`int_0^1(f(x))/(f(x)+f(1-x))dx=`

If f(x) is integrable on [0,a], then int_0^(a) (f(x))/(f(x)+ f(a-x)) dx =

f(x)>0AAx in R and is bounded. If lim_(n->oo)[int_0^a(f(x)dx)/(f(x)+f(a-x))+aint_a^(2a)(f(x)dx)/(f(x)+f(3a-x)) +a^2int_(2a)^(3a)(f(x)dx)/(f(x)+f(5a-x))+...+a^(n-1)int_((n-1)a)^(n a)(f(x)dx)/(f(x)+f[(2n-1)a-x]]] =7//5 (where a<1), then a is equal to

f(x)>0AAx in R and is bounded. If lim_(n->oo)[int_0^a(f(x)dx)/(f(x)+f(a-x))+aint_a^(2a)(f(x)dx)/(f(x)+f(3a-x)) +a^2int_(2a)^(3a)(f(x)dx)/(f(x)+f(5a-x))+...+a^(n-1)int_((n-1)a)^(n a)(f(x)dx)/(f(x)+f[(2n-1)a-x]]] =7//5 (where a<1), then a is equal to

Consider a function f: [0, 1] -> R satisfying int_0 ^1 (f(x)(x-f(x))dx = 1/12 , then, f'(1) is k where [1/k] is

If (X) = f(X), f(0) = 1 then int (dx)/(f(x) + f(-x))=

If f(x)+f(2-x)=0 , then int_(0)^(2)(1)/(1+2^(f(x)))dx=

If f(x)+f(2-x)=0 , then int_(0)^(2)(1)/(1+2^(f(x)))dx=

If int_(0)^(1)f(x)dx=1, int_(0)^(1)x f(x)dx=a and int_(0)^(1)x^(2)f(x)dx=a^(2) , then : int_(0)^(1)(a-x)^(2)f(x)dx=

Let f(x) be a continous function on R. If int_(0)^(1)[f(x)-f(2x)]dx=5 and int_(0)^(2)[f(x)-f(4x)]dx=10 then the value of int_(0)^(1)[f(x)-f(8x)]dx=