Let f(x) be a positive continuous function on [0, 1] . Let I₁ = ₀^n x f( x (1- x )) dx and I₂ = ₀^1 f(x(1-x)) dx , where n is a natural number and x denotes the fractional part of x . If I₁ = 72 I₂ , then the value of n is equal to

Step-by-step solution

We can split the integral I₁ into intervals of length 1: I₁ = _ k=0 ^ n-1 _k^ k+1 x f( x (1- x )) dx Let x = k + t . Then dx = dt , and as x goes from k to k+1 , t goes from 0 to 1 . Also, the fractional part x = k+t = t . Substituting this into the integral: I₁ = _ k=0 ^ n-1 ₀^1 (k + t) f(t(1-t)) dt I₁ = _ k=0 ^ n-1 ( k ₀^1 f(t(1-t)) dt + ₀^1 t f(t(1-t)) dt ) Notice that ₀^1 f(t(1-t)) dt = I₂ . Let J = ₀^1 t f(t(1-t)) dt . Applying the property ₀^a g(t) dt = ₀^a g(a-t) dt : J = ₀^1 (1-t) f((1-t)(1-(1-t))) dt = ₀^1