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

If | int_a ^b f(x) dx| = int_a ^b |f(x)| dx, a lt b , then f(x) = 0 has

If int_(a)^(b)f(x)dx=int_(a)^(b)phi(x)dx , then-

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

int e^x {f(x)-f'(x)}dx= phi(x) , then int e^x f(x) dx is

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)

The integral of the from int e^(x) [ phi (x)+phi' (x)]dx is computed using the substitution-

int (f(x)phi'(x)-phi(x)f'(x))/(f(x)phi(x)) {logphi(x) - logf(x)} dx is

Let inte^(x){f(x)-f'(x)}dx=phi(x) . Then inte^(x)*f(x)dx is a) phi(x)+e^(x)f(x) b) phi(x)-e^(x)f(x) c) 1/2{phi(x)+e^(x)f(x)} d) 1/2{phi(x)+e^(x)f'(x)}