`int_a ^b f(x) dx = phi(b) - phi(a)`

int_a^b[d/dx(f(x))]dx

By substitution: Theorem: If int f(x)dx=phi(x) then int f(ax+b)dx=(1)/(a)phi(ax+b)dx

If f(a+b-x)=f(x) , then int_a^b x f(x)dx is equal to (A) (a-b)/2int_a^b f(a+b-x)dx (B) (a+b)/2int_a^b f(b-x)dx (C) (a+b)/2int_a^b f(x)dx (D) (b-a)/2int_a^b f(x)dx

if x=phi(t) and int f(x)dx=F(x) then int f(phi(t))phi'(t)dt=(A)phi(x)(B)F(t)(C)F(x)(D)F'(x)

int_(a)^(b)f(x)dx=int_(a)^(b)f(a+b-x)dx. Hence evaluate : int_(a)^(b)(f(x))/(f(x)+f(a+b-x))dx.

If |int_(a)^(b)f(x)dx|=int_(a)^(b)|f(x)|dx,a

If f(0)=2, f'(x) =f(x), phi (x) = x+f(x) then int_(0)^(1) f(x) phi (x) dx is

If phi(x)=int(phi(x))^(-2)dx and phi(1)=0 then phi(x) is

Let inte^(x){f(x)-f'(x)}dx=phi(x) . then, inte^(x)f(x)dx is equal to