Show that `int_a^b(|x|)/x dx=|b|-|a|dot`

int_a^b logx/x dx

int_(a)^(b) x dx

int_a^b logx/x^2dx

int_(a)^(b)cos x dx

int_(a)^(b) sin x dx

STATEMENT 1: If f(x) is continuous on [a , b] , then there exists a point c in (a , b) such that int_a^bf(x)dx=f(c)(b-a) STATEMENT 2: For a < b , if ma n dM are, respectively, the smallest and greatest values of f(x)on[a , b] , then m(b-a)lt=int_a^bf(x)dxlt=(b-a)Mdot

If m is the least value (global minimum) and M is the greatest value (global maximum) of the function f(x) on the interval [a , b] (estimation of an integral), then m(b-a)lt=int_a^bf(x)dxlt=M(b-a)dot

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

int b^(x+a) " dx "

Let f(x) = ln ((1-sinx)/(1+sinx)) , then show that int_(a)^(b) f(x)dx = int_(b)^(a) ln((1+sinx)/(1-sinx))dx .