`int_(-a)^(a)f(x)*dx=int_(0)^(a)f(x)+f(-x)*dx`

int_(0)^(a)f(x)dx

Prove that int_(0)^(2a)f(x)dx=int_(0)^(a)[f(a-x)+f(a+x)]dx

Prove that int_(0)^(a)f(x)g(a-x)dx=int_(0)^(a)g(x)f(a-x)dx .

Which of the following is incorrect? int_(a+ c)^(b+c)f(x)dx=int_a^bf(x+c)dx int_(ac)^(b c)f(x)dx=cint_a^bf(c x)dx int_(-a)^af(x)dx=1/2int_(-a)^a(f(x)+f(-x)dx None of these

Statement-1: int_(0)^(pi//2) (1)/(1+tan^(3)x)dx=(pi)/(4) Statement-2: int_(0)^(a) f(x)dx=int_(0)^(a) f(a+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

int_(m)^(0)f(x)dx is

Let a gt 0 and f(x) is monotonic increase such that f(0)=0 and f(a)=b, "then " int_(0)^(a) f(x) dx +int_(0)^(b) f^(-1) (x) dx is equal to

Prove that int_(a)^(b)f(x)dx=(b-a)int_(0)^(1)f((b-a)x+a)dx

STATEMENT-1 : int_(0)^(2)[x+[x+[x]]]dx=3 and STATEMENT-2 : int_(a)^(b)f(x)dx=int_(a)^(b)f(a+b-x)dx