27 x ^9 + 24 x ^6 + 3 x ^3 ) (3 x ^6 + 4 x ^3 + 1 )^ 1/3 dx , where x denotes the fractional part of x , is equal to

Step-by-step solution

Let I = ₀^5 (27 x ^9 + 24 x ^6 + 3 x ^3 ) (3 x ^6 + 4 x ^3 + 1 )^ 1/3 dx Since the fractional part function x is periodic with period 1 , we can use the property ₀^ nT f(x) dx = n ₀^T f(x) dx . I = 5 ₀^1 (27x^9 + 24x^6 + 3x^3) (3x^6 + 4x^3 + 1)^ 1/3 dx Factor out x from the first bracket and push it into the cube root as x^3 : I = 5 ₀^1 (27x^8 + 24x^5 + 3x^2) x (3x^6 + 4x^3 + 1)^ 1/3 dx I = 5 ₀^1 (27x^8 + 24x^5 + 3x^2) (3x^9 + 4x^6 + x^3)^ 1/3 dx Let t = 3x^9 + 4x^6 + x^3 Differentiating with respect to x : dt = (2