Theorem: Let f be continuous on [a,b]. The function g defined on [a,b] by http://tutorial.math.lamar.edu/AllBrowsers/2413/DefnofDefiniteIntegral_files/eq0051M.gif [Broken] is continuous on [a,b], differentiable on (a,b), and has derivative g'(x)=f(x) for all x in (a,b) 1) Given that g is defined by g(x)=definite integral of sin (pi t) from 0 to x for all real numbers x, determine g'(-2), if it exists. [In this case a=0, according to the above theorem g'(x)=f(x) for all x in (a,b), but the theorem doesn't say that g'(x)=f(x) for x<a, and in this case -2<0=a. Does it mean that g'(-2) does not exist? ] 2) Let g be defined by g(x)=definite integral of 1/ (1+t^2) from 0 to x for all real numbers x. 2a) Show that if x<0, then g(x)<0 using properties of integrals. [Sorry, I really have no clue about part a and I need some help on it...] 2b) Find the intervals on which g is increasing or decreasing. [According to the theorem on the top g'(x)=1/ (1+x^2) is true only for all x in (a,b), in this case a=0, then how can I find g'(x) for the part x<0 ?] I hope someone would be nice enough to help me out! Thanks a lot!