Property 11: `int_0 ^(nT) f(x)= n int_0 ^T f(x) dx` where T is the period of the function and n is integer

Property 12:int_(a)^(a+nT)f(x)dx=n int_(0)^(T)f(x)dx where T is period of f(x)

Property 14:int_(a+nT)^(b+nT)f(x)dx=int_(a)^(b)f(x)dx where T is the period of the f(x) and n is integer

Property 13:int_(mT)^(nT)f(x)dx=(n-m)int_(0)^(T)f(x)dx where T is period of f(x) and m and n are integer

Statement 1: int_(0)^(10)(2^(x))/(2^([x]))dx=(10)/(ln2) (where [.] denotes the greatest integer function) because Statement 2: int_(0)^(nT)f(x)dx=n int_(0)^(T)f(x)dx if T is the fundamental period of f(x), n is rational number.

Prove that int_0^t f(x)g(t-x)dx=int_0^t g(x)f(t-x)dx

Statement-1: int_(0)^(npi+v)|sin x|dx=2n+1-cos v where n in N and 0 le v lt pi . Stetement-2: If f(x) is a periodic function with period T, then (i) int_(0)^(nT) f(x)dx=n int_(0)^(T) f(x)dx , where n in N and (ii) int_(nT)^(nt+a) f(x)dx=int_(0)^(a) f(x) dx , where n in N

f(x)=int_0^x e^t f(t)dt+e^x , f(x) is a differentiable function on x in R then f(x)=

If n in N , the value of int_(0)^(n) [x] dx (where [x] is the greatest integer function) is

int_(-a)^(a)f(x)dx= 2int_(0)^(a)f(x)dx, if f is an even function 0, if f is an odd function.

For the function f(x)=int_(0)^(x)(sin t)/(t)dt where x>0