The problems in the file submitted are from an undergraduate course in real Analysis

(a) Answer: 0
From the definition of g(x), we know, when 0<=x<1/3, g(x)=0 and
thus dg=0; when 1/3<=x<=2/3, g(x)=1 and dg=0; when 2/3<x<=1, g(x)=2
and dg=0. So dg=0 almost everywhere in the interval [0,1], thus
the integral of x^2dg over [0,1] is 0.
(b) Answer: 0
Let int(f,a,b) denote the integral of f over (a,b).
When x is in (-pi,0), sinx<0, then d|sinx|=d(-sinx)=-cosx dx, thus
the integral of cosx d|sinx| on (-pi,0) is
int(-(cosx)^2,-pi,0)=int(-(1+cos2x)/2,-pi,0)=-pi/2
When x is in (0,pi), sinx>0, then d|sinx|=cosx dx, thus
the integral of cosx d|sinx| on (0,pi) is
int((cosx)^2, 0, pi)=int((1+cos2x)/2,0,pi)=pi/2
therefore, the integral of cosx d|sinx| over (-pi, pi) is
-pi/2+pi/2=0.